Mixing
If $W_t$ is a Wiener process, what is $ operatorname{Var}(W_t+W_s)$?
$begingroup$
Doing some questions from the book '150 Most Frequently Asked Questions on Quant Interviews', I try to solve the following:
If $W_t$ is a Wiener process, what is $ operatorname{Var}(W_t+W_s)$?
It seemed pretty straightforward to me, and solved as follows, for $t$ and $s$ positive:
$begin{align*}
operatorname{Var}(W_t+W_s) &= operatorname{Var}(W_t)+ operatorname{Var}(W_s)+2 operatorname{Cov}(Wt,Ws)\
&= E[W_t^2]-(E[W_t])^2+E[W_s^2]-(E[W_s])^2+2(E[W_tWs]-E[W_t]E[Ws])\
&= E[W_t^2] + E[W_s^2] + 2E[W_tWs]\
&= t+s+2min(s,t)
end{align*}$
However, the solution in the book is:
$ operatorname{Var}(W_t+W_s) =max(s,t)+3min(s,t)$
Is my solution wrong? If yes, what part of it is wrong?
stochastic-processes stochastic-calculus brownian-motion
edited Jan 26 at 20:15
Bernard
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
$endgroup$
add a comment |
$begingroup$
Doing some questions from the book '150 Most Frequently Asked Questions on Quant Interviews', I try to solve the following:
If $W_t$ is a Wiener process, what is $ operatorname{Var}(W_t+W_s)$?
It seemed pretty straightforward to me, and solved as follows, for $t$ and $s$ positive:
$begin{align*}
operatorname{Var}(W_t+W_s) &= operatorname{Var}(W_t)+ operatorname{Var}(W_s)+2 operatorname{Cov}(Wt,Ws)\
&= E[W_t^2]-(E[W_t])^2+E[W_s^2]-(E[W_s])^2+2(E[W_tWs]-E[W_t]E[Ws])\
&= E[W_t^2] + E[W_s^2] + 2E[W_tWs]\
&= t+s+2min(s,t)
end{align*}$
However, the solution in the book is:
$ operatorname{Var}(W_t+W_s) =max(s,t)+3min(s,t)$
Is my solution wrong? If yes, what part of it is wrong?
stochastic-processes stochastic-calculus brownian-motion
edited Jan 26 at 20:15
Bernard
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
$endgroup$
add a comment |
$begingroup$
Doing some questions from the book '150 Most Frequently Asked Questions on Quant Interviews', I try to solve the following:
If $W_t$ is a Wiener process, what is $ operatorname{Var}(W_t+W_s)$?
It seemed pretty straightforward to me, and solved as follows, for $t$ and $s$ positive:
$begin{align*}
operatorname{Var}(W_t+W_s) &= operatorname{Var}(W_t)+ operatorname{Var}(W_s)+2 operatorname{Cov}(Wt,Ws)\
&= E[W_t^2]-(E[W_t])^2+E[W_s^2]-(E[W_s])^2+2(E[W_tWs]-E[W_t]E[Ws])\
&= E[W_t^2] + E[W_s^2] + 2E[W_tWs]\
&= t+s+2min(s,t)
end{align*}$
However, the solution in the book is:
$ operatorname{Var}(W_t+W_s) =max(s,t)+3min(s,t)$
Is my solution wrong? If yes, what part of it is wrong?
stochastic-processes stochastic-calculus brownian-motion
edited Jan 26 at 20:15
Bernard
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
$endgroup$
Doing some questions from the book '150 Most Frequently Asked Questions on Quant Interviews', I try to solve the following:
If $W_t$ is a Wiener process, what is $ operatorname{Var}(W_t+W_s)$?
It seemed pretty straightforward to me, and solved as follows, for $t$ and $s$ positive:
$begin{align*}
operatorname{Var}(W_t+W_s) &= operatorname{Var}(W_t)+ operatorname{Var}(W_s)+2 operatorname{Cov}(Wt,Ws)\
&= E[W_t^2]-(E[W_t])^2+E[W_s^2]-(E[W_s])^2+2(E[W_tWs]-E[W_t]E[Ws])\
&= E[W_t^2] + E[W_s^2] + 2E[W_tWs]\
&= t+s+2min(s,t)
end{align*}$
However, the solution in the book is:
$ operatorname{Var}(W_t+W_s) =max(s,t)+3min(s,t)$
Is my solution wrong? If yes, what part of it is wrong?
