Mixing
How to prove the following inequality? (related to no-arbitrage conditions)
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I'm working through a practice book on mathematical finance, but struggling to prove part of a question on no-arbitrage conditions.
In the problem, I'm first given $K_1 < K_2 < K_3 $. Then, the book claims that the following relationship: $x_1 (S - K_1) - x_2 (S - K_2) + x_3 (S- K_3) geq 0, forall S geq 0$,
holds if and only if the two conditions are satisfied:
$x_1 - x_2 + x_3 geq 0$ and $x_1 (K_3 - K_1) - x_2 (K_3 - K_2) geq 0$.
For the life of me I can't seem to prove why this is the case! I know it should be basic algebra but I'm really struggling for some reason. Anyone have any tips?
inequality finance
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
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add a comment |
$begingroup$
I'm working through a practice book on mathematical finance, but struggling to prove part of a question on no-arbitrage conditions.
In the problem, I'm first given $K_1 < K_2 < K_3 $. Then, the book claims that the following relationship: $x_1 (S - K_1) - x_2 (S - K_2) + x_3 (S- K_3) geq 0, forall S geq 0$,
holds if and only if the two conditions are satisfied:
$x_1 - x_2 + x_3 geq 0$ and $x_1 (K_3 - K_1) - x_2 (K_3 - K_2) geq 0$.
For the life of me I can't seem to prove why this is the case! I know it should be basic algebra but I'm really struggling for some reason. Anyone have any tips?
inequality finance
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
$endgroup$
$begingroup$
@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
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– John Omielan
Jan 20 at 22:00
$begingroup$
@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
$endgroup$
– RRL
Jan 20 at 22:22
$begingroup$
@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
$endgroup$
– John Omielan
Jan 20 at 22:28
$begingroup$
@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
$endgroup$
– RRL
Jan 20 at 22:32
add a comment |
$begingroup$
I'm working through a practice book on mathematical finance, but struggling to prove part of a question on no-arbitrage conditions.
In the problem, I'm first given $K_1 < K_2 < K_3 $. Then, the book claims that the following relationship: $x_1 (S - K_1) - x_2 (S - K_2) + x_3 (S- K_3) geq 0, forall S geq 0$,
holds if and only if the two conditions are satisfied:
$x_1 - x_2 + x_3 geq 0$ and $x_1 (K_3 - K_1) - x_2 (K_3 - K_2) geq 0$.
For the life of me I can't seem to prove why this is the case! I know it should be basic algebra but I'm really struggling for some reason. Anyone have any tips?
inequality finance
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
$endgroup$
I'm working through a practice book on mathematical finance, but struggling to prove part of a question on no-arbitrage conditions.
In the problem, I'm first given $K_1 < K_2 < K_3 $. Then, the book claims that the following relationship: $x_1 (S - K_1) - x_2 (S - K_2) + x_3 (S- K_3) geq 0, forall S geq 0$,
holds if and only if the two conditions are satisfied:
$x_1 - x_2 + x_3 geq 0$ and $x_1 (K_3 - K_1) - x_2 (K_3 - K_2) geq 0$.
For the life of me I can't seem to prove why this is the case! I know it should be basic algebra but I'm really struggling for some reason. Anyone have any tips?
inequality finance
inequality finance
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
asked Jan 20 at 3:11
David LounganiDavid Loungani
183
183
$begingroup$
@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
$endgroup$
– John Omielan
Jan 20 at 22:00
$begingroup$
@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
$endgroup$
– RRL
Jan 20 at 22:22
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@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
$endgroup$
– John Omielan
Jan 20 at 22:28
$begingroup$
@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
$endgroup$
– RRL
Jan 20 at 22:32
add a comment |
$begingroup$
@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
$endgroup$
– John Omielan
Jan 20 at 22:00
$begingroup$
@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
$endgroup$
– RRL
Jan 20 at 22:22
$begingroup$
@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
$endgroup$
– John Omielan
Jan 20 at 22:28
$begingroup$
@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
$endgroup$
– RRL
Jan 20 at 22:32
$begingroup$
@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
$endgroup$
– John Omielan
Jan 20 at 22:00
$begingroup$
@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
$endgroup$
– John Omielan
Jan 20 at 22:00
$begingroup$
@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
$endgroup$
– RRL
Jan 20 at 22:22
$begingroup$
@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
$endgroup$
– RRL
Jan 20 at 22:22
$begingroup$
@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
$endgroup$
– John Omielan
Jan 20 at 22:28
$begingroup$
@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
$endgroup$
– John Omielan
Jan 20 at 22:28
$begingroup$
@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
$endgroup$
– RRL
Jan 20 at 22:32
$begingroup$
@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
$endgroup$
– RRL
Jan 20 at 22:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.
