Question

If for a Poisson variate \( X \), \( P(X = k) = P(X = k + 1) \), then the variance of \( X \) is:

• A. \( k - 1 \)
• B. \( k \)
• C. \( k + 1 \)
• D. \( k + 2 \)

Hint:

In a Poisson distribution, the mean and variance are both equal to \( \lambda \). If probabilities of consecutive values are equal, solve for \( \lambda \) by equating their probabilities.

Answer 1

The Correct Option is C

For a Poisson distribution, the probability mass function is given by: \[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}. \] Given that \( P(X = k) = P(X = k + 1) \), we equate: \[ \frac{\lambda^k e^{-\lambda}}{k!} = \frac{\lambda^{k+1} e^{-\lambda}}{(k+1)!}. \] Simplifying, we get: \[ \frac{1}{k!} = \frac{\lambda}{(k+1)!} \quad \Rightarrow \quad \lambda = k + 1. \] For a Poisson distribution, the variance is equal to \( \lambda \). Thus, the variance of \( X \) is: \[ {Variance} = k + 1. \]