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

Step 1: Describe the random variable. 
The random variable $X$ can take only two possible values: $1$ and $2$.
Let the probabilities be defined as:

\[ P(X=1)=p \quad \text{and} \quad P(X=2)=1-p \]

The expected value of $X$ is given as $E[X]=\frac{10}{7}$.
Using the definition of expectation:

\[ E[X] = 1\cdot p + 2\cdot(1-p) = 2 - p \]

Equating this to the given mean:

\[ 2 - p = \frac{10}{7} \]

Solving for $p$:

\[ p = 2 - \frac{10}{7} = \frac{4}{7} \]

Hence,

\[ P(X=1)=\frac{4}{7}, \quad P(X=2)=\frac{3}{7} \]

Step 2: Construct the moment generating function.
The moment generating function (MGF) of $X$ is defined as:

\[ M_X(t)=E[e^{tX}] \]

Substituting the values of $X$ and their probabilities:

\[ M_X(t)=\frac{4}{7}e^{t}+\frac{3}{7}e^{2t} \]

Step 3: Differentiate the MGF four times.
Taking successive derivatives:

\[ M_X'(t)=\frac{4}{7}e^{t}+\frac{6}{7}e^{2t} \]

\[ M_X''(t)=\frac{4}{7}e^{t}+\frac{12}{7}e^{2t} \]

\[ M_X^{(3)}(t)=\frac{4}{7}e^{t}+\frac{24}{7}e^{2t} \]

\[ M_X^{(4)}(t)=\frac{4}{7}e^{t}+\frac{48}{7}e^{2t} \]

Now evaluate the fourth derivative at $t=0$:

\[ M_X^{(4)}(0)=\frac{4}{7}+\frac{48}{7}=\frac{52}{7} \]

Step 4: Final result.
The value of the fourth derivative of the moment generating function at zero is:

\[ \boxed{\frac{52}{7}} \]