2
$\begingroup$

Let $Z \sim \mathcal{N}(0,1)$ and $Y|Z \sim \mathcal{N}(Z, 1)$.

Show that $f_{Z|Y}(z|y)$ is a normal density, and find the parameters of this density.

What I have so far: \begin{align*} f_Z(z) &= \phi(z)\\ f_{Y|Z}(y|z) &= \frac{1}{\sigma} \phi\left( \frac{y - \mu}{\sigma} \right) = \phi\left( y - z \right)\\ f(y,z) &= f_Z(z) \cdot f_{Y|Z}(y|z) = \phi(z) \cdot \phi\left( y - z \right) \end{align*} From the above, we can use Bayes' theorem express $f_{Z|Y}(z|y)$: \begin{align*} f_{Z|Y}(z|y) &= \frac{ f_Z(z) \cdot f_{Y|Z}(y|z)}{ f_Y(y) }\\ &= \frac{ \phi(x) \cdot \phi(y-z) }{ \int_{-\infty}^\infty f(y,z)\,\mathrm{d}z }\\ &= \frac{ \phi(x) \cdot \phi(y-z) }{ \int_{-\infty}^\infty\phi(z) \cdot \phi\left( y - z \right)\,\mathrm{d}z }\\ \end{align*} However, I am stuck at this point. I'm not exactly sure how to compute the integral in the denominator, and then express $f_{Z|Y}(z|y)$ as a normal distribution function. Is there a better way to approach this? Thanks!

$\endgroup$
2
$\begingroup$

$$ f_{Z|Y}(z|y) = \frac{f_Z(z)f_{Y|Z}(y|z)}{\int_{-\infty}^{+\infty}{f_Z(a)f_{Y|Z}(y|a)da}} = \frac{\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-z)^2}{2}}} {\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-a)^2}{2}}da}} = \frac{e^{-z^2+yz-\frac{y^2}{2}}}{\int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da}} $$

Now we'll calculate the integral in the denominator: $$ \int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da} = \int_{-\infty}^{+\infty}{e^{-\frac{(\sqrt{2}a - \frac{\sqrt{2}}{2}y)^2 + \frac{y^2}{2}}{2}}da} = e^{-\frac{y^2}{4}} \int_{-\infty}^{+\infty}{e^{-\frac{(\sqrt{2}a - \frac{\sqrt{2}}{2}y)^2}{2}}da} $$ Defining $b=\sqrt{2}a$ and multiplying and dividing by $\sqrt{\pi}$ we get: $$ e^{-\frac{y^2}{4}}\sqrt{\pi} \int_{-\infty}^{+\infty} {\frac{e^{-\frac{(b - \frac{\sqrt{2}}{2}y)^2}{2}}}{\sqrt{2\pi}}db} $$

Notice that we have the pdf of the normal distribution $\mathcal{N}\left(\frac{\sqrt{2}}{2}y, 1\right)$ inside the new integral. Therefore it equals 1.

Substituting $ \int_{-\infty}^{+\infty}{e^{-\frac{2a^2-2ya+y^2}{2}}da} = e^{-\frac{y^2}{4}}\sqrt{\pi} $ into the initial formula we have:

$$ f_{Z|Y}(z|y) = \frac{1}{\sqrt{\pi}}e^{-z^2+yz-\frac{y^2}{2}+\frac{y^2}{4}} = \frac{1}{\sqrt{\pi}}e^{-(z-\frac{y}{2})^2} = \frac{1}{\sqrt{2\pi} \frac{1}{\sqrt{2}}}e^{-\frac{(z-\frac{y}{2})^2}{2(\frac{1}{\sqrt{2}})^2}} $$

That is, $$ Z|Y \sim \mathcal{N}\left(\frac{Y}{2}, \frac{1}{\sqrt{2}}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.