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Let $W$ be Brownian motion, let $\mathcal{F_t}$ be the generated filtration, and let $T > t$ be given.

I want $E [ W^4 |\mathcal{F_t}]$.

I'd love to give you my attempt but I don't have one. I have no clue where to start.

2 Answers 2

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First method :

$$E(W_T^4|\mathcal{F_t})=E((W_T-W_t+W_t)^4|\mathcal{F_t})=E((W_T-W_t)^4|\mathcal{F_t})+4E((W_T-W_t)^3W_t|\mathcal{F_t})+6E((W_T-W_t)^2W_t^2|\mathcal{F_t})E(W_t^4|\mathcal{F_t})+4E((W_T-W_t)W_t^3|\mathcal{F_t})+E(W_t^4|\mathcal{F_t})$$

Using the independence of $W_T-W_t$ with $\mathcal{F_t}$, we have

$$E(W_T^4|\mathcal{F_t})=E((W_T-W_t)^4)+4W_tE((W_T-W_t)^3)+6W_t^2E((W_T-W_t)^2)+4W_t^3E((W_T-W_t))+W_t^4$$

We know that for $n \in \mathbb{N}$, and $Y$ a standard normal variable $$E(Y^{2n+1})=0$$ and $$E(Y^{2n})=\frac{(2n)!}{n!2^n}$$

Thus, $$E(W_T^4|\mathcal{F_t})=3(T-t)^2+6W_t^2(T-t)+W_t^4$$

With Ito's lemma :

let $X_t=W_{t}^4$, by applying Ito's lemma on $X$, we have $$dX_t=4W_t^3dW_t+6W_t^2dt$$

We integrate the equation from $t$ to $T$, we have

$$X_T=X_t+4\int_{t}^{T}{W_u^3dW_u}+6\int_{t}^{T}{W_u^2du}$$

Taking the conditional expectation of $X_T$, we have $$E(X_T|\mathcal{F_t})=E(X_t|\mathcal{F_t})+4E(\int_{t}^{T}{W_u^3dW_u}|\mathcal{F_t})+6E(\int_{t}^{T}{W_u^2du}|\mathcal{F_t})$$

By Tonelli's theorem, $$E(\int_{t}^{T}{W_u^2du}|\mathcal{F_t})=\int_{t}^{T}{E(W_u^2|\mathcal{F_t})du}$$

and by independence between $W_u-W_t$ and $\mathcal{F_t}$ $$E(W_u^2|\mathcal{F_t})=E((W_u-W_t+W_t)^2|\mathcal{F_t})=u-t+W_t^2$$

Thus,

$$E(\int_{t}^{T}{W_u^2du}|\mathcal{F_t})=\frac{(T-t)^2}{2}+W_t^2(T-t)$$

Moreover , by construction of the Ito integral, $$E(\int_{t}^{T}{W_u^3dW_u}|\mathcal{F_t})=0$$

Finally, $$E(X_T|\mathcal{F_t})=W_t^4+6W_t^2(T-t)+3(T-t)^2$$

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Conditional on $\mathcal{F}_t,$ you know what $W_t$ is. This is Brownian motion, so given the value of $W_t,$ $W_T$ is conditionally distributed as a normal with mean $W_t$ and variance $T-t$ (i.e. standard deviation $\sqrt{T-t}$). So $E(W_T^4|\mathcal{F}_t)$ is the expected value of $X^4$ where $X$ has distribution $N(W_t,T-t)$ So $$ \int_{-\infty}^\infty x^4\frac{1}{\sqrt{2\pi(T-t)}}e^{-\frac{(x-W_t)^2}{2(T-t)}}dx.$$