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

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

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2 Answers 2

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$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.

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    $\begingroup$ But it is "s" in the integral and not t. You can not take it outside ...maybe it was edited after your answer $\endgroup$
    – Richi Wa
    Oct 9, 2018 at 9:23
  • $\begingroup$ Yes the question was changed $\endgroup$ Oct 9, 2018 at 13:50
  • $\begingroup$ Why can't you take t outside? I'm looking over this problem for review and this part didn't make sense to me. $\endgroup$
    – Jessie
    May 31, 2023 at 16:39
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    $\begingroup$ @Jessie you can, but I'm answering the original question, not the edited question. $\endgroup$ May 31, 2023 at 21:18
  • $\begingroup$ Many thanks! Would you be able to provide a hint on how t^2 and 3t^2 were derived? I believe Itô's isometry needs to be applied here but unsure how. $\endgroup$
    – Jessie
    Jun 2, 2023 at 17:23
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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.

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