Hi: I am learning stochastic calculus and encounter a question.
Consider the 1-d Ornstein Uhlenbeck process $dX_{t}=-X_{t}dt+dW_{t}$ with $X_{0}=0$.
Find the value function $f(x,t)=\mathbb{E}[X_{T}^4|X_t=x]$ for $t<T$ and therefore $\mathbb{E}[f(X_{t},t)],$which should be independent of t.
(Hint: for the value function, seek for a polynomial solution to the backward equation).
So far, this is what I got:
First of all, the solution to this O-U process is $X_{t}=\int_{0}^{t}e^{-(t-s)}dW_{s}$ with $\mathbb{E}[dX_{t}]=-X_{t}dt$ and $\mathbb{E}[dX_{t}^2]=1$
Next, I follow the hint to look for backward equation: $\partial tf+Lf=0$ where L is the generator. Plug in the $\mathbb{E}[dX_{t}]$ and $\mathbb{E}[dX_{t}^2]$, I got the equation: $\partial t f-x\partial x f +\frac{1}{2}\partial x^2 f=0$ And it should be equipped with the boundary condition: $f(x,T)=x^4$.
And I don't quite know how to solve such equation. I tried using separation of variable, writing $f=X(x)T(t)$ and end up with the equation $X(x)T'(t)-xX'(x)T(t)+\frac{1}{2}X''(x)T(t)=0$ Then I am stuck here because no trick I learned so far can help solving this...
I am still working on this, and looking for some help.
Thanks in advance for your help.
Not a complete answer but too long for a comment. Instead of looking for $g(x) h(t)$ (I don't like $X$ and $T$ since they already denote the solution and the time horizon), look for $$ f(x,t) = \sum_{k=0}^4 x^k h_k(t). $$ Then, your equation reads, $$ \sum_{k=0}^4 x^k \Bigl(h_k'(t) - k\, h_{k}(t) + \frac{(k+2)(k+1)}2 h_{k+2}(t) \Bigr) = 0, $$ where $h_5(t) = h_6(t) = 0$.
Now equating all expressions near the powers of $x$ to zero, you will get a system of equations, which is not too hard to solve. Start with $k=4$ and $k=3$ and use the conditions that $h_4(T) = 1$, $h_k(T) = 0$, $k=0,1,2,3$.