# Finding as expressions for the $E[X^2_T]$ for a particular stochastic process.

Suppose that we have the following stochastic process and we want to find an expression for the expected value of this at time $$t$$ of the process at some future time $$T$$ of this process squared, $$E[X^2_T]$$.

$$dX_s=(\alpha X_s) ds + \sigma e^{-s} dW_s,\ X_t=x$$ In order to find an explicit solution for $$E_{t,x}[X_T^2]$$ I thought I should use Itô's lemma applied to the function $$z(t,x)=x^2$$. $$dz(t,X_s)=\left( (\alpha X_s)(2X_s) + \frac{1}{2}\sigma^2e^{-2s}(2)\right)ds+\sigma e^{-s}(2X_s)dW_s$$ Now we integrate and simplify a bit, plugging in the boundary condition $$X_t=x$$. $$z(T,X_T)= z(t,x) + \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds+ \int_t^T \sigma e^{-s}(2X_s)dW_s$$ We now take the expectation, which gives us the expression we are looking for, ie $$E_{t,x}[z(T,X_T)] = E_{t,x}[X_T^2]$$. $$E_{t,x}[z(T,X_T)]= z(t,x) + E_{t,x}\left[ \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds \right]+ E_{t,x}\left[ \int_t^T \sigma e^{-s}(2X_s)dW_s \right]$$ As usual we take the expectation of the stochastic term to be zero. $$E_{t,x}[z(T,X_T)]= z(t,x) + E_{t,x}\left[ \int_t^T \left( 2 \alpha X_s^2 + \sigma^2e^{-2s} \right)ds \right]$$

However, I am seeing that I have not really simplified anything since we sill have to take a expectation of $$X^2_s$$ in order to evaluate the integral. Am I missing something here or should I use a different technique besides Itô's lemma?

• First question: you are right, we should include $x$ but it doesn't really makes a difference. Second question: I have seen that you edited the SDE. Now the SDE is different to what you wrote before and clearly changes the problem. My answer will look wrong to anyone that check this question. I should have asked you before if there was a typo in your original SDE. I will delete my answer.
– UBM
Commented Dec 4, 2020 at 15:44
• Yes I was wondering if that is what the issue was, sorry about that, I forgot I have fixed that typo by time I left my comment. I see, after considering your technique, I was thinking maybe I should use Itô's lemma on $z(t,x)=x$, getting an expression for $z(T,X_T)=X_T$ in this way, however directly taking expectations you still have to deal with an $E[X_s]$ inside the integral. Commented Dec 4, 2020 at 15:54

Assume we have proven that $$\mathbb{E}_{t,x}[X_T^2] = x^2 + \mathbb{E}_{t,x}[\int_t^T 2\alpha X_s^2 + \sigma^2e^{-2s} ds]$$.
(Note: this can be easily done using the fact that both the coefficients of the original SDE are lipschitz in $$x$$ uniformly wrt $$t$$, and thus $$\mathbb{E}[\sup\limits_{0\leq s\leq T} X_s^2]$$ is bounded)
Now, our equation can be rewritten as $$\mathbb{E}_{t,x}[X_T^2] = x^2 + 2\alpha\mathbb{E}_{t,x}[\int_t^T X_s^2 ds] + \frac{1}{2}\sigma^2(e^{-2t}-e^{-2T})$$. This means, since everything is positive, that $$\mathbb{E}_{t,x}[X_T^2] = x^2 + 2\alpha\int_t^T\mathbb{E}_{t,x}[ X_s^2 ] ds + \frac{1}{2}\sigma^2(e^{-2t}-e^{-2T}).$$
We now pose $$f(T) = \mathbb{E}_{t,x}[X_T^2]$$ for every $$T\geq t$$. We have $$f(t)=x^2$$ and $$f'(T) = 2\alpha f(T) + \sigma^2e^{-2T}$$ for every $$T\geq t$$.
• I like the first step there, I had thought about doing that. In the end I first used the solution for $X_T$, which is a drift term plus stochastic integral, taking expectation for $E[X_T]$ you also lose the stochastic integral, then calculating $Var(X_T)=(X_T-E[X_T])^2$ seems to get to the same conclusion as you, although have not checked exactly. Commented Dec 7, 2020 at 11:20