The payoff of this option portfolio at expiration for underlying price $S$ is
$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$
where $(S-K_j)^+ = max(S-K_j,0)$ the payoff of a standard European call option.
For $0 leqslant S leqslant K_1$ we have $V(s) = 0$.
For $K_1 leqslant S leqslant K_2$ we have $V(S) = x_1(S - K_1) geqslant 0$.
For $K_2 leqslant S leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(,K_2,,x_1(K_2 - K_1),)$ and $(,K_3,,x_1(K_3-K_1)- x_2(K_3-K_2),)$.
Consequently we have $V(S) geqslant 0$ for $K_2 leqslant S leqslant K_3$ if and only if
$$tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) geqslant 0$$
For $S geqslant K_3$ we have
$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$
and assuming that inequality (*) holds we have $V(S) geqslant 0$ for all $S geqslant K_3$ if and only if
$$tag{**} x_1 - x_2 + x_3 geqslant 0$$
answered Jan 20 at 6:06
RRLRRL
52.1k42573
$endgroup$
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
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– David Loungani
Jan 20 at 7:04
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
add a comment |
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Note what you state seems to require that $K_3 le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.
Using this, note that
$$x_1left(S - K_1right) - x_2left(S - K_2right) + x_3left(S - K_3right) ge 0 tag{1}label{eq1}$$
becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that
$$Sleft(x_1 - x_2 + x_3right) - x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{2}label{eq2}$$
To show the "if" part, if
$$x_1 - x_2 + x_3 ge 0 tag{3}label{eq3}$$
then
$$Sleft(x_1 - x_2 + x_3right) ge 0 tag{4}label{eq4}$$
based on $S ge 0$. With the second condition,
$$x_1left(K_3 - K_1right) - x_2left(K_3 - K_2right) ge 0 tag{5}label{eq5}$$
expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives
$$-x_1 K_1 + x_2 K_2 + K_3left(x_1 - x_2right) ge 0 tag{6}label{eq6}$$
From eqref{eq3}, and assuming that $K_3 le 0$, we get
$$x_1 - x_2 ge -x_3 Rightarrow K_3left(x_1 - x_2right) le -K_3 x_3 tag{7}label{eq7}$$
Reversing the inequality shows this can be substituted into eqref{eq6} to give that
$$-x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{8}label{eq8}$$
Putting eqref{eq4} and eqref{eq8} together gives eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 le 0$, note that if $x_1 = x_2 = 0$, then eqref{eq3} requires that $x_3 ge 0$. Setting $S = 0$ causes eqref{eq2} to become
$$0 - 0 + 0 - x_3 K_3 ge 0 tag{9}label{eq9}$$
but if $x_3 gt 0$, then $K_3 leq 0$. As the OP has stated in a comment to this answer that
the $K$'s represent different "strike prices" which would in fact generally be greater than $0$
there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.
For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of eqref{eq3} or eqref{eq5} don't hold, then eqref{eq2} doesn't hold either.
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
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Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
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– David Loungani
Jan 20 at 4:05
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@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
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– John Omielan
Jan 20 at 4:25
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Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
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– David Loungani
Jan 20 at 4:31
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@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
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– John Omielan
Jan 20 at 4:35
1
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@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
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– John Omielan
Jan 20 at 7:27
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.
The payoff of this option portfolio at expiration for underlying price $S$ is
$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$
where $(S-K_j)^+ = max(S-K_j,0)$ the payoff of a standard European call option.
For $0 leqslant S leqslant K_1$ we have $V(s) = 0$.
For $K_1 leqslant S leqslant K_2$ we have $V(S) = x_1(S - K_1) geqslant 0$.
