Question

use poisson distribution to show that
$e(x) = \frac{1}{\lambda}$
$var(x) = \frac{1}{\lambda}$

Explanation:

Step1: Recall Poisson PMF

The probability mass function of a Poisson random variable \(X\) with parameter \(\lambda > 0\) is:
$$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}, \quad k=0,1,2,...$$

Step2: Calculate expected value

By definition of expected value:

$$\begin{align*} E(X) &= \sum_{k=0}^{\infty}k \cdot \frac{e^{-\lambda}\lambda^k}{k!} \\ &= e^{-\lambda}\lambda \sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!} \\ &= e^{-\lambda}\lambda e^{\lambda} = \lambda \end{align*}$$

Step3: Calculate \(E(X^2)\)

Use the identity \(E(X^2) = E[X(X-1)] + E(X)\):

$$\begin{align*} E[X(X-1)] &= \sum_{k=0}^{\infty}k(k-1) \cdot \frac{e^{-\lambda}\lambda^k}{k!} \\ &= e^{-\lambda}\lambda^2 \sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{(k-2)!} \\ &= e^{-\lambda}\lambda^2 e^{\lambda} = \lambda^2 \end{align*}$$

Thus, \(E(X^2) = \lambda^2 + \lambda\)

Step4: Calculate variance

Use the variance formula \(Var(X) = E(X^2) - [E(X)]^2\):
$$Var(X) = (\lambda^2 + \lambda) - \lambda^2 = \lambda$$

Answer:

For a Poisson random variable \(X\):

1. \(E(X) = \lambda\)
2. \(Var(X) = \lambda\)