If$ X\sim N(\mu,\sigma^2),Y\sim N(X,\tau^2)$, how can I know the distrubution of $Y$?

Question

It seems to be a simple computation of condiontional distrubution function.But when I really work it out: \begin{align*} P(Y\leq y)&=\int^{\infty}_{-\infty} f_X(x)P(Y\leq y|X=x) \mathrm{d}x\\ &=\int^{\infty}_{-\infty}\int^y_{-\infty} \dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\dfrac{1}{\tau\sqrt{2\pi}}e^{-\frac{(t-x)^2}{2\tau^2}}\mathrm{d}t\mathrm{d}x, \end{align*} I find it difficult to summarize. What I'm concering is that if $Y\sim N(X,\tau^2)$,then$Y-X \sim N(0,\tau^2)$ ,and this implies $Y=(Y-X)+X\sim N(\mu,\sigma^2+\tau^2)$.So my question is:Is this conclusion correct?Whether right or not, how can I summarize it in the direct computaion of conditional probobality?

Answer 1

Actually we just need to combine the alike terms. Note that $ e^{-\frac{(x-\mu)^2}{2\sigma^2}}e^{-\frac{(x-t)^2}{2\tau^2}}=e^{-\frac{1}{2}((\frac{1}{\sigma^2}+\frac{1}{\tau^2})x^2-2(\frac{\mu}{\sigma^2}+\frac{t}{\mu^2})x+(\frac{\mu^2}{\sigma^2}+\frac{t^2}{\tau^2}))},$

and we assume all of the integral commutes.Let $z=\frac{\sigma^2+\tau^2}{\sigma\tau}x$, then $\mathrm{d}x=\frac{\sigma\tau}{\sigma^2+\tau^2}\mathrm{d}z$,and original integral becomes \begin{align}\label{eq} P(Y\leq y)&=\frac{1}{\sigma\tau 2\pi}\int_{-\infty}^ye^{-\frac{1}{2}(\frac{\mu^2}{\sigma^2}+\frac{t^2}{\tau^2})}\mathrm{d}t\int_{-\infty}^{+\infty}\frac{\sigma\tau}{(\sigma^2+\tau^2)}e^{\frac{1}{-2(\sigma^2+\tau^2)}\quad(z^2-2(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau})z)}\mathrm{d}z\\ &=\frac{1}{(\sigma^2+\tau^2) 2\pi}\int_{-\infty}^ye^{-\frac{1}{2}((\frac{\mu^2}{\sigma^2}+\frac{t^2}{\tau^2})-\frac{1}{\sigma^2+\tau^2}(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau})^2)}\mathrm{d}t\int_{-\infty}^{+\infty}e^{-\frac{1}{2(\sigma^2+\tau^2)}\quad(z-(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau}))^2}\mathrm{d}z. \end{align} Note that for the normal distribution $N(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau},\sigma^2+\tau^2)$,we have \begin{equation} \frac{1}{\sqrt{\sigma^2+\tau^2}\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-\frac{1}{2(\sigma^2+\tau^2)}\quad(z-(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau}))^2}\mathrm{d}z=1, \end{equation} thus the distribution function becomes \begin{align} P(Y\leq y)&=\frac{1}{\sqrt{\sigma^2+\tau^2} \sqrt{2\pi}}\int_{-\infty}^ye^{-\frac{1}{2}((\frac{\mu^2}{\sigma^2}+\frac{t^2}{\tau^2})-\frac{1}{\sigma^2+\tau^2}(\mu\frac{\tau}{\sigma}+t\frac{\sigma}{\tau})^2)}\mathrm{d}x\\ &=\frac{1}{\sqrt{\sigma^2+\tau^2} \sqrt{2\pi}}\int_{-\infty}^y e^{-\frac{1}{2}(\frac{t^2}{\tau^2}-\frac{1}{\sigma^2+\tau^2}\frac{\sigma^2}{\tau^2}t^2-\frac{2\mu t}{\sigma^2+\tau^2}-\frac{\tau^2}{\sigma^2+\tau^2}\frac{\mu^2}{\sigma^2}+\frac{\mu^2}{\sigma^2})}\mathrm{d}t\\ &=\frac{1}{\sqrt{\sigma^2+\tau^2} \sqrt{2\pi}}\int_{-\infty}^y e^{-\frac{1}{2}(\frac{t^2}{\sigma^2+\tau^2}-\frac{2\mu t}{\sigma^2+\tau^2}+\frac{\sigma^2}{\sigma^2+\tau^2}\frac{\mu^2}{\sigma^2})}\mathrm{d}t\\ &=\frac{1}{\sqrt{\sigma^2+\tau^2} \sqrt{2\pi}}\int_{-\infty}^y e^{-\frac{1}{2(\sigma^2+\tau^2)}(t-\mu)^2}\mathrm{d}t, \end{align} which is exactly the distribution function of the nomral distribution $N(\mu,\sigma^2+\tau^2)$.