This is my first question, so have mercy on me :) I want to calculate the variance of the Itô integral: $$\int_0^T (W_t+t)^2 \;\mathrm{dW_t}$$ I know that $Var[I] = E[I^2]$, and that we can use the Itô Isometry Therefore, it should be $Var[I] = $$\int_0^T E[(W_t+t)^4] \;\mathrm{dW_t}$. This is were I get in trouble. I want to expand the expression and separate the integrals, which give me the answer $ T^5/5 + T^4t + 3T^3t^2 + 2T^2t^3 $, but I know that the answer is $T^3 + 3/2T^4 + T^5/5$
Does anyone have any ideas on how I can solve this??
Let $$ I_T=\int_{0}^T (W_t+t)^2 \mathrm{d}W_t $$
We have, by applying Itô isometry, \begin{align} \mathbb{E}[I_t^2]& = \int_{0}^T \mathbb{E}(W_t+t)^4 \mathrm{d}t\\ &=\int_{0}^T (3t^2+6t^3+t^4)\mathrm{d}t\\ &=T^3+3/2T^4+T^5/5 \end{align}