Suppose that $W$ has an exponential distribution with mean $1$, and suppose that $X|W ∼N(0,W^{-1})$.

Write down the probability density of $W$ conditional on $X$.

My workings so far: $$f_W(w)=e^{-w}, \ w\geq 0$$ $$f_{X\mid W=w}(x,w)=\frac{\sqrt{w}}{\sqrt{2\pi}}e^{-\frac{wx^2}2}$$ $$f_{X,W}(w,x)=\frac{\sqrt{w}}{\sqrt{2\pi}}e^{-w-\frac{wx^2}2}$$

Now in order to get $f_{W\mid X=x}(w,x)$ I need to divide $f_{X,W}(w,x)$ by $f_X(x)$. To get $f_X(x)$ I took the limit of $f_{X,W}(w,x)$ as $w$ tends to $\infty$, which gives me $0$, so I'm dividing by $0$. Can somebody show me where I've made my mistake.


1 Answer 1


It is not the limit that you have to take when you want to find a marginal. It is the integral of the joint distribution. In this case you have to evaluate the integral below $$f_X(x)=\int_0^{\infty}f_{X,W}(w,x)\ dw=\int_0^{\infty}\frac{\sqrt{w}}{\sqrt{2\pi}}e^{-w-\frac{wx^2}2}\ dw= $$$$=\frac1{(x^2+2)^{\frac32}}.$$

Then you have to divide $f_{X,W}(x,w)$ by $f_X(x)$.


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