Fourth moment of a itos integral

$$I(t)=\int_0^t \sqrt sdW_s$$

What is $$E(I(t)^4)$$

2 Answers

$$I(t)=\int_0^t \sqrt tdW_s=\sqrt t \int_0^t dW_s =\sqrt t W_t$$ and then $$E(I(t)^4)=E(t^2 W_t^4)=t^2 \cdot 3t^2=3t^4$$ using the 4th moment of the $$N(0,\sigma^2=t)$$ distribution.

• But it is "s" in the integral and not t. You can not take it outside ...maybe it was edited after your answer – Ric Oct 9 '18 at 9:23
• Yes the question was changed – Bjørn Kjos-Hanssen Oct 9 '18 at 13:50

Note that because the integrand is deterministic this Itô integral is normally distributed with parameters (cf. Itô isometry) $$I_t := \int_0^t \sqrt{s} dW_s \sim N(0, t^2/2)$$ Now you can just use the results that apply for the moments of a Gaussian variable.