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

Explanation:

Step1: Define the Poisson probability mass function

$$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \dots$$

Step2: Calculate the expected value $E[X]$

$$E[X] = \sum_{k=0}^{\infty} k \frac{e^{-\lambda} \lambda^k}{k!} = \sum_{k=1}^{\infty} \frac{e^{-\lambda} \lambda^k}{(k-1)!}$$

Step3: Simplify the expectation summation

$$E[X] = \lambda e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!} = \lambda e^{-\lambda} e^{\lambda} = \lambda$$

Step4: Calculate the second factorial moment

$$E[X(X-1)] = \sum_{k=2}^{\infty} k(k-1) \frac{e^{-\lambda} \lambda^k}{k!} = \lambda^2 e^{-\lambda} \sum_{k=2}^{\infty} \frac{\lambda^{k-2}}{(k-2)!} = \lambda^2$$

Step5: Determine the variance $Var(X)$

$$Var(X) = E[X(X-1)] + E[X] - (E[X])^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$$

Answer:

For a Poisson distribution with parameter $\lambda$, the expected value is $E[X] = \lambda$ and the variance is $Var(X) = \lambda$. Note: The handwritten prompt in the image appears to contain a notation error suggesting $1/\lambda$; however, by definition of the Poisson distribution, both mean and variance are equal to $\lambda$.