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.