Solving this equation, we get:
$$\frac{\partial \log L}{\partial \lambda} = \sum_{i=1}^{n} \frac{x_i}{\lambda} - n = 0$$
Suppose we have a sample of size $n$ from a Poisson distribution with parameter $\lambda$. Find the MLE of $\lambda$. theory of point estimation solution manual
$$\hat{\lambda} = \bar{x}$$
$$\hat{\mu} = \bar{x}$$
$$L(\lambda) = \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!}$$
$$L(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right)$$ Solving this equation
Solving these equations, we get: