The objective is an explicit calculation of $\mathbb{E}\left[ {\left( {\int\limits_0^T {{e^{at}}dW(t)} } \right)\left( {\int\limits_0^T {{e^{bt}}dW(t)} } \right)} \right]$. I'm trying to compare two approaches to this: via Ito's isometry and a 'statistical' approach.
Ito's isometry quite clearly says that
$\mathbb{E}\left[ {\left( {\int\limits_0^T {{e^{at}}dW(t)} } \right)\left( {\int\limits_0^T {{e^{bt}}dW(t)} } \right)} \right] = \mathbb{E}\left[ {\int\limits_0^T {{e^{(a + b)t}}dt} } \right] = \frac{{{e^{(a + b)T}} - 1}}{{a + b}}$. Let's call this result 1.
However, because $\int\limits_0^T {f(t)dW(t)} \sim N\left( {0,\int\limits_0^T {{f^2}(t)dt} } \right)$, for a deterministic function $f$, I can also write the initial problem as $\mathbb{E}\left[ {\left( {\int\limits_0^T {{e^{at}}dW(t)} } \right)\left( {\int\limits_0^T {{e^{bt}}dW(t)} } \right)} \right] = \mathbb{E}\left[ {\sqrt {\int\limits_0^T {{e^{2at}}} } \varepsilon \sqrt {\int\limits_0^T {{e^{2bt}}} } \varepsilon } \right] = \sqrt {\int\limits_0^T {{e^{2at}}} } \sqrt {\int\limits_0^T {{e^{2bt}}} } \mathbb{E}\left[ {{\varepsilon ^2}} \right]$, where $\varepsilon \sim N(0,1)$. This implies $\mathbb{E}\left[ {{\varepsilon ^2}} \right] = 1$ (as ${\varepsilon ^2}$ is chi-squared with one d.f.). After explicit integration this gives $\mathbb{E}\left[ {\left( {\int\limits_0^T {{e^{at}}dW(t)} } \right)\left( {\int\limits_0^T {{e^{bt}}dW(t)} } \right)} \right] = \frac{1}{2}\sqrt {\frac{{({e^{2aT}} - 1)({e^{2bT}} - 1)}}{{ab}}}$. This is result 2.
Comparing result 1 and result 2 I get $\frac{{{e^{(a + b)T}} - 1}}{{a + b}} \ne \frac{1}{2}\sqrt {\frac{{({e^{2aT}} - 1)({e^{2bT}} - 1)}}{{ab}}} $, unless $a=b$. When substituting some reasonably small $a \ne b$ and some $T$, then both results are numerically 'close' yet still not similar.
My questions are:
- Which of the results is exact?
- Why each result is different, although hypothetically correct in its own world?
My intuition about the difference points towards the nature of the exponential integrand. Maybe it does not satisfy some technical condition needed by Ito's isometry or by the other approach?
The first calculation (via Itô's isometry) is correct and the "statistical" approach is not.
If $X \sim N(0,\sigma_1^2)$, $Y \sim N(0,\sigma_2^2)$ are two normal random variables, then the covariance
$$\mathbb{E}(X Y)$$
depends on the joint distribution of $X$ and $Y$. For instance if $X$ and $Y$ are independent, then
$$\mathbb{E}(XY) = \mathbb{E}(X) \mathbb{E}(Y)=0.$$
In contrast, if we choose $Y=X$, then
$$\mathbb{E}(X Y) = \mathbb{E}(X^2) = \sigma_1^2.$$
In particular, it does (in general) not hold true that
$$\mathbb{E}(XY) = \mathbb{E}\big((\sigma_1 \epsilon) \cdot (\sigma_2 \epsilon) \big)$$
for $\epsilon \sim N(0,1)$ (... and that's where your statistical approach fails).