Expectation of Brownian function.

Question

$(W_t)t≥0$ be a standard one-dimensional Brownian motion

Let $f : R → R$ be a given function and $0 ≤ s ≤ t$. Write down an expression for $E(f(W_t)|W_s = x)$ in terms of $ϕ(x) = Φ′(x)$

Where $Φ(x)$ denotes the cumulative distribution function of a standard normal random variable

So what I did but I think it is wrong:

$E(f(W_t)|W_s = x) = E(f(W_t - W_s + W_s)|W_s = x)$

$E(f(W_t - W_s + x))$

$E(f(W_t - W_s + x))$

$E(f(\frac{(W_t - W_s)(t-s)^{0.5}}{(t-s)^{0.5}} + x))$

I'm really not sure from here I tried normalizing it but I'm not sure

Answer 1

Since $\operatorname{Corr}(W_t,W_s)=\sqrt{s/t}$, $$ W_t\mid W_s=x\sim N(x,t-s). $$ Thus, $$ \mathsf{E}[f(W_t)\mid W_s=x]=\int_{-\infty}^{\infty} \frac{f(w)}{\sqrt{t-s}}\phi\!\left(\frac{w-x}{\sqrt{t-s}}\right)\, dw. $$