Question

Let \( (X, Y)^T \) follow a bivariate normal distribution with \[ E(X) = 2, \, E(Y) = 3, \, {Var}(X) = 16, \, {Var}(Y) = 25, \, {Cov}(X, Y) = 14. \] Then \[ 2\pi \left( \Pr(X>2, Y>3) - \frac{1}{4} \right) \] equals _________ (rounded off to two decimal places).

Hint:

For joint probabilities involving bivariate normal distributions, standardize the variables and use the correlation coefficient to calculate the probability.

Answer 1

Step 1: Standardize the bivariate normal distribution.
We need to compute \( \Pr(X>2, Y>3) \) for a bivariate normal distribution. To do this, we first standardize the variables. Let \( Z_X = \frac{X - E(X)}{\sigma_X} \) and \( Z_Y = \frac{Y - E(Y)}{\sigma_Y} \), where \( \sigma_X = \sqrt{16} = 4 \) and \( \sigma_Y = \sqrt{25} = 5 \). \[ Z_X = \frac{X - 2}{4}, \quad Z_Y = \frac{Y - 3}{5}. \] Step 2: Compute the joint probability.
We need to compute \( \Pr(X>2, Y>3) \) or equivalently \( \Pr(Z_X>0, Z_Y>0) \). The joint probability for a bivariate normal distribution is given by the standard bivariate normal CDF: \[ \Pr(Z_X>0, Z_Y>0) = 1 - \Phi(0, 0, \rho), \] where \( \rho = \frac{{Cov}(X, Y)}{\sigma_X \sigma_Y} = \frac{14}{4 \times 5} = 0.7 \). From standard tables or software, we find that \( \Phi(0, 0, 0.7) \approx 0.758 \). Therefore: \[ \Pr(Z_X>0, Z_Y>0) = 1 - 0.758 = 0.242. \] Step 3: Compute the final expression.
Now, compute the desired value: \[ 2\pi \left( 0.242 - \frac{1}{4} \right) = 2\pi \times 0.242 - 2\pi \times 0.25. \] Using \( \pi \approx 3.1416 \), we get: \[ 2\pi \times 0.242 \approx 1.522, \quad 2\pi \times 0.25 \approx 1.5708. \] Thus: \[ 1.522 - 1.5708 = -0.0488. \] The final result is approximately \( \boxed{-0.05} \).