Let $X_1,\dots,X_n$ and $Y_1,\dots,Y_m$ be two independent samples from a Poisson distribution with parameter $\lambda$. Let $a,b$ be two positive numbers. Consider the following estimator for $\lambda$: $\hat{\lambda} = a \dfrac{X_1 + \cdots + X_n}{n} + b\dfrac{Y_1 + \cdots + Y_m}{m}$
(a) What condition is needed on $a$ and $b$ so that $\hat{Y}$ is unbiased?
(b) What is the MSE of $\hat{\lambda}$?
Unbiased means $E[\hat{\lambda}] = \lambda$. So $$E[\hat{\lambda}] = \frac{a}{n}\sum_{i=1}^n E[X_i] + \frac{b}{m} \sum_{i=1}^{m} E[Y_i] = a\lambda + b\lambda = (a+b) \lambda.$$
So $\hat{\lambda}$ is unbiased as long as $a+b=1$.
The MSE is defined as: $$MSE= E[(\hat{\lambda} - \lambda)^2] = Var[\hat{\lambda}]$$ where the last follows since $\hat{\lambda}$ is unbiased. Hence (here we strongly need the independence of $X_i$ and $Y_i$) we have $$Var[\hat{\lambda}] = \frac{a^2}{n}\lambda + \frac{b^2}{m}\lambda = \left(\frac{a^2}{n}+\frac{b^2}{m}\right)\lambda.$$
Observe that $Var[\hat{\lambda}] \to 0$ as $n,m\to \infty$, that means $\hat{\lambda}$ is a consistent estimator, which is good. In other words, this implies (by Chebyshev's inquality) that $\hat{\lambda} \to \lambda$ a.s. as $n\to\infty$.