Question

If X is a Poisson random variable with variance \(P(X = 2) = 9 P(X = 4)\) then the mean and variance respectively are

• A. \(\dfrac{2}{\sqrt{3}}\) and \(\dfrac{2}{\sqrt{3}}\)
• B. \(\dfrac{1}{\sqrt{3}}\) and \(\dfrac{2}{\sqrt{3}}\)
• C. 1 and 2
• D. 2 and 2

Hint:

For a Poisson distribution with parameter \(\lambda\), the probability mass function is \(P(X = k) = \dfrac{e^{-\lambda}\lambda^{k}}{k!}\), and a special property is that both the mean and the variance equal \(\lambda\).

Answer 1

The Correct Option is A

Concept:
For a Poisson distribution with parameter \(\lambda\), the probability mass function is \(P(X = k) = \dfrac{e^{-\lambda}\lambda^{k}}{k!}\), and a special property is that both the mean and the variance equal \(\lambda\).

Step 1:
Write the given condition \(P(X = 2) = 9\,P(X = 4)\) using the formula:
\[\frac{e^{-\lambda}\lambda^{2}}{2!} = 9 \cdot \frac{e^{-\lambda}\lambda^{4}}{4!}.\]

Step 2:
Cancel \(e^{-\lambda}\) from both sides and simplify the factorials (\(2! = 2\), \(4! = 24\)):
\[\frac{\lambda^{2}}{2} = \frac{9\lambda^{4}}{24} = \frac{3\lambda^{4}}{8}.\] Multiply both sides by \(8\) and divide by \(\lambda^{2}\): \(4 = 3\lambda^{2}\), so \(\lambda^{2} = \dfrac{4}{3}\) and \(\lambda = \dfrac{2}{\sqrt{3}}\).

Step 3:
Since mean = variance = \(\lambda\) for a Poisson variable, both equal \(\dfrac{2}{\sqrt{3}}\).

Answer: Option (1) — \(\dfrac{2}{\sqrt{3}}\) and \(\dfrac{2}{\sqrt{3}}\).