Question

Let \( X \) and \( Y \) be two random variables such that the moment generating function of \( X \) is \( M(t) \) and the moment generating function of \( Y \) is \[ H(t) = \left( \frac{3}{4} e^{2t} + \frac{1}{4} \right) M(t), \] where \( t \in (-h, h), h>0 \). If the mean and the variance of \( X \) are \( \frac{1}{2} \) and \( \frac{1}{4} \), respectively, then the variance of \( Y \) (in integer) is equal to ________

Hint:

To find the mean and variance of a random variable using its MGF, differentiate the MGF and evaluate at \( t = 0 \).

Answer 1

Correct Answer: 1

The moment generating function (MGF) of a random variable \( Z \) is defined as \( M_Z(t) = E[e^{tZ}] \). The MGF of \( Y \) is given as: \[ H(t) = \left( \frac{3}{4} e^{2t} + \frac{1}{4} \right) M_X(t), \] where \( M_X(t) \) is the MGF of \( X \). The mean and variance of \( X \) are the first and second moments of the MGF of \( X \). From the properties of MGFs: \[ E[X] = M_X'(0), \quad \text{Var}(X) = M_X''(0) - \left( M_X'(0) \right)^2. \] We are given \( E[X] = \frac{1}{2} \) and \( \text{Var}(X) = \frac{1}{4} \). Now, using the MGF of \( Y \), we calculate the mean and variance of \( Y \). First, differentiate \( H(t) \) to find \( E[Y] \) and \( \text{Var}(Y) \). Using the chain rule: \[ E[Y] = H'(0), \quad \text{Var}(Y) = H''(0) - (H'(0))^2. \] After calculating the derivatives and substituting the known values for \( E[X] \) and \( \text{Var}(X) \), we find that the variance of \( Y \) is \( \boxed{1} \).