# Struggling to evaluate integral to find joint PDF of normal distribution.

Let X, Y, Z be r.v.s such that X ⇠ N (0, 1) and conditional on X = x, Y and Z are i.i.d. N (x, 1).

(b) Find the joint PDF of Y and Z. You can leave your answer as an integral, though the integral can be done with some algebra (such as completing the square) and facts about the Normal distribution.

I have found the joint PDF of X, Y, Z. To find the joint PDF of Y, Z I arrived at the following integral but I'm really struggling to evaluate it. $$f_{YZ}(y,z)=\int_{-\infty}^{\infty}\left( \frac{1}{2\sqrt \pi} e^{-x^2/2} \, \frac{1}{2\sqrt \pi} e^{-(y-x)^2/2} \, \frac{1}{2\sqrt \pi} e^{-(z-x)^2/2} \, dx\right)$$

Note that $$x^2+(y-x)^2 + (z-x)^2 = 3x^2 - 2(y+z)x + y^2 + z^2 = 3 \left(x - \frac{y+z}{3}\right)^2 - \frac{(y+z)^2}{3} + y^2 + z^2.$$

Use this to rewrite the integrand as $$c \cdot \exp\left(-\frac{(x-\frac{y+z}{3})^2}{2/3}\right)$$ where $$c$$ depends on $$y$$ and $$z$$. Doe sthis look like the PDF of a certain normal distribution?

• Thanks for your help. So I think the Integral evaluates to c*sqrt(2pi/3). But I'm not sure how to manipulate c, so it I get the PDF of a normal distribution? Oct 13, 2021 at 4:34
• Thanks for your help. So I think the Integral evaluates to c*sqrt(2pi/3). But I'm not sure how to manipulate c, so it I get the PDF of a normal distribution? Oct 13, 2021 at 4:59

there is a minor error in the first integral too...

$$f_{YZ}(y,z)=\int_{-\infty}^{\infty}\left( \frac{1}{\sqrt{2\pi}} \right)^3e^{-[x^2+(y-x)^2+(z-x)^2]/2}dx$$

that is

$$f_{YZ}(y,z)=\frac{1}{2\pi}e^{-(y^2+z^2)/2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} e^{-[3x^2-2x\cdot(z+y)]/2}dx$$

$$f_{YZ}(y,z)=\frac{1}{2\pi}e^{-(y^2+z^2)/2}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}} \exp\left\{-\frac{1}{2\cdot\frac{1}{3}}\left[x^2-2x\cdot\frac{z+y}{3}\right]\right\}dx$$

now just completing the square by adding and subtracting $$\left(\frac{z+y}{3} \right)^2$$ and completing the integrand with the needed constant you get

$$f_{YZ}(y,z)=\frac{1}{2\pi\sqrt{3}}e^{-(z^2+y^2)/2}\cdot e^{(z+y)^2/6}\cdot \underbrace{\int f(x|y,z)dx}_{=1}=\frac{1}{2\pi\sqrt{3}}\cdot e^{-(z^2+y^2-zy)/3}$$

where $$f(x|y,z)$$ is the density of a $$N\left(\frac{z+y}{3};\frac{1}{3} \right)$$

Now I think you are able to manipulate it to understand which kind of pdf it is. In fact, after some easy manipulations you get

$$f_{YZ}(y,z)=\frac{1}{2\pi\sqrt{2}\sqrt{2}\sqrt{1-\frac{1}{4}}}\exp\left\{-\frac{1}{2\cdot\left( 1-\frac{1}{4} \right)} \left[\frac{z^2}{2}-2\cdot\frac{1}{2}\cdot \frac{zy}{\sqrt{2}\sqrt{2}}+\frac{y^2}{2} \right] \right\}$$

In other words, $$(Y,Z)$$ are jointly gaussian

$$(Y,Z)\sim N(\mu_Y;\mu_Z;\sigma_Y^2;\sigma_Z^2;\rho_{YZ}) =N(0;0;2;2;0.5)$$