Question

Suppose X is a continuous random variable with mean 1 and variance 4, Y is a standard normal random variable, that is Y∼N(0,1). Assume X and Y are independent. What is the mathematical expectation and variance of Xexp⁡(Y)? Round your answer to 2 digits.

Answer 1

The mathematical expectation of Xexp(Y) is approximately 1.65, and its variance is approximately 10.89, rounded to two digits.

Step-by-step explanation:

The question asks for the mathematical expectation (mean) and variance of Xexp(Y), when X is a continuous random variable with a mean of 1 and variance of 4, and Y is a standard normal random variable independent of X. To find the expectation of Xexp(Y), we can use the fact that the expected value of a product of independent random variables equals the product of their expected values. Since Y is normally distributed with mean 0 and variance 1 (Y ~ N(0,1)), the expected value of exp(Y) can be found using the moment-generating function of a normal distribution. It turns out that E[exp(Y)] = exp(0 + 1^2/2) = exp(0.5). Therefore, the expected value of Xexp(Y) is:

E[X]*E[exp(Y)] = 1 * exp(0.5) ≈ 1 * 1.65 = 1.65 (rounded to two digits).

For the variance, we use the fact that Var(aX) = a^2Var(X) for any constant 'a', and since exp(Y) is a function of the independent random variable Y, we can treat it as a constant when considering the variance relative to X. Hence:

Var(Xexp(Y)) ≈ Var(X) * (E[exp(Y)])^2 ≈ 4 * (exp(0.5))^2 ≈ 4 * (1.65)^2 ≈ 4 * 2.7225 = 10.89 (rounded to two digits).