Question

Let \(X\) and \(Y\) have a bivariate normal distribution with parameters \(\mu_{1}=\) \(20, \mu_{2}=40, \sigma_{1}^{2}=9, \sigma_{2}^{2}=4\), and \(\rho=0.6 .\) Find the shortest interval for which \(0.90\) is the conditional probability that \(Y\) is in the interval, given that \(X=22\).

Short answer

The shortest interval for which 0.90 is the conditional probability that Y is in the interval, given that X=22 is approximately (38.6 , 43.01).

Step-by-step solution

Understand conditional distribution

Firstly, if (X,Y) follows a bivariate normal distribution, then the conditional distribution of Y given X is normal distribution itself. To find the required probabilities, let's calculate the mean and standard deviation for the conditional distribution.

Calculate Mean of Conditional Distribution

The mean of the conditional distribution of Y|X=22, denoted as \( \mu_{Y|X=22}\), can be computed using the formula: \[ \mu_{Y|X=22} = \mu_{2} + \rho * \frac{\sigma_{2}}{\sigma_{1}} * (X - \mu_{1}) \] Substitute the given values, we have: \[ \mu_{Y|X=22} = 40 + 0.6 * \frac{2}{3} * (22 -20) = 40.8 \]

Calculate Variance of Conditional Distribution

The variance of the conditional distribution of Y|X=22, denoted as \( \sigma_{Y|X=22}^2\), can be computed using the formula: \[\sigma_{Y|X=22}^2 = \sigma_{2}^2 * (1 - \rho^2)\] Substitute the given values, we have: \[\sigma_{Y|X=22}^2 = 4 * (1 - 0.6^2) = 2.24 \] Therefore, the standard deviation \( \sigma_{Y|X=22} = \sqrt{2.24} \approx 1.5 \]

Determine the interval

Since Y|X=22 is normally distributed, approximately 90% of data lies within ±1.645 standard deviations around the mean. Therefore, the interval for which 0.90 is the conditional probability can be calculated as: \[(\mu_{Y|X=22} - 1.645*\sigma_{Y|X=22}, \mu_{Y|X=22} + 1.645*\sigma_{Y|X=22})\] Substitute the values, we have: \[(40.8 - 1.645*1.5, 40.8 + 1.645*1.5) = (38.6 , 43.01)\]