For $K_2 leqslant S leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(,K_2,,x_1(K_2 - K_1),)$ and $(,K_3,,x_1(K_3-K_1)- x_2(K_3-K_2),)$.
Consequently we have $V(S) geqslant 0$ for $K_2 leqslant S leqslant K_3$ if and only if
$$tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) geqslant 0$$
For $S geqslant K_3$ we have
$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$
and assuming that inequality (*) holds we have $V(S) geqslant 0$ for all $S geqslant K_3$ if and only if
$$tag{**} x_1 - x_2 + x_3 geqslant 0$$
answered Jan 20 at 6:06
RRLRRL
52.1k42573
$endgroup$
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
add a comment |
$begingroup$
This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.
The payoff of this option portfolio at expiration for underlying price $S$ is
$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$
where $(S-K_j)^+ = max(S-K_j,0)$ the payoff of a standard European call option.
For $0 leqslant S leqslant K_1$ we have $V(s) = 0$.
For $K_1 leqslant S leqslant K_2$ we have $V(S) = x_1(S - K_1) geqslant 0$.
For $K_2 leqslant S leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(,K_2,,x_1(K_2 - K_1),)$ and $(,K_3,,x_1(K_3-K_1)- x_2(K_3-K_2),)$.
Consequently we have $V(S) geqslant 0$ for $K_2 leqslant S leqslant K_3$ if and only if
$$tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) geqslant 0$$
For $S geqslant K_3$ we have
$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$
and assuming that inequality (*) holds we have $V(S) geqslant 0$ for all $S geqslant K_3$ if and only if
$$tag{**} x_1 - x_2 + x_3 geqslant 0$$
answered Jan 20 at 6:06
RRLRRL
52.1k42573
$endgroup$
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
add a comment |
$begingroup$
This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.
The payoff of this option portfolio at expiration for underlying price $S$ is
$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$
where $(S-K_j)^+ = max(S-K_j,0)$ the payoff of a standard European call option.
For $0 leqslant S leqslant K_1$ we have $V(s) = 0$.
For $K_1 leqslant S leqslant K_2$ we have $V(S) = x_1(S - K_1) geqslant 0$.
For $K_2 leqslant S leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(,K_2,,x_1(K_2 - K_1),)$ and $(,K_3,,x_1(K_3-K_1)- x_2(K_3-K_2),)$.
Consequently we have $V(S) geqslant 0$ for $K_2 leqslant S leqslant K_3$ if and only if
$$tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) geqslant 0$$
For $S geqslant K_3$ we have
$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$
and assuming that inequality (*) holds we have $V(S) geqslant 0$ for all $S geqslant K_3$ if and only if
$$tag{**} x_1 - x_2 + x_3 geqslant 0$$
answered Jan 20 at 6:06
RRLRRL
52.1k42573
$endgroup$
This will make sense for a portfolio of three call options with common expiry and with strike prices $0 < K_1 < K_2 < K_3$, where we are long $x_1 > 0$ options struck at $K_1$, short $x_2 > 0$ options struck at $K_2$ and long $x_3> 0$ options struck at $K_3$.
The payoff of this option portfolio at expiration for underlying price $S$ is
$$V(S) = x_1(S-K_1)^+ - x_2(S - K_2)^+ + x_3(S - K_3)^+$$
where $(S-K_j)^+ = max(S-K_j,0)$ the payoff of a standard European call option.
For $0 leqslant S leqslant K_1$ we have $V(s) = 0$.
For $K_1 leqslant S leqslant K_2$ we have $V(S) = x_1(S - K_1) geqslant 0$.
For $K_2 leqslant S leqslant K_3$ we have $V(S) = x_1(S - K_1) - x_2(S- K_2)$. This is a linear function joining the points $(,K_2,,x_1(K_2 - K_1),)$ and $(,K_3,,x_1(K_3-K_1)- x_2(K_3-K_2),)$.
