Probability Density of x

Question

Suppose that I have the function $$x(t)=e^{-b(W(t))^2} \ \ \ \ \ \ \ \ (1)$$ where W(t) is just a Wiener process (i.e. a Gaussian in general). I want to know what the probability density for $x$, $P(x)$, is. I started off by just assuming that I want to measure the expectation value of an observable $f(x)$, so $$<f(x)>=\int_{W=0}^{W=t}{P(W)f(g(W))dW} \ \ ,\ \ x=g(W) $$ Then I transformed variables from $W$ to $t$ and I got $$<f(x)>=\int_{x=g(0)}^{x=g(t)}P(g^{-1}(x))f(x)(\frac{dW}{dx})dx=\int_{x=g(0)}^{x=g(t)}\frac{P(g^{-1}(x))}{g'(g^{-1}(x))}f(x)dx$$ so I just assume that the probability for x is $$P(x)=\frac{P(g^{-1}(x))}{g'(g^{-1}(x))}$$ Since $x=g(W)$, then $W=g^{-1}(x)$, but from (1) I get $$\frac{dW}{dx}=\frac{1}{2x\sqrt{blnx}}$$ and assuming that $\sigma^2=V=t$, I get $$P(x)=\frac{e^{lnx/{2bt}}2\sqrt{blnx}}{x\sqrt{2{\pi}t}}$$ Is this right? Shouldn't I get a Gaussian? Also, was I right to take the values of $x$ to be $[0,t]$ and not $(-\infty,+\infty)$ ? Thank you in advance for reading my question :)

Answer 1

Hint:

The distribution of a (monotone) transformation $Y=g(X)$ of a random variable $X$ is given by

$$F_Y(y)=F_X(g^{-1}(y))$$ from where you can deduce that the density is given by

$$f_Y(y)=\frac{d}{dy}F_X(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)|=f_X(g^{-1}(y))\cdot |\frac{d}{dy}g^{-1}(y)|.$$

In your case the random variable $X=W(t)$ is Gaussian with variance $t$, hence you know what $f_X$ is. Your transformation is not monotone, hence you have to divide the domain of $g$ into two parts ($(-\infty,0)$ and $[0,\infty)$) and then add up the two parts.