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Suppose we have a filtration $(\mathcal{F}_t)$ satisfying the usual conditions. Let $W$ be a Brownian Motion with respect to that filtration. We define the two processes

$X_t:=W^2_t$ and $Y_t:=\int_0^tX_sdX_s$

Then I want to prove:

  1. $Y_t$ is a submartingale
  2. $E[Y_t]=\frac{t^2}{2}$

This is exercise from an old exam. There are two hints: Using Itô to write $X$ in differential form and rewrite $Y$ accordingly. And if $U$ is standard normal distributed, then $E[U^6]=15$.

Using Itô I can write: $X_t=f(W_t)$ with $f(x)=x^2$. Hence Itô implies

$$dX_t=2W_tdW_t + dt $$

Therefore $Y_t=2\int_0^tW_s^3 dW_s + \int_0^tW_s^2ds$. Unfortunately I do not see how to proceed for both question. Some Help would be appreciated.

cheers

math

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2  
The hint about $\mathsf E[U^6]$ is scary –  Ilya Jan 21 '13 at 13:10
    
@Ilya me too! But it is really the hint, no typo :) –  math Jan 21 '13 at 13:33

1 Answer 1

up vote 4 down vote accepted

Well, you know that the Ito integral is a martingale, so that $Y_t = M_t + \int_0^t W^2_s\mathrm ds$ and we only have to show that the latter integral is a submartingale. By definition we have $$ \mathsf E\left[\left.\int_0^t W^2_s\mathrm ds\right|\mathscr F_u\right] = \mathsf E\left[\left.\int_0^t W^2_s\mathrm ds\right|\mathscr F_u\right] = \int_0^u W^2_s\mathrm ds + \mathsf E\left[\left.\int_u^t W^2_s\mathrm ds\right|\mathscr F_u\right] \geq \int_0^u W^2_s\mathrm ds $$ since $\mathsf E\left[\left.\int_u^t W^2_s\mathrm ds\right|\mathscr F_u\right]\geq 0$ as an integral of a non-negative function. Furthermore you have $$ \mathsf E[Y_t] = \mathsf E[M_t]+\mathsf E\left[\int_0^t W^2_s\mathrm ds\right] = 0+\int_0^t\mathsf E[W^2_s]\mathrm ds = \dots $$

In order to address your point about the integrability, consider $$ \int_{[0,t]\times\Omega } |W^n|\mathrm d(\lambda\otimes \mathsf P). $$ We know that for even $n$ it is finite, so for odd $n$ we have $$ \int_{[0,t]\times\Omega } |W^n|\mathrm d(\lambda\otimes \mathsf P) = $$ $$ \int_{[0,t]\times\Omega } |W^n|\cdot1(|W^n|\leq 1)\mathrm d(\lambda\otimes \mathsf P)+\int_{[0,t]\times\Omega } |W^n|\cdot1(|W^n|>1)\mathrm d(\lambda\otimes \mathsf P) $$ $$ \leq t+\int_{[0,t]\times\Omega } |W^{n+1}|\cdot1(|W^n|>1)\mathrm d(\lambda\otimes \mathsf P) $$ $$ \leq t+\int_{[0,t]\times\Omega } |W^{n+1}|\mathrm d(\lambda\otimes \mathsf P)<\infty. $$

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Is it in general true that $E[\int_0^t W^p ds]=\int_0^TE[W^p]ds$ for $p\in\mathbb{N}$. Here it is obvious, since for $p=2$ we have non negativity and can use Tonelli to interchange the two integrals. What if $p$ is odd? Can something be said for $E[\int_0^t f(W_s) ds]$ for a deterministic and say continuous function $f$? Which means can we then also always interchang the integrals: $E[\int_0^t f(W_s) ds]=\int_0^t E[f(W_s) ]ds$? –  math Jan 21 '13 at 13:32
    
@math: I think you can just do Fubini which requires only integrability –  Ilya Jan 21 '13 at 13:34
1  
For that we need product integrability. Why is this here that obvious? –  math Jan 21 '13 at 13:38
3  
@math: added. You judge, how obvious it is :) –  Ilya Jan 21 '13 at 15:17
1  
@Did: please, don't hesitate to use "@" stuff - otherwise me or math may miss the comment –  Ilya Jan 21 '13 at 15:34

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