Consequently we have $V(S) geqslant 0$ for $K_2 leqslant S leqslant K_3$ if and only if
$$tag{*}x_1(K_3-K_1)- x_2(K_3-K_2) geqslant 0$$
For $S geqslant K_3$ we have
$$V(S) = x_1(S- K_1) - x_2(X- K_2) + x_3(S- K_3) \ = x_1(K_3-K_1)- x_2(K_3-K_2) +(x_1 - x_2 + x_3)(S - K_3),$$
and assuming that inequality (*) holds we have $V(S) geqslant 0$ for all $S geqslant K_3$ if and only if
$$tag{**} x_1 - x_2 + x_3 geqslant 0$$
answered Jan 20 at 6:06
RRLRRL
52.1k42573
answered Jan 20 at 6:06
RRLRRL
52.1k42573
answered Jan 20 at 6:06
RRLRRL
52.1k42573
answered Jan 20 at 6:06
RRLRRL
52.1k42573
52.1k42573
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
add a comment |
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
1
1
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
$begingroup$
Yes!!!!!!!!! This all makes sense now! You're certainly right about the payoffs, it's obvious now to me that it's a typo. Thank you!
$endgroup$
– David Loungani
Jan 20 at 7:04
1
1
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
$begingroup$
@DavidLoungani: You're very welcome. Glad it is clear now.
$endgroup$
– RRL
Jan 20 at 8:17
add a comment |
$begingroup$
Note what you state seems to require that $K_3 le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.
Using this, note that
$$x_1left(S - K_1right) - x_2left(S - K_2right) + x_3left(S - K_3right) ge 0 tag{1}label{eq1}$$
becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that
$$Sleft(x_1 - x_2 + x_3right) - x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{2}label{eq2}$$
To show the "if" part, if
$$x_1 - x_2 + x_3 ge 0 tag{3}label{eq3}$$
then
$$Sleft(x_1 - x_2 + x_3right) ge 0 tag{4}label{eq4}$$
based on $S ge 0$. With the second condition,
$$x_1left(K_3 - K_1right) - x_2left(K_3 - K_2right) ge 0 tag{5}label{eq5}$$
expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives
$$-x_1 K_1 + x_2 K_2 + K_3left(x_1 - x_2right) ge 0 tag{6}label{eq6}$$
From eqref{eq3}, and assuming that $K_3 le 0$, we get
$$x_1 - x_2 ge -x_3 Rightarrow K_3left(x_1 - x_2right) le -K_3 x_3 tag{7}label{eq7}$$
Reversing the inequality shows this can be substituted into eqref{eq6} to give that
$$-x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{8}label{eq8}$$
Putting eqref{eq4} and eqref{eq8} together gives eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 le 0$, note that if $x_1 = x_2 = 0$, then eqref{eq3} requires that $x_3 ge 0$. Setting $S = 0$ causes eqref{eq2} to become
$$0 - 0 + 0 - x_3 K_3 ge 0 tag{9}label{eq9}$$
but if $x_3 gt 0$, then $K_3 leq 0$. As the OP has stated in a comment to this answer that
the $K$'s represent different "strike prices" which would in fact generally be greater than $0$
there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.
For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of eqref{eq3} or eqref{eq5} don't hold, then eqref{eq2} doesn't hold either.
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
$endgroup$
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
1
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
add a comment |
$begingroup$
Note what you state seems to require that $K_3 le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.
Using this, note that
$$x_1left(S - K_1right) - x_2left(S - K_2right) + x_3left(S - K_3right) ge 0 tag{1}label{eq1}$$
becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that
$$Sleft(x_1 - x_2 + x_3right) - x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{2}label{eq2}$$
To show the "if" part, if
$$x_1 - x_2 + x_3 ge 0 tag{3}label{eq3}$$
then
$$Sleft(x_1 - x_2 + x_3right) ge 0 tag{4}label{eq4}$$
based on $S ge 0$. With the second condition,
$$x_1left(K_3 - K_1right) - x_2left(K_3 - K_2right) ge 0 tag{5}label{eq5}$$
expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives
$$-x_1 K_1 + x_2 K_2 + K_3left(x_1 - x_2right) ge 0 tag{6}label{eq6}$$
From eqref{eq3}, and assuming that $K_3 le 0$, we get
$$x_1 - x_2 ge -x_3 Rightarrow K_3left(x_1 - x_2right) le -K_3 x_3 tag{7}label{eq7}$$
Reversing the inequality shows this can be substituted into eqref{eq6} to give that
$$-x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{8}label{eq8}$$
Putting eqref{eq4} and eqref{eq8} together gives eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 le 0$, note that if $x_1 = x_2 = 0$, then eqref{eq3} requires that $x_3 ge 0$. Setting $S = 0$ causes eqref{eq2} to become
$$0 - 0 + 0 - x_3 K_3 ge 0 tag{9}label{eq9}$$
but if $x_3 gt 0$, then $K_3 leq 0$. As the OP has stated in a comment to this answer that
the $K$'s represent different "strike prices" which would in fact generally be greater than $0$
there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.
