Find the solution $f(x, t)$ to the partial differential equation for $0 \leq t \leq 1$ $$ \frac{x^{2}}{2} \frac{\partial^{2}}{\partial x^{2}} f(x, t)+\frac{\partial}{\partial t} f(x, t)=f(x, t), \quad f(x, 1)=x^{2} $$ by using the Probability method.
My attempt
I am not sure how to go about this question but I gave it a shot as best as I could:
Recall Ito's formula: $d f\left(x_{t}, t\right)=\frac{\partial f\left(x_{t,} t\right)}{\partial t} d t+f^{\prime}\left(x_{t}, t\right) d X_{t}+\frac{1}{2} f^{\prime \prime}\left(x_{t},t\right)\left(d x_{t}\right)^{2}$
We introduce the constants $\mu$ and $\sigma$
Now to make $\frac{x^{2}}{2} \frac{\partial^{2}}{\partial x^{2}} f(x, t)+\frac{\partial}{\partial t} f(x, t)$ = $ \sigma \frac{1}{2} \frac{\partial^{2}}{\partial x^{2}} f(x, t)+ \mu \frac{\partial}{\partial t} f(x, t)$
we make the $\mu = 1$ and $\sigma = x^2$
Thus, our stochastic equation $d X_{t}=\mu\left(x_{t}\right) d t+\sigma\left(x_{t}\right) d B_{t}$ will now be: $d X_{t}=X_{t}^{2} d B_{t}$.
I am assuming I need to solve $X_t$ but I am bit lost on how to go about solving it. Please help. Thank you!
The Black-Scholes equation in $t \in [0,T]$ $$\frac{\partial f}{\partial t}+\mu x \frac{\partial f}{\partial x}+\frac{1}{2}\sigma^2x^2\frac{\partial^2 f}{\partial x^2}=rf$$ with terminal condition $f(x,T)=h(x)$ has solution given by the Feynman-Kac formula $$f(x,t)=e^{-r(T-t)}\mathbb{E}[h(X_T)|X_t=x]$$ where the expectation is with respect to the lognormal distribution with Normal location $\ln(x)+(\mu-0.5\sigma^2)(T-t)$ and Normal variance $\sigma^2(T-t)$. In your case: $\mu=0$, $r=1$, $\sigma^2=1$, $T=1$, $h(x)=x^2$. The solution is $$f(x,t)=e^{-(1-t)}\mathbb{E}[X_T^2|X_t=x]=e^{-(1-t)}(\mathbb{V}[X_T|X_t=x]+\mathbb{E}[X_T|X_t=x]^2)$$ where $$\mathbb{V}[X_T|X_t=x]=(e^{\sigma^2(T-t)}-1)e^{2\ln(x)+2\mu(T-t)}=(e^{\sigma^2(T-t)}-1)x^2e^{2\mu(T-t)}$$ and $$\mathbb{E}[X_T|X_t=x]^2=x^2e^{2\mu(T-t)}$$ So ultimately we get $$f(x,t)=e^{-(1-t)}x^2e^{2\mu(1-t)+\sigma^2(1-t)}$$