I am a beginner in stochastics, studying it by myself using the book by Kurt Jacobs (Introduction to stochastic processes). In there is the following exercise:
$x(t)= \mathrm{e}^{-b (W(t))^2}$ where $W(t)=\int^t_0 dW$ is a Wiener process.
a) Calculate the probability density for x, P(x).
b) Determine $\langle x^2 \rangle$ using P(W)
c) Determine $\langle x^2 \rangle$ using P(x)$\\$
I also found related threads on here, e.g., but none of them from which i could extract an answer.
My answer for b) is: I use that $W_t$ is normally distributed as $N_W(0,t)$ and get
$$\langle x^2 \rangle = \int_{-\infty}^{\infty} x(W)^2 P(W) \mathrm{d}W = \int\mathrm{d}w \ \mathrm{e}^{-2b W^2} \frac{1}{\sqrt{2 \pi t^2}}\mathrm{e}^{-\frac{-W^2}{2t^2}} = \int\mathrm{d}W \ \frac{1} {\sqrt{2 \pi t^2}}\mathrm{exp}\left[-W^2 (2b+\frac{1}{2t^2})\right]=\frac{1}{\sqrt{4bt^2+1}}$$
I would be kind of okay with that result, but from now on it gets a bit hairy. For a) i tried to apply a coordinate transformation with the idea of holding a expection value fixed:
$x=g(w)=\mathrm{e}^{-bW^2}$ and $w=g^{-1}(x)$ then
$$\langle f(x) \rangle = \int_{-\infty}^{\infty} P_w(W) \ f(x) \ \mathrm{d}W = \int P_w(W) \ f(g(w)) \ \mathrm{d}W = \int P_w(g^{-1}(x)) \ f(x) \ \frac{\mathrm{d}W}{\mathrm{d}x}\mathrm{d}x
$$
which has to equal
$$\langle f(x) \rangle = \int_0^{\infty} P_x(x) \ f(x) \ \mathrm{d}x
$$
Also,
$$
g^{-1}(x) = \sqrt{-ln(x) \ b^{-1}}
$$
and
$$
\frac{\mathrm{d}W}{\mathrm{d}x} = \frac{\mathrm{d}g^{-1}(x)}{\mathrm{d}x} = \frac{1}{2x \sqrt{-b \ ln(x)}}
$$
So we get $$ P_x(x) = P_w(g^{-1}(x)) \ \frac{\mathrm{d}W}{\mathrm{d}x} = P_w \left(\sqrt{-ln(x) b^{-1}} \right) \frac{1}{2x \sqrt{-ln(x)b}} $$ which just does not look good to me. When I then try to compute $\langle x^2 \rangle$ with this $P(x)$ (exercise c), my integral diverges.
Could you tell me where I did something wrong?
Thanks in advance!
