Question

Let \( X \) be a random variable taking only two values, 1 and 2. Let \( M_X(t) \) be the moment generating function of \( X \). If the expectation of \( X \) is \( \frac{10}{7} \), then the fourth derivative of \( M_X(t) \) evaluated at 0 equals

• A. \( \frac{48}{7} \)
• B. \( \frac{67}{7} \)
• C. \( \frac{52}{7} \)
• D. \( \frac{60}{7} \)

Hint:

For a discrete random variable, the moment generating function is the weighted sum of \( e^{tx} \) for all possible values of \( x \), where the weights are the probabilities of the corresponding values.

Answer 1

The Correct Option is C

To solve the problem, we need to find the fourth derivative of the moment-generating function \( M_X(t) \) evaluated at \( t = 0 \) for the random variable \( X \) taking values 1 and 2, where the expectation of \( X \) is given as \( \frac{10}{7} \). Let's proceed step-by-step:

1. Define the probabilities associated with the values of the random variable \( X \). Let \( P(X=1) = p \) and \( P(X=2) = 1 - p \). We know that the expected value \( E(X) \) is calculated by: \(E(X) = 1 \cdot p + 2 \cdot (1-p)\).
2. Given that \( E(X) = \frac{10}{7} \), substituting into the expectation formula gives: \(p + 2(1 - p) = \frac{10}{7}\), which simplifies to: \(p + 2 - 2p = \frac{10}{7}\).
3. Simplify further: \(2 - p = \frac{10}{7}\), \(p = 2 - \frac{10}{7}\), \(p = \frac{4}{7}\).
4. So, \( P(X=1) = \frac{4}{7} \) and \( P(X=2) = \frac{3}{7} \).
5. The moment-generating function \( M_X(t) \) is given by: \(M_X(t) = E(e^{tX}) = e^t \cdot \frac{4}{7} + e^{2t} \cdot \frac{3}{7}\).
6. To find the fourth derivative of \( M_X(t) \) at \( t = 0 \), we need:
   • \( M_X'(t) = \frac{4}{7}e^t + \frac{3}{7}(2e^{2t}) \)
   • \( M_X''(t) = \frac{4}{7}e^t + \frac{3}{7}(4e^{2t}) \)
   • \( M_X'''(t) = \frac{4}{7}e^t + \frac{3}{7}(8e^{2t}) \)
   • \( M_X^{(4)}(t) = \frac{4}{7}e^t + \frac{3}{7}(16e^{2t}) \)
7. Evaluate the fourth derivative at \( t=0 \): \(M_X^{(4)}(0) = \frac{4}{7}\cdot1 + \frac{3}{7}(16\cdot1) = \frac{4}{7} + \frac{48}{7} = \frac{52}{7}\).

The final result, \(\frac{52}{7}\), matches option one. 

Answer 2

Step 1: Understanding the random variable \( X \). 
The random variable \( X \) takes only two values, 1 and 2. Let the probability mass function of \( X \) be: \[ P(X = 1) = p \quad \text{and} \quad P(X = 2) = 1 - p \] The expectation of \( X \), \( E[X] \), is given as \( \frac{10}{7} \). Using the formula for expectation: \[ E[X] = 1 \cdot p + 2 \cdot (1 - p) = p + 2(1 - p) = 2 - p \] Given \( E[X] = \frac{10}{7} \), we have the equation: \[ 2 - p = \frac{10}{7} \] Solving for \( p \): \[ p = 2 - \frac{10}{7} = \frac{14}{7} - \frac{10}{7} = \frac{4}{7} \] So, \( p = \frac{4}{7} \) and \( P(X = 1) = \frac{4}{7} \), and \( P(X = 2) = \frac{3}{7} \). 
Step 2: Moment generating function of \( X \). 
The moment generating function \( M_X(t) \) of \( X \) is defined as: \[ M_X(t) = E[e^{tX}] = e^{t \cdot 1} \cdot P(X = 1) + e^{t \cdot 2} \cdot P(X = 2) \] Substituting the values for \( P(X = 1) \) and \( P(X = 2) \): \[ M_X(t) = e^t \cdot \frac{4}{7} + e^{2t} \cdot \frac{3}{7} \] Step 3: Finding the fourth derivative of \( M_X(t) \). 
To find the fourth derivative of \( M_X(t) \) at \( t = 0 \), we differentiate \( M_X(t) \) four times: \[ M_X'(t) = \frac{4}{7} e^t + \frac{6}{7} e^{2t} \] \[ M_X''(t) = \frac{4}{7} e^t + \frac{12}{7} e^{2t} \] \[ M_X^{(3)}(t) = \frac{4}{7} e^t + \frac{24}{7} e^{2t} \] \[ M_X^{(4)}(t) = \frac{4}{7} e^t + \frac{48}{7} e^{2t} \] Now, evaluate \( M_X^{(4)}(t) \) at \( t = 0 \): \[ M_X^{(4)}(0) = \frac{4}{7} e^0 + \frac{48}{7} e^0 = \frac{4}{7} + \frac{48}{7} = \frac{52}{7} \] Step 4: Conclusion. 
The fourth derivative of \( M_X(t) \) evaluated at 0 is \( \frac{52}{7} \), hence the correct answer is \( \boxed{C} \).