Question

Let 𝑋 be a random variable having a probability mass function 𝑝(𝑥) which is positive only for non-negative integers. If
𝑝(𝑥+1)=(\(\frac{ ln\,\, 3}{𝑥+1}\)) 𝑝(𝑥), 𝑥=0, 1, 2, … , 
then 𝑉𝑎𝑟(𝑋) equals

• A. ln 3
• B. ln 6
• C. ln 9
• D. ln 18

Answer 1

The Correct Option is A

Given the probability mass function \( p(x) \) for a random variable \( X \), which is positive only for non-negative integers, we have the relation:

\(p(x+1) = \left(\frac{\ln 3}{x+1}\right) p(x), \quad x = 0, 1, 2, \ldots\) 

To find the variance \( \text{Var}(X) \), we first need to determine the probability mass function explicitly.

Let's start with an initial assumption \( p(0) = c \), where \( c \) is a normalizing constant. Using the recursive relation, we find:

1. \(p(1) = \left(\frac{\ln 3}{1}\right) p(0) = c \ln 3\)
2. \(p(2) = \left(\frac{\ln 3}{2}\right) p(1) = \frac{(c \ln 3)^2}{2!}\)
3. \(p(3) = \left(\frac{\ln 3}{3}\right) p(2) = \frac{(c \ln 3)^3}{3!}\)
4. \(\ldots\)

Generalizing, we have:

\(p(x) = \frac{(c \ln 3)^x}{x!}\)

To determine \( c \), use the normalization condition:

\(\sum_{x=0}^{\infty} p(x) = 1\)

This leads to:

\(\sum_{x=0}^{\infty} \frac{(c \ln 3)^x}{x!} = e^{c \ln 3} = 1\)

This implies, \(c = \frac{1}{\ln 3}\)

Thus, the probability mass function is:

\(p(x) = \frac{(\ln 3)^x}{x! \, e^{\ln 3}} = \frac{1}{3}\frac{(\ln 3)^x}{x!}\)

This is the probability mass function of a Poisson distribution with rate parameter \(\lambda = \ln 3\).

The variance of a Poisson distribution with parameter \(\lambda\) is equal to \(\lambda\). Therefore:

The variance \(\text{Var}(X) = \ln 3\).

This matches the correct answer choice:

• \(\ln 3\)
• \(\ln 6\)
• \(\ln 9\)
• \(\ln 18\)

Conclusion: Therefore, the variance \( \text{Var}(X) \) of the random variable is \(\ln 3\).