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

Explanation:

Step1: Define the Poisson probability mass function

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

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

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

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

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

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 $Var(X) = \lambda$. The expressions $1/\lambda$ shown in the image typically refer to the Exponential distribution's mean and variance (where $Var(X)=1/\lambda^2$), or there is a notation mismatch in the prompt's premise. Under standard Poisson definitions, both equal $\lambda$.