stochastic-processes stochastic-calculus brownian-motion
stochastic-processes stochastic-calculus brownian-motion
edited Jan 26 at 20:15
Bernard
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
edited Jan 26 at 20:15
Bernard
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
edited Jan 26 at 20:15
Bernard
123k741117
edited Jan 26 at 20:15
Bernard
123k741117
edited Jan 26 at 20:15
Bernard
123k741117
123k741117
asked Jan 26 at 19:58
Zizou23Zizou23
524315
asked Jan 26 at 19:58
Zizou23Zizou23
524315
asked Jan 26 at 19:58
Zizou23Zizou23
524315
524315
add a comment |
add a comment |
3 Answers
3
active
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votes
$begingroup$
I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then
$$
begin{aligned}
t + s + 2 min(s, t)
&= max(s, t) + min(s, t) + 2 min(s, t) \
&= max(s, t) + 3 min(s, t).
end{aligned}
$$
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
$endgroup$
add a comment |
$begingroup$
Your solution is correct - and so is theirs. This is because
$$t+s+2mathrm{min}(s,t) = mathrm{max}(s,t) + 3mathrm{min}(s,t).$$
What I'm wondering about is how they got to that form.
answered Jan 26 at 20:07
DahnDahn
2,42811937
$endgroup$
add a comment |
$begingroup$
Nothing is wrong -- since $t+s = max(s,t) + min(s,t)$ your solution equals the book's.
It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then
$$
begin{aligned}
t + s + 2 min(s, t)
&= max(s, t) + min(s, t) + 2 min(s, t) \
&= max(s, t) + 3 min(s, t).
end{aligned}
$$
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
$endgroup$
add a comment |
$begingroup$
I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then
$$
begin{aligned}
t + s + 2 min(s, t)
&= max(s, t) + min(s, t) + 2 min(s, t) \
&= max(s, t) + 3 min(s, t).
end{aligned}
$$
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
$endgroup$
add a comment |
$begingroup$
I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then
$$
begin{aligned}
t + s + 2 min(s, t)
&= max(s, t) + min(s, t) + 2 min(s, t) \
&= max(s, t) + 3 min(s, t).
end{aligned}
$$
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
$endgroup$
I was staring at this for a while before realizing your solutions are equivalent. For example, if $s < t$, then
$$
begin{aligned}
t + s + 2 min(s, t)
&= max(s, t) + min(s, t) + 2 min(s, t) \
&= max(s, t) + 3 min(s, t).
end{aligned}
$$
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
answered Jan 26 at 20:08
Artem MavrinArtem Mavrin
1,9631713
1,9631713
add a comment |
add a comment |
$begingroup$
Your solution is correct - and so is theirs. This is because
$$t+s+2mathrm{min}(s,t) = mathrm{max}(s,t) + 3mathrm{min}(s,t).$$
What I'm wondering about is how they got to that form.
answered Jan 26 at 20:07
DahnDahn
2,42811937
$endgroup$
add a comment |
$begingroup$
Your solution is correct - and so is theirs. This is because
$$t+s+2mathrm{min}(s,t) = mathrm{max}(s,t) + 3mathrm{min}(s,t).$$
What I'm wondering about is how they got to that form.
answered Jan 26 at 20:07
DahnDahn
2,42811937
$endgroup$
add a comment |
$begingroup$
Your solution is correct - and so is theirs. This is because
$$t+s+2mathrm{min}(s,t) = mathrm{max}(s,t) + 3mathrm{min}(s,t).$$
What I'm wondering about is how they got to that form.
answered Jan 26 at 20:07
DahnDahn
2,42811937
$endgroup$
Your solution is correct - and so is theirs. This is because
$$t+s+2mathrm{min}(s,t) = mathrm{max}(s,t) + 3mathrm{min}(s,t).$$
What I'm wondering about is how they got to that form.
answered Jan 26 at 20:07
DahnDahn
2,42811937
answered Jan 26 at 20:07
DahnDahn
2,42811937
answered Jan 26 at 20:07
DahnDahn
2,42811937
answered Jan 26 at 20:07
DahnDahn
2,42811937
2,42811937
add a comment |
add a comment |
$begingroup$
Nothing is wrong -- since $t+s = max(s,t) + min(s,t)$ your solution equals the book's.
It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
$begingroup$
Nothing is wrong -- since $t+s = max(s,t) + min(s,t)$ your solution equals the book's.
It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
$endgroup$
add a comment |
$begingroup$
Nothing is wrong -- since $t+s = max(s,t) + min(s,t)$ your solution equals the book's.
It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
$endgroup$
Nothing is wrong -- since $t+s = max(s,t) + min(s,t)$ your solution equals the book's.
It may simplify the calculation to assume WLOG that $t<s$ -- then $W_t+W_s=2W_t+(W_s-W_t)$, and $W_t$ and $W_s-W_t$ are independent.
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
answered Jan 26 at 20:08
Henning MakholmHenning Makholm
242k17308550
242k17308550
add a comment |
add a comment |
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