For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of eqref{eq3} or eqref{eq5} don't hold, then eqref{eq2} doesn't hold either.
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
$endgroup$
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
1
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
add a comment |
$begingroup$
Note what you state seems to require that $K_3 le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.
Using this, note that
$$x_1left(S - K_1right) - x_2left(S - K_2right) + x_3left(S - K_3right) ge 0 tag{1}label{eq1}$$
becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that
$$Sleft(x_1 - x_2 + x_3right) - x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{2}label{eq2}$$
To show the "if" part, if
$$x_1 - x_2 + x_3 ge 0 tag{3}label{eq3}$$
then
$$Sleft(x_1 - x_2 + x_3right) ge 0 tag{4}label{eq4}$$
based on $S ge 0$. With the second condition,
$$x_1left(K_3 - K_1right) - x_2left(K_3 - K_2right) ge 0 tag{5}label{eq5}$$
expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives
$$-x_1 K_1 + x_2 K_2 + K_3left(x_1 - x_2right) ge 0 tag{6}label{eq6}$$
From eqref{eq3}, and assuming that $K_3 le 0$, we get
$$x_1 - x_2 ge -x_3 Rightarrow K_3left(x_1 - x_2right) le -K_3 x_3 tag{7}label{eq7}$$
Reversing the inequality shows this can be substituted into eqref{eq6} to give that
$$-x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{8}label{eq8}$$
Putting eqref{eq4} and eqref{eq8} together gives eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 le 0$, note that if $x_1 = x_2 = 0$, then eqref{eq3} requires that $x_3 ge 0$. Setting $S = 0$ causes eqref{eq2} to become
$$0 - 0 + 0 - x_3 K_3 ge 0 tag{9}label{eq9}$$
but if $x_3 gt 0$, then $K_3 leq 0$. As the OP has stated in a comment to this answer that
the $K$'s represent different "strike prices" which would in fact generally be greater than $0$
there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.
For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of eqref{eq3} or eqref{eq5} don't hold, then eqref{eq2} doesn't hold either.
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
$endgroup$
Note what you state seems to require that $K_3 le 0$. I later found that this is due to the question statement not quite being correct. Please read RRL's answer for a solution to the question as intended. Nonetheless, the following does prove part of the conditions for options, but not arbitrage.
Using this, note that
$$x_1left(S - K_1right) - x_2left(S - K_2right) + x_3left(S - K_3right) ge 0 tag{1}label{eq1}$$
becomes after expanding the terms, collecting all those using $S$ and factoring it out to get that
$$Sleft(x_1 - x_2 + x_3right) - x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{2}label{eq2}$$
To show the "if" part, if
$$x_1 - x_2 + x_3 ge 0 tag{3}label{eq3}$$
then
$$Sleft(x_1 - x_2 + x_3right) ge 0 tag{4}label{eq4}$$
based on $S ge 0$. With the second condition,
$$x_1left(K_3 - K_1right) - x_2left(K_3 - K_2right) ge 0 tag{5}label{eq5}$$
expanding it, collecting the terms using $K_3$ and then factoring out $K_3$, gives
$$-x_1 K_1 + x_2 K_2 + K_3left(x_1 - x_2right) ge 0 tag{6}label{eq6}$$
From eqref{eq3}, and assuming that $K_3 le 0$, we get
$$x_1 - x_2 ge -x_3 Rightarrow K_3left(x_1 - x_2right) le -K_3 x_3 tag{7}label{eq7}$$
Reversing the inequality shows this can be substituted into eqref{eq6} to give that
$$-x_1 K_1 + x_2 K_2 - x_3 K_3 ge 0 tag{8}label{eq8}$$
Putting eqref{eq4} and eqref{eq8} together gives eqref{eq2}. Note I didn't use anything regarding the relationship of $K_1$ or $K_2$ in terms of $K_3$ here. As for the requirement for $K_3 le 0$, note that if $x_1 = x_2 = 0$, then eqref{eq3} requires that $x_3 ge 0$. Setting $S = 0$ causes eqref{eq2} to become
$$0 - 0 + 0 - x_3 K_3 ge 0 tag{9}label{eq9}$$
but if $x_3 gt 0$, then $K_3 leq 0$. As the OP has stated in a comment to this answer that
the $K$'s represent different "strike prices" which would in fact generally be greater than $0$
there must be some other conditions or restrictions I'm not aware of, or possibly one of the statements is not presented correctly. For now, this is the best I can do here.
