I had some homework exercises from stochastic processes and have some difficult with expression.
So, I need to compute $\mathbb{E}\left[\left(X_1 + 2X_2 + 3X_3\right)^2\right]$, where $X_t$ is a Wiener process.
[Updated]
What I had do: 1. Set $D_2$ = $X_2$ - $X_1$, then $D_3$ = $X_3$ - $X_2$
- Calculate:
$$ \begin{split} \mathbb{E}\left[\left(X_1 + 2X_2 + 3X_3\right)^2\right] &= \mathbb{E}\left[\left(X_1 + 2(X_1+D_2) + 3(X_2 + D_3)\right)^2\right]\\ &= \mathbb{E}\left[\left(3X_1 + 2D_2 + 3X_2 + 3D_3\right)^2\right] \\ &= ... ? \end{split} $$
- Stop on this, should I expand binomial coefficient next?
Let's do a simpler example. Let $D_2 = X_2 - X_1$, then $D_2$ and $X_1$ are independent because of independent increments property of the Wiener process.
You have $$ \begin{split} \mathbb{E}\left[\left(X_1 + 2X_2\right)^2\right] &= \mathbb{E}\left[\left(X_1 + 2(X_1+D_2)\right)^2\right] \\ &= \mathbb{E}\left[\left(3X_1 + 2D_2\right)^2\right] \\ &= \mathbb{E}\left[9X_1^2 + 4D_2^2+12X_1D_2\right] \\ &= 9 \mathbb{E}\left[X_1^2\right] + 4 \mathbb{E}\left[D_2^2\right] + 12 \mathbb{E}[X_1 D_2] \\ &= 9 \mathbb{E}\left[X_1^2\right] + 4 \mathbb{E}\left[D_2^2\right] + 12 \mathbb{E}[X_1] \cdot \mathbb{E}[D_2] \end{split} $$ where the last step is due to independence of $X_1$ and $D_2$.
Can you finish the example and apply to your problem?
The alternative approach is to first expand $$ \begin{split} \mathbb{E}\left[\left(X_1 + 2X_2 + 3X_3\right)^2\right] &= \mathbb{E}\left[X_1^2\right] + 4\mathbb{E}\left[X_2^2\right] + 9\mathbb{E}\left[X_3^2\right]\\ &+ 2 \mathbb{E}[X_1 \cdot 2X_2] + 2 \mathbb{E}[X_1 \cdot 3X_3] + 2 \mathbb{E}[2X_2 \cdot 2X_3] \end{split} $$ and then make the substitutions $X_2 = X_1 + D_2$ and $X_3 = X_1 + D_2 + D_3$.