# Doob-Meyer Decomposition for $(W_{t}^{2}-t)^{2}$

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.

• I edited this.Thank you. May 8, 2020 at 18:15

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