Question

For the discrete distribution which is correct?

• A. $\beta_2 \geq 1$
• B. $\beta_2 \leq 2$
• C. $\beta_2 \le \beta_1$
• D. $\beta_2 = \beta_1$

Hint:

Remember the general boundary rule for Pearson's coefficients:
- \(\beta_2 \ge \beta_1 + 1\) is a fundamental inequality that applies to all probability distributions.
- Because \(\beta_1\) is always \(\ge 0\), \(\beta_2\) can never be less than 1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In statistics, the shape of a probability distribution is characterized by its moments.
Kurtosis (\(\beta_2\)) is a measure of the "tailedness" of a probability distribution.
Karl Pearson defined the coefficients of skewness (\(\beta_1\)) and kurtosis (\(\beta_2\)) using central moments.

Step 2: Key Formula or Approach:
The coefficients \(\beta_1\) and \(\beta_2\) are defined as:
\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3} \] \[ \beta_2 = \frac{\mu_4}{\mu_2^2} \] Where \(\mu_2\) (variance), \(\mu_3\), and \(\mu_4\) are the second, third, and fourth central moments.
For any real probability distribution, the central moments must satisfy the mathematical inequality:
\[ \mu_4 \mu_2 - \mu_3^2 - \mu_2^3 \ge 0 \]

Step 3: Detailed Explanation:
Let us derive the relationship between \(\beta_2\) and \(\beta_1\):
1. We begin with the inequality:
\[ \mu_4 \mu_2 \ge \mu_3^2 + \mu_2^3 \] 2. Divide both sides of the inequality by \(\mu_2^3\):
\[ \frac{\mu_4 \mu_2}{\mu_2^3} \ge \frac{\mu_3^2}{\mu_2^3} + \frac{\mu_2^3}{\mu_2^3} \] 3. Simplifying the terms:
\[ \frac{\mu_4}{\mu_2^2} \ge \frac{\mu_3^2}{\mu_2^3} + 1 \] 4. Substituting Pearson's coefficients:
\[ \beta_2 \ge \beta_1 + 1 \] 5. Since the coefficient of skewness \(\beta_1\) is a squared ratio, it is always non-negative (\(\beta_1 \ge 0\)).
6. Applying this to our inequality:
\[ \beta_2 \ge 0 + 1 \implies \beta_2 \ge 1 \] Therefore, for any probability distribution (including discrete distributions), the kurtosis coefficient \(\beta_2\) must be greater than or equal to 1.

Step 4: Final Answer:
The mathematically correct relationship is \(\beta_2 \ge 1\), matching Option (A).