1. **Problem Statement:** Find the mean and variance of a Poisson distribution with parameter $\lambda$. 2. **Recall the probability generating function (p.g.f) of a Poisson distribution:** $$P(s) = e^{\lambda(s-1)}$$ This function generates probabilities $p_k = P(X=k)$ for $k=0,1,2,...$ where $X \sim \text{Poisson}(\lambda)$. 3. **Mean of Poisson distribution:** The mean $\mu$ is given by the first derivative of the p.g.f evaluated at $s=1$: $$\mu = P'(1) = \frac{d}{ds} e^{\lambda(s-1)} \bigg|_{s=1} = \lambda e^{\lambda(1-1)} = \lambda$$ 4. **Variance of Poisson distribution:** The variance $\sigma^2$ is given by: $$\sigma^2 = P''(1) + P'(1) - (P'(1))^2$$ Calculate second derivative: $$P''(s) = \frac{d^2}{ds^2} e^{\lambda(s-1)} = \lambda^2 e^{\lambda(s-1)}$$ Evaluate at $s=1$: $$P''(1) = \lambda^2$$ Therefore, $$\sigma^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$$ 5. **Interpretation:** The mean and variance of a Poisson distribution are both equal to the parameter $\lambda$. **Final answer:** $$\boxed{\text{Mean} = \lambda, \quad \text{Variance} = \lambda}$$