Question

Suppose X is an exponential random variable with a mean of 6.1. Another random variable Y is defined as Y = X + (-7.3)X. Calculate the covariance between X and Y, rounding your answer to two decimal places.

Answer 1

The covariance between the exponential random variable X with mean 6.1 and Y (defined as Y = -6.3X) is calculated as -234.42 after rounding to two decimal places.

Step-by-step explanation:

Since Y is defined as Y = X + (-7.3)X, we can simplify this expression to Y = -6.3X. The covariance between a random variable X and a linear transformation of itself (aX, where a is a constant), such as Y in this case, is directly proportional to the variance of X. The variance of an exponential random variable is the square of its mean, so for X, which has a mean of 6.1, the variance (Var(X)) is 6.1^2 = 37.21.

The covariance between X and Y is then calculated by multiplying the variance of X by the constant factor that relates Y to X: Cov(X, Y) = -6.3 * Var(X). Plugging in the values, we get Cov(X, Y) = -6.3 * 37.21, which equals -234.423. When rounded to two decimal places, the covariance between X and Y is -234.42.