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

Explanation:

Step1: Define the Poisson PMF

The probability mass function for $X \sim \text{Poisson}(\lambda)$ is: $$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \dots$$

Step2: Calculate the expected value $E(X)$

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

Step3: Simplify the expectation sum

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

Step4: Calculate the second factorial moment

$$E(X(X-1)) = \sum_{k=0}^{\infty} k(k-1) \frac{e^{-\lambda} \lambda^k}{k!} = \lambda^2 e^{-\lambda} \sum_{k=2}^{\infty} \frac{\lambda^{k-2}}{(k-2)!} = \lambda^2$$

Step5: Determine the variance $\text{Var}(X)$

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

Answer:

For a Poisson distribution with parameter $\lambda$, the mean is $E(X) = \lambda$ and the variance is $\text{Var}(X) = \lambda$. (Note: The handwritten prompt appears to contain a notation error writing $1/\lambda$ instead of $\lambda$).