Bivariate normal distribution when $\rho$ is 0

1.1k Views Asked by At

What happens to the bivariate normal distribution when $\rho$ is 0?The bi-variate normal reduces to a simpler distribution, but what is it? and how do you calculate the cdf then?

What I have tried:

When $\rho$ is 0, the bivariate normal distribution becomes independent normal cdf. Cdf then be calculated by normal cdf. Is this right?

1

There are 1 best solutions below

0
On BEST ANSWER

Let $X$ and $Y$ be the random variables that follow the normal distribution. Let the joint PDF of $X$ and $Y$ be the bivariate normal distribution. $$ f_{X,Y}(x,y)=\frac{\exp\left(-\frac{1}{2\left(1-\rho^2\right)}\left[\frac{(x-\mu_x)^2}{\sigma_x^2}+\frac{(y-\mu_y)^2}{\sigma_y^2}-\frac{2\rho(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}\right]\right)}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}, $$ where $\rho$ is the correlation between $X$ and $Y$. The definition of the correlation is $$ \rho_{X,Y}=\frac{\text{Cov}[X,Y]}{\sigma_x\sigma_y}. $$ If $\rho_{X,Y}=0$, then $\text{Cov}[X,Y]=0$, which means $X$ and $Y$ is independent. Hence, the joint PDF of $X$ and $Y$ can be written as $$ f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y). $$ Proof :

$$ \begin{align} f_{X,Y}(x,y)&=\frac{\exp\left(-\frac{1}{2\left(1-0^2\right)}\left[\frac{(x-\mu_x)^2}{\sigma_x^2}+\frac{(y-\mu_y)^2}{\sigma_y^2}-\frac{2(0)(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}\right]\right)}{2\pi\sigma_x\sigma_y\sqrt{1-0^2}}\\ &=\frac{1}{2\pi\sigma_x\sigma_y}\exp\left(-\frac{1}{2}\left[\frac{(x-\mu_x)^2}{\sigma_x^2}+\frac{(y-\mu_y)^2}{\sigma_y^2}\right]\right)\\ &=\frac{1}{\sqrt{2\pi}\sigma_x}\exp\left(-\frac{(x-\mu_x)^2}{2\sigma_y^2}\right)\cdot\frac{1}{\sqrt{2\pi}\sigma_y}\exp\left(-\frac{(y-\mu_y)^2}{2\sigma_y^2}\right)\\ &=f_X(x)\cdot f_Y(y).&\quad\blacksquare \end{align} $$ The distribution of $X$ and $Y$ is still bivariate normal distribution, more details the independent bivariate normal distribution. The CDF of $X$ and $Y$ can be calculated as follows $$ \begin{align} \text{Pr}[X\le x,Y\le y]=F_{X,Y}(x,y)&=\int_{-\infty}^y\int_{-\infty}^x f_{X,Y}(u,v)\ du\,dv\\ &=\int_{-\infty}^x f_{X}(u)\ du\int_{-\infty}^yf_{Y}(v)\ dv, \end{align} $$ where $u$ and $v$ are just the dummy variables that substitute $x$ and $y$.

$$\\$$

$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$