Question

Let the m.g.f. of a random variable \(X\) be of the form \(M_X(t)=\exp\!\left\{3\left(e^t-1\right)\right\}\). What is the value of the kurtosis coefficient \(\beta_2\)?

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{\sqrt{3}}\)
• C. \(\frac{10}{3}\)
• D. \(\frac{10}{\sqrt{3}}\)

Hint:

For a Poisson distribution, the excess kurtosis \(\gamma_2\) is always \(\frac{1}{\lambda}\).
Since \(\lambda = 3\), we get:
\[ \beta_2 = 3 + \gamma_2 = 3 + \frac{1}{3} = \frac{10}{3} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given moment generating function (MGF) has the form \(M_X(t) = e^{\lambda(e^t - 1)}\).
This is the MGF of a Poisson distribution with parameter \(\lambda = 3\).
We can compute the kurtosis coefficient \(\beta_2\) using the moments or cumulants of the Poisson distribution.
Key Formula or Approach:
For a Poisson distribution with parameter \(\lambda\):
- Mean, \(\mu = \lambda\)
- Variance, \(\mu_2 = \lambda\)
- Third central moment, \(\mu_3 = \lambda\)
- Fourth central moment, \(\mu_4 = \lambda + 3\lambda^2\)
The coefficient of kurtosis \(\beta_2\) is defined as:
\[ \beta_2 = \frac{\mu_4}{\mu_2^2} \]

Step 2: Detailed Explanation:
From the given MGF, we identify the parameter \(\lambda = 3\).
Now, calculate the central moments needed for the kurtosis coefficient:
- Second central moment (variance):
\[ \mu_2 = \lambda = 3 \]
- Fourth central moment:
\[ \mu_4 = \lambda + 3\lambda^2 = 3 + 3(3^2) = 3 + 3(9) = 3 + 27 = 30 \]
Now, substitute these moments into the formula for \(\beta_2\):
\[ \beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{30}{3^2} = \frac{30}{9} = \frac{10}{3} \]
Alternatively, we can use the relation for the excess kurtosis of a Poisson distribution:
\[ \gamma_2 = \beta_2 - 3 = \frac{1}{\lambda} \]
\[ \beta_2 = 3 + \frac{1}{\lambda} = 3 + \frac{1}{3} = \frac{10}{3} \]

Step 3: Final Answer:
The correct option is (C).