Let $X \sim N(\mu, 1)$. $Y=X^2$. What is the pdf of $Y$?
I use the usual transformation to get my answer:
$f_Y (y) = \frac{1}{2 \sqrt{2 \pi y}} (e^{-\frac 12 (-\sqrt{y}- \mu)^2} + e^{-\frac 12 (\sqrt{y}- \mu)^2})$
I hope my above attempt is correct.
Now the question is: the pdf of $Y=X^2$ can be expressed as the marginal pdf of $Y$ obtained from the joint distribution of the pair $(Y, J)$, with $J$ being discrete, as stated below:
$Y|J = j \sim \chi_{1+2j}^2$
$J \sim Poisson (\lambda /2)$, where $\lambda = \mu^2$
I want to get the density of $Y$ by using the above mixture. How to do this?
The formula $\frac{(2j)!}{j!} = \frac{2^{2j} \Gamma (j+\frac 12) }{\sqrt{\pi}}$ is given
You just use Bayes rule to find the joint distribution and then marginalize over $j$: for any $y > 0$: \begin{align*} f_Y(y) = \sum_{j = 0}^\infty f_{Y, J}(y, j) = \sum_{j = 0}^\infty f_Y(y \ | \ j)f_J(j) &= \sum_{j = 0}^\infty \frac{y^{j - 1/2}e^{-y/2}}{2^{j + 1/2}\Gamma(j + 1/2)} \frac{(\mu^2/2)^je^{-\mu^2/2}}{j!} \\ &= \frac{e^{-(y + \mu^2)/2}}{\sqrt{2y}} \sum_{j = 0}^\infty \frac{(\sqrt{y}\mu)^{2j}}{ \underbrace{ 2^{2j}\Gamma(j + 1/2)j!}_{\text{use given identity here}} } \\ &= \frac{e^{-(y + \mu^2)/2}}{\sqrt{2\pi y}} \underbrace{ \sum_{j = 0}^\infty \frac{(\sqrt{y}\mu)^{2j}}{(2j)!} }_{\text{series of }\cosh(\sqrt{y}\mu)} \\ &= \frac{e^{-(y + \mu^2)/2}}{\sqrt{2\pi y}} \Big(\frac{e^{\sqrt{y}\mu} + e^{-\sqrt{y}\mu}}{2}\Big) \\ &= \frac{e^{-(\sqrt{y} + \mu)^2/2} + e^{-(\sqrt{y} - \mu)^2/2}}{2\sqrt{2\pi y}} \end{align*} and of course $f_Y(y) = 0, y \leq 0$. This is exactly what you found earlier through the other method.