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Given a stochastic process $M=(M_{t})_{t\geq0} ,M_{t}=W_{t}^{2}-t,(W_{t}$ is Βrownian Motion).Find the Doob-Meyer decomposition of the $M^{2}$.

Attemp:

I firstly proved that $M^{2}$ is a submartingale

for $s\leq t$

$\mathbb{E}[M_{t}^{2}|\mathcal{F}_{s}]\overset{Jensen's\,inequality}{\geq}\left(\mathbb{E}[M_{t}|\mathcal{F}_{s}]\right)^{2}=M_{s}^{2}$ because $f(X)=X^{2}$ is a curve function and $W_{t}^{2}-t$ is a Martingale

Since $M^{2}$ is a submartingale there is a unique Doob-Meyer decomposition

$M_{t}^{2}=X_{t}+A_{t}$

$X_{t}$ must be a Martingale and $A_{t}$ an increasing predictable process

We already Know that $W_{t}^{2}-t$ is a Martingale and I tried to use this for my proof but I didn't find the solution.

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  • $\begingroup$ I edited this.Thank you. $\endgroup$
    – kawrik
    May 8, 2020 at 18:15

1 Answer 1

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Using the It$\hat{o}$ formula for the function $f(X_{t},t)=(W_{t}^{2}-t)^{2}$ we have:

$f(x,t)=(x^{2}-t)^{2}$

$f_{x}(x,t)=4x(x^{2}-t)$

$f_{xx}(x,t)=4(3x^{2}-t)$

$f_{t}(x,t)=-2(x^{2}-t)$

$f(t,W_{t})=f(0,W_{0})+\int_{0}^{t}f_{t}(u,W_{u})du+\int_{0}^{t}f_{x}(u,W_{u})dW_{u}+\frac{1}{2}\int_{0}^{t}f_{xx}(u,W_{u})d[W]_{u}$ $f(t,W_{t})=0+\int_{0}^{t}-2(W_{u}^{2}-u)du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}+\frac{1}{2}\int_{0}^{t}4W_{u}(3W_{u}^{2}-u)d[W]_{u}$

but $[W]_{u}=u$ so

$f(t,W_{t})=\int_{0}^{t}-2(W_{u}^{2}-u)du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}+\frac{1}{2}\int_{0}^{t}4(3W_{u}^{2}-u)du$

$f(t,W_{t})=\int_{0}^{t}\left(-2(W_{u}^{2}-u)+2W_{u}(3W_{u}^{2}-u)\right)du+\int_{0}^{t}2(W_{u}^{2}-u)dW_{u}$

$f(t,W_{t})=M_{t}^{2}=\int_{0}^{t}4W_{u}^{2}du+\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}$

But the process $\int_{0}^{t}4W_{u}^{2}du$ is increasing.In addition is predictable.

$\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}$ is a Martingale because for $s\leq t$

$\mathbb{E}\left[\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\mathbb{E}\left[\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]+\mathbb{E}\left[\int_{s}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]$

$\mathbb{E}\left[\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$because $\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$ $\mathcal{F}_{s}$-measurable

$ dW_{u}=W_{u+du}-W_{u} , W_{u+du}-W_{u}\sim N\left(0,du\right)$ and

$4W_{u}(W_{u}^{2}-u)$ independent of $W_{u+du}-W_{u} $

$\mathbb{E}\left[\int_{s}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{s}^{t}\left(\mathbb{E}\left[4W_{u}(W_{u}^{2}-u)\right]\mathbb{E}\left[dW_{u}\right]\right)|\mathcal{F}_{s}=0$

so $\mathbb{E}\left[\int_{0}^{t}4W_{u}(W_{u}^{2}-u)dW_{u}|\mathcal{F}_{s}\right]=\int_{0}^{s}4W_{u}(W_{u}^{2}-u)dW_{u}$ is Martingale

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