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

Step 1: Write the Poisson probabilities.
With mean \(\lambda\), \(P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}\).

Step 2: Plug in the given condition.
\(P(X=2)=9P(X=4)\) becomes \[\frac{e^{-\lambda}\lambda^2}{2!}=9\cdot\frac{e^{-\lambda}\lambda^4}{4!}.\]
Step 3: Cancel the common parts.
The \(e^{-\lambda}\) cancels. Divide both sides by \(\lambda^2\): \[\frac{1}{2}=9\cdot\frac{\lambda^2}{24}.\]
Step 4: Solve for \(\lambda^2\).
\[\frac12=\frac{9\lambda^2}{24}=\frac{3\lambda^2}{8}\;\Rightarrow\;\lambda^2=\frac{8}{6}=\frac{4}{3}.\]
Step 5: Take the square root.
\[\lambda=\sqrt{\tfrac43}=\frac{2}{\sqrt3}.\]
Step 6: State mean and variance.
For a Poisson, both equal \(\lambda\), so both are \(\dfrac{2}{\sqrt3}\). This is option 1.
\[ \boxed{\text{mean}=\text{variance}=\dfrac{2}{\sqrt3}} \]