- #1
lelouch_v1
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- Homework Statement
- Given ##x(t)=\exp{-b(W(t))^2}##, where ##W(t)## is a Wiener process, solve the below questions:
i)What values can x take?
ii)What is the probability density for x?
- Relevant Equations
- ##P(W)=\frac{\exp{-W^2/(2t)}}{\sqrt{2\pi t}}##
Suppose that 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 :)
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