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I need to calculate the variance of this double stochastic integration: $$ I=\int_{t-m}^t \int_{s-m}^se^{-k(s-u)}dW(u)ds $$

Note that variable $s$ is in the integrand as well as the boundaries of the integral. I have this solution by changing the order of stochastic integration and then apply Ito Isometry to calculate the variance.

Solution A $$I=\int_{t-2m}^t \int_{t-m}^{u+m}e^{-k(s-u)}dsdW(u)+\int_{t-m}^t \int_{u}^te^{-k(s-u)}dsdW(u)$$ hence $$I=-\frac{1}{k}(J+K)$$ with $$ J=\int_{t-2m}^{t-m} [e^{-km}-e^{-k(t-m-u)}]dW(u)$$ and $$K=\int_{t-m}^t [e^{-k(t-u)}-1]dW(u) $$ Apply Ito Isometry to the integration $$ Var\{J\} =\int_{t-2m}^{t-m} [e^{-km}-e^{-k(t-m-u)}]^2du =\frac{1}{2k}-\frac{2}{k} e^{-km}+(m+\frac{3}{2k}) e^{-2km} $$ $$ Var\{K\}=\int_{t-m}^{t} [e^{-k(t-u)}-1]^2du=m-\frac{3}{2k}+\frac{2}{k} e^{-km}-\frac{1}{2k}e^{-2km} $$

However, my colleague suggested another solution which is quite different from above approach

Solution B $$ I=\int_{t-m}^tds\int_{s-m}^se^{-k(s-u)}dW(u) $$

Set $y=u-(s-m)$ then $u=y+(s-m)$ substitute $u$ by $y$. $$ I=\int_{t-m}^tds\,L\qquad\text{with}\qquad L=\int_0^me^{-k(m-y)}dW(y) $$ hence $$ Var\{L\}=\int_{0}^m e^{-2k(m-y)}dy=\frac{1}{2k}(1-e^{-2km}) $$ Would appreciate if you can provide some fresh opinions on both solutions.

Did
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Wu David
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  • Would appreciate if you can provide some opinions on both solutions and check what might be missing on both solutions – Wu David Nov 02 '15 at 11:58
  • Where is the computation of the variance? – Did Nov 02 '15 at 12:22
  • Please see the revised one with variance calculation by applying Ito Isometry – Wu David Nov 02 '15 at 14:12
  • The Ito Isometry is straightforward. But is it really true that we can separate the double integration according to solution B when the boundaries are functions of the variables. I think Solution A by changing the order of the integration should be the right direciton – Wu David Nov 02 '15 at 14:15
  • Here is a somewhat more systematic approach: to be computed is $E(X^2)$ with $$X=\int g(u)dW(u)\qquad g(u)=\int\mathbb 1_{t-m<s<t}\mathbb 1_{s-m<u<s}e^{-k(s-u)}ds,$$ The function $g$ is deterministic hence $$E(X^2)=\int g(u)^2du=\iint!!!!!\int\mathbb 1_{s-m<u<s}\mathbb 1_{r-m<u<r}\mathbb 1_{t-m<s<t}\mathbb 1_{t-m<r<t}e^{-k(s+r-2u)}dsdrdu,$$ or, equivalently, $$E(X^2)=\iint\left(\int\mathbb 1_{\max(s,r)-m<u<\min(s,r)}e^{2ku}du\right)\mathbb 1_{t-m<s<t}\mathbb 1_{t-m<r<t}e^{-k(s+r)}dsdr.$$ Performing the inner integration and using $2\max(s,r)-(s+r)=|s-r|=s+r-\min(s,r)$, ... – Did Nov 02 '15 at 14:54
  • ... one gets $$2kE(X^2)=\iint(e^{-k|s-r|}-e^{k|s-r|-2km})\mathbb 1_{t-m<s<t}\mathbb 1_{t-m<r<t}dsdr=m^2h(km),$$ with $$h(a)=\iint_{[0,1]^2}(e^{-a|s-r|}-e^{a|s-r|-2a})dsdr=2e^{-a}\int_0^1(e^{aw}-e^{-aw})wdw,$$ which surely you can finish. – Did Nov 02 '15 at 14:54
  • Thanks Did, by following your solution I have derived the same answer as my solution A. In solution A I divided the integration domain into two triangles which is then a bit more easy to solve. A stupid question: how did you get the last equation? – Wu David Nov 03 '15 at 17:53
  • I cheated... More seriously, I happen to know that if $(S,R)$ is uniform on the unit square then $V=|S-R|$ has density $2(1-v)$ on the unit interval hence $W=1-|S-R|$ has density $2w$. I should have included this explanation into my comments but I was tired. Sorry. – Did Nov 03 '15 at 17:56
  • Thanks for your quick response. Your systematic solution is very helpful to confirm that my solution A. Do you think solution B is not quite right even based on instinct – Wu David Nov 03 '15 at 18:02
  • The trouble there is that $$\int_{s-m}^se^{-k(s-u)}dW(u)\ne\int_0^me^{-k(m-y)}dW(y)$$ but $$\int_{s-m}^se^{-k(s-u)}dW(u)\ne\int_0^me^{-k(m-y)}dW^s(y),$$ where each $W^s$ is a Brownian motion, defined by $$W^s(y)=W(s-m+y),$$ and when you will regroup all these different processes $W^s$ to integrate over $s$ as if these were all equal to $W$, funny things will ensue – Did Nov 03 '15 at 18:08
  • *Typo: The second $\ne$ in my comment should be $=$ and it might be neater to define each $W^s$ by $$W^s(y)=W(s-m+y)-W(s-m),$$ so that each $W^s$ is a standard Brownian motion starting from $W^s(0)=0$. – Did Nov 08 '15 at 17:05
  • Thanks for the clarification – Wu David Nov 16 '15 at 11:29

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