Question

The random variables \(X\) and \(Y\) are said to have a bivariate normal distribution if their joint density function is given by $$ \begin{aligned} f(x, y)=& \frac{1}{2 \pi \sigma_{x} \sigma_{y} \sqrt{1-\rho^{2}}} \exp \left\\{-\frac{1}{2\left(1-\rho^{2}\right)}\right.\\\ &\left.\times\left[\left(\frac{x-\mu_{x}}{\sigma_{x}}\right)^{2}-\frac{2 \rho\left(x-\mu_{x}\right)\left(y-\mu_{y}\right)}{\sigma_{x} \sigma_{y}}+\left(\frac{y-\mu_{y}}{\sigma_{y}}\right)^{2}\right]\right\\} \end{aligned} $$ for \(-\infty0, \sigma_{y}>0,-\infty<\mu_{x}<\infty,-\infty<\mu_{y}<\infty\) (a) Show that \(X\) is normally distributed with mean \(\mu_{x}\) and variance \(\sigma_{x}^{2}\), and \(Y\) is normally distributed with mean \(\mu_{y}\) and variance \(\sigma_{y}^{2}\). (b) Show that the conditional density of \(X\) given that \(Y=y\) is normal with mean \(\mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{x}^{2}\left(1-\rho^{2}\right)\) The quantity \(\rho\) is called the correlation between \(X\) and \(Y\). It can be shown that $$ \begin{aligned} \rho &=\frac{E\left[\left(X-\mu_{x}\right)\left(Y-\mu_{y}\right)\right]}{\sigma_{x} \sigma_{y}} \\ &=\frac{\operatorname{Cov}(X, Y)}{\sigma_{x} \sigma_{y}} \end{aligned} $$

Short answer

To summarize, we integrated the joint PDF of X and Y to find their marginal PDFs and verified that X and Y are normally distributed with means \(\mu_x\) and \(\mu_y\) and variances \(\sigma_x^2\) and \(\sigma_y^2\), respectively. Furthermore, we found the conditional PDF of X given Y=y and verified that it is normally distributed with mean \(\mu_X^{*} = \mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{X^{*}}^{2} = \sigma_{x}^{2}\left(1-\rho^{2}\right)\).

Step-by-step solution

(Marginal PDF of X and Y)

: First, we will find the marginal PDFs for X and Y by integrating their joint PDF. Marginal PDF of X can be found by integrating the joint PDF with respect to y: \(f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \)

(Verify Normal Distribution of X)

: To show that X is normally distributed with mean \(\mu_x\) and variance \(\sigma_x^2\), we need to verify that the marginal PDF of X is of the normal distribution form: \(f_X(x) = \frac{1}{\sqrt{2\pi\sigma_x^2}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu_x}{\sigma_x}\right)^2\right\}\) Similarly, Marginal PDF of Y can be found by integrating the joint PDF with respect to x: \(f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx \)

(Verify Normal Distribution of Y)

: To show that Y is normally distributed with mean \(\mu_y\) and variance \(\sigma_y^2\), we need to verify that the marginal PDF of Y is of the normal distribution form: \(f_Y(y) = \frac{1}{\sqrt{2\pi\sigma_y^2}}\exp\left\{-\frac{1}{2}\left(\frac{y-\mu_y}{\sigma_y}\right)^2\right\}\)

(Conditional PDF of X given Y=y)

: Now, we will find the conditional PDF of X given that Y=y. By the definition of conditional probability, we can find the conditional PDF of X given Y=y as: \(f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)}\)

(Verify Normal Distribution of X given Y=y)

: To show that the conditional PDF of X given that Y=y is normal with mean \(\mu_X^{*} = \mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{X^{*}}^{2} = \sigma_{x}^{2}\left(1-\rho^{2}\right)\), we need to verify that the conditional PDF of X given Y=y is of the normal distribution form: \(f_{X|Y}(x|y) = \frac{1}{\sqrt{2\pi\sigma_{X^{*}}^2}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu_{X^{*}}}{\sigma_{X^{*}}}\right)^2\right\}\)