I am currently learning Random Processes and have encountered one such exercise about Wiener process that I can't figure out why the answer is like this. The question is as follows:
Suppose W(t) is a Wiener Process such that $R_{w}(t_1,t_2) = \alpha\, min(t_1, t_2)$. $R_w(t_1,t_2)$ should stand for auto-correlation function of W(t).
Let the event B be $W(4) = 1$. Give the conditional PDF $f_{W(t)}(w|B)$ for $t > 4$.
And the answer for it is $f_{W(t)}(w|B) = \frac{1}{\sqrt{2\pi\alpha(t-4)}}e^{\frac{-(w-1)^2}{2\alpha(t-4)}}$
Can someone tell me why it is $N(1, \alpha(t-4))$? Thanks a lot.
Best Answer
I have asked my professor about this and got the answer. Basically the method is to obtain the derivative of cdf for W(t)|B to get the pdf. Then one can easily obtain that the cdf is of the form $P(Z \leq w-1)$, where Z ~ $N(0,\alpha(t-4))$.