For the "only if" part, I believe the easiest way to show it would possibly be to provide an example where one of eqref{eq3} or eqref{eq5} don't hold, then eqref{eq2} doesn't hold either.
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
edited Jan 20 at 22:30
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
answered Jan 20 at 3:42
John OmielanJohn Omielan
3,4801215
3,4801215
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
1
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
add a comment |
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
1
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
Thank you for your response! Unfortunately, I don't think we can assume the K's are less than 0. It's not explicitly stated in the problem, but the K's represent different "strike prices" which would in fact generally be greater than 0.
$endgroup$
– David Loungani
Jan 20 at 4:05
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
@DavidLoungani Thanks for the additional information. I edited the answer to give an explicit example that shows what you're asking only works in general if $K_3 leq 0$. Thus, as I stated, there must be some other conditions or restrictions, or something is wrong. Please check to see if that is the case. Regardless, you could likely use similar techniques to what I've shown to solve this, so perhaps you can do it on your own. Good luck with it.
$endgroup$
– John Omielan
Jan 20 at 4:25
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
Thanks John, I follow your argument and I think I'm going to actually reach out to the textbook authors to see what they have to say. As you mentioned, I'm guessing there is some implicit assumption which is being left unstated.
$endgroup$
– David Loungani
Jan 20 at 4:31
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
$begingroup$
@DavidLoungani You are welcome. Note that nothing is stated about the $x_1$, $x_2$ and $x_3$, so it seems strange that they are used unless they have some specific meaning. I have a feeling that there are some additional conditions on these values which will cause everything to work properly.
$endgroup$
– John Omielan
Jan 20 at 4:35
1
1
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
$begingroup$
@DavidLoungani I see now why my work didn't solve your problem appropriately. With the appropriate additional conditions that RRL states, everything will work then, as shown in RRL's answer.
$endgroup$
– John Omielan
Jan 20 at 7:27
add a comment |
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@RRL I tried to answer the question as it's stated, but was only able to prove the "if" part always holding in the case where $K_3 le 0$. Now I understand why I wasn't able to succeed as I know nothing about arbitrage so I didn't realize the correct form of using $left(S - K_jright)^{+}$ instead. Since I didn't answer the intended question, and I don't think there's much value for anybody to read my solution (as you say, "this is nonsense"), I am deleting it.
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– John Omielan
Jan 20 at 22:00
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@JohnOmielan: Nonsense means there is no such arbitrage condition for forward contracts with nonnegative forward price. The question itself is not nonsense. It still takes some effort to prove the conditions. Others may find this useful -- although it is entirely up you. Undelete and I will give it an upvote.
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– RRL
Jan 20 at 22:22
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@RRL Thanks for the comment & support. I only proved part of it, but perhaps it might be of some use for a few people. As you requested, I have undeleted my solution. Also, I am going to add a comment at the top that it's trying to answer the question as stated, not as it was intended.
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– John Omielan
Jan 20 at 22:28
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@JohnOmielan: I would not get too concerned. There are relatively few questions here with the finance tag and many are elementary (compound interest, etc.) QuantitativeFinance.SE is usually the best place to ask, but I am always happy when something interesting comes up here.
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– RRL
Jan 20 at 22:32