Mixing
Current price of a European call option
$begingroup$
I have the following situation:
Consider a European call option on a non-dividend paying stock,
with strike $K$, time to expiry $T$ and current stock price $S_0$. Assume that
the risk free rate is given by $r$ (continuous compounding). Let $c$ denote the
current price of the option.
How can I prove that $c geq max{S_0 − Ke^{-rT}, 0}$ ?
stochastic-processes stochastic-calculus finance
asked Jan 31 at 14:35
LorLor
283
$endgroup$
add a comment |
$begingroup$
I have the following situation:
Consider a European call option on a non-dividend paying stock,
with strike $K$, time to expiry $T$ and current stock price $S_0$. Assume that
the risk free rate is given by $r$ (continuous compounding). Let $c$ denote the
current price of the option.
How can I prove that $c geq max{S_0 − Ke^{-rT}, 0}$ ?
stochastic-processes stochastic-calculus finance
asked Jan 31 at 14:35
LorLor
283
$endgroup$
$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38
add a comment |
$begingroup$
I have the following situation:
Consider a European call option on a non-dividend paying stock,
with strike $K$, time to expiry $T$ and current stock price $S_0$. Assume that
the risk free rate is given by $r$ (continuous compounding). Let $c$ denote the
current price of the option.
How can I prove that $c geq max{S_0 − Ke^{-rT}, 0}$ ?
stochastic-processes stochastic-calculus finance
asked Jan 31 at 14:35
LorLor
283
$endgroup$
I have the following situation:
Consider a European call option on a non-dividend paying stock,
with strike $K$, time to expiry $T$ and current stock price $S_0$. Assume that
the risk free rate is given by $r$ (continuous compounding). Let $c$ denote the
current price of the option.
How can I prove that $c geq max{S_0 − Ke^{-rT}, 0}$ ?
stochastic-processes stochastic-calculus finance
stochastic-processes stochastic-calculus finance
asked Jan 31 at 14:35
LorLor
283
asked Jan 31 at 14:35
LorLor
283
edited Jan 31 at 15:47
Lor
asked Jan 31 at 14:35
LorLor
283
asked Jan 31 at 14:35
LorLor
283
asked Jan 31 at 14:35
LorLor
283
283
$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38
add a comment |
$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38
$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You use the no free lunch principle. You describe a strategy that will guarantee a profit if $c$ is less than either number. The $0$ comes from noting that if somebody will pay you to to take the option you can collect the payment and let the option expire. The other term comes from selling a share and buying an option, then exercising at expiration. If the option price is below this value, you make a guaranteed profit.
A good way to simplify it is to assume $r=0$, which is not far wrong today. Then it just says that if the strike price is below the stock price, the option has to cost at least the difference. If it didn't, instead of buying the stock you would buy the option and exercise it. The effect of interest is to reduce the value of the $K$ you will have to pay in the future, so the option price has to be a little higher yet to avoid this arbitrage.
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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oldest
votes
$begingroup$
You use the no free lunch principle. You describe a strategy that will guarantee a profit if $c$ is less than either number. The $0$ comes from noting that if somebody will pay you to to take the option you can collect the payment and let the option expire. The other term comes from selling a share and buying an option, then exercising at expiration. If the option price is below this value, you make a guaranteed profit.
A good way to simplify it is to assume $r=0$, which is not far wrong today. Then it just says that if the strike price is below the stock price, the option has to cost at least the difference. If it didn't, instead of buying the stock you would buy the option and exercise it. The effect of interest is to reduce the value of the $K$ you will have to pay in the future, so the option price has to be a little higher yet to avoid this arbitrage.
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
add a comment |
$begingroup$
You use the no free lunch principle. You describe a strategy that will guarantee a profit if $c$ is less than either number. The $0$ comes from noting that if somebody will pay you to to take the option you can collect the payment and let the option expire. The other term comes from selling a share and buying an option, then exercising at expiration. If the option price is below this value, you make a guaranteed profit.
A good way to simplify it is to assume $r=0$, which is not far wrong today. Then it just says that if the strike price is below the stock price, the option has to cost at least the difference. If it didn't, instead of buying the stock you would buy the option and exercise it. The effect of interest is to reduce the value of the $K$ you will have to pay in the future, so the option price has to be a little higher yet to avoid this arbitrage.
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
add a comment |
$begingroup$
You use the no free lunch principle. You describe a strategy that will guarantee a profit if $c$ is less than either number. The $0$ comes from noting that if somebody will pay you to to take the option you can collect the payment and let the option expire. The other term comes from selling a share and buying an option, then exercising at expiration. If the option price is below this value, you make a guaranteed profit.
A good way to simplify it is to assume $r=0$, which is not far wrong today. Then it just says that if the strike price is below the stock price, the option has to cost at least the difference. If it didn't, instead of buying the stock you would buy the option and exercise it. The effect of interest is to reduce the value of the $K$ you will have to pay in the future, so the option price has to be a little higher yet to avoid this arbitrage.
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
$endgroup$
You use the no free lunch principle. You describe a strategy that will guarantee a profit if $c$ is less than either number. The $0$ comes from noting that if somebody will pay you to to take the option you can collect the payment and let the option expire. The other term comes from selling a share and buying an option, then exercising at expiration. If the option price is below this value, you make a guaranteed profit.
A good way to simplify it is to assume $r=0$, which is not far wrong today. Then it just says that if the strike price is below the stock price, the option has to cost at least the difference. If it didn't, instead of buying the stock you would buy the option and exercise it. The effect of interest is to reduce the value of the $K$ you will have to pay in the future, so the option price has to be a little higher yet to avoid this arbitrage.
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
edited Feb 1 at 15:35
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
answered Jan 31 at 15:19
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
add a comment |
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Thanks, I think I am getting closer to understanding it. I guess I am having trouble understanding what the term $S_0 − Ke^{-rT}$ is, where does it come from? Shouldn't it just be K?
$endgroup$
– Lor
Jan 31 at 16:54
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Today you can sell the stock and buy an option. You get $S_0-c$ for that. You deposit the money in the bank, getting interest, so at the end of the option you have $(S_0-c)e^{rT}$ in cash and are short one share. Now you exercise the option, pay $K$, deliver the share, and have closed all your positions. If you have any money left over, that is free money because you were not at risk of any market changes.
$endgroup$
– Ross Millikan
Jan 31 at 17:04
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Ok thanks. Is it the same in the case of an American call option? I don't see why it shouldn't be but I am not 100% sure
$endgroup$
– Lor
Feb 1 at 14:13
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Yes, it would. Both options allow this course of action. I recall that you should never exercise a call early unless to capture a dividend, so they are effectively the same for a non-dividend stock.
$endgroup$
– Ross Millikan
Feb 1 at 15:03
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
$begingroup$
Great, thank you so much
$endgroup$
– Lor
Feb 1 at 15:38
add a comment |
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$begingroup$
Is your formula correct? As we let $Tto infty$ your bound goes to $0$ (since $Ke^{rt}to infty$), when clearly the call should increase in value. Did you possibly mean $Ke^{-rT}$?
$endgroup$
– lulu
Jan 31 at 15:20
$begingroup$
You're right, it's a mistake. I did mean $Ke^{-rT}$
$endgroup$
– Lor
Jan 31 at 15:38