Q.6.50 - A First Course in Probability

Question

Let $Z_1$ and $Z_2$ be independent standard normal random variables. Show that X, Y has a bivariate normal distribution when $X =Z_1, Y = Z_1 + Z_2$.

Step-by-Step Solution

Answer

The probability density function of a bivariate normal distribution is $\frac{1}{2\mathrm{\pi }}{e}^{\frac{-(2x^2+2y+y^2)}{2}}$

Step 1 : Bivariate normal distribution :

The statistical distribution with a probability density function $\left(1\right)$ is known as the bivariate normal distribution.

Step 2 : Explanation :

Let $Z_1$ and $Z_2$ be standard normal random variables that are independent.

$Z_1$ and $Z_2$ are independent based on the information provided, and their covariance is zero.

The two standard normal random variables' probability density function is.

$$\begin{gathered} f\left(Z_1\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{e}^{-t_1^2/2} \\ f\left(Z_2\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{e}^{-z_2^2/2} \end{gathered}$$

Since, $Z_1$ and $Z_2$ are two independent variables so by independence joint distribution of them will be product of their marginal density function is $f(Z_1,Z_2)=\frac{1}{2\mathrm{\pi }}{e}^{\frac{1}{2}({\theta }^2+t_2^2)}$.

Based on the information provided $X=Z_1$and

$$\begin{gathered} Y= Z_1+Z_2 \\ \Rightarrow Z_2=Y-X \end{gathered}$$

Using the Jacobian transformation ,

$$\begin{gathered} J=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right| \\ =\left|\begin{array}{cc}1& 0\\ -1& 1\end{array}\right| \\ =1 \end{gathered}$$

So, the joint density of X and Y will be,

$$\begin{gathered} f_{xy}(x,y)=\frac{1}{2\mathrm{\pi }}{e}^{-\frac{1}{2}(z^2+a_i^2)}\left|J\right| \\ =\frac{1}{2\mathrm{\pi }}{e}^{\frac{-(x^2+{(y-x)}^2)}{2}}\left|1\right| \\ =\frac{1}{2\mathrm{\pi }}{e}^{\frac{-(2x^2+2y+y^2)}{2}} \end{gathered}$$

Hence, this is probability density function of a bivariate normal distribution.