Question

The standard deviation of a Poisson distribution is $\sigma$. If $P(X = r) = K$, then $P(X = r + 2) =$

• A. \(\dfrac{4K}{r(r + 2)}\)
• B. \(\dfrac{16K}{(r + 2)(r + 3)}\)
• C. \(\dfrac{4K}{(r + 2)(r + 1)}\)
• D. \(\dfrac{15K}{(r + 1)(r + 2)}\)

Hint:

Express Poisson probabilities using factorial form and reduce ratio of terms using \(\lambda\) relationships.

Answer 1

The Correct Option is C

In a Poisson distribution, the probability mass function is given by:

$$P(X = r) = \frac{\lambda^r e^{-\lambda}}{r!}$$

where \( \lambda \) is the average rate of occurrence and \( r \) is the actual number of occurrences.

Given \( P(X = r) = K \), we have:

$$\frac{\lambda^r e^{-\lambda}}{r!} = K$$

We need to find \( P(X = r + 2) \), which is:

$$P(X = r + 2) = \frac{\lambda^{r+2} e^{-\lambda}}{(r+2)!}$$

Dividing \( P(X = r + 2) \) by \( P(X = r) \), we get:

$$\frac{P(X = r + 2)}{P(X = r)} = \frac{\frac{\lambda^{r+2} e^{-\lambda}}{(r+2)!}}{\frac{\lambda^r e^{-\lambda}}{r!}}$$

This simplifies to:

$$\frac{\lambda^{r+2}}{(r+2)!} \times \frac{r!}{\lambda^r} = \frac{\lambda^2}{(r+2)(r+1)}$$

Therefore:

$$P(X = r + 2) = \frac{\lambda^2 K}{(r+2)(r+1)}$$

Given \(\sigma = \sqrt{\lambda}\), we substitute \(\lambda = \sigma^2\) in the above equation:

$$P(X = r + 2) = \frac{(\sigma^2)^2 K}{(r+2)(r+1)} = \frac{\sigma^4 K}{(r+2)(r+1)}$$

Thus:

$$P(X = r + 2) = \frac{4K}{(r+2)(r+1)}$$

The correct answer is:

\(\frac{4K}{(r + 2)(r + 1)}\)