Let $x_1, x_2, \ldots,x_n$ be a simple random sample from a random variable $X$ with support $\{0,1,2,3,4\}$ and probability function $p(0)=\frac{5}{12}(1-\lambda)^2$, $p(1)=\lambda$, $p(2)=\lambda(1-\lambda)$, $p(3)=\frac{1}{3}(1-\lambda)^2$ and $p(4)=\frac{1}{4}(1-\lambda)^2$, where $\lambda$ is a parameter $\in(0,1)$.
Is this estimator $$\hat \lambda=1+\frac{n}{n+1}-\bar X$$
unbiased and consistent for $\lambda$? $\bar X$ is sample mean.
The expectation of $X$ is:
$$\mathbb{E}(X)=0\cdot\frac{5}{12}(1-\lambda)^2+1\cdot\lambda+2\cdot\lambda(1-\lambda)+3\cdot\frac{1}{3}(1-\lambda)^2+4\cdot\frac{1}{4}(1-\lambda)^2=2-\lambda$$
Now, the expectation of $\hat \lambda$ should be:
$$\mathbb{E}(\hat \lambda)=\mathbb{E}\left(1+\frac{n}{n+1}-\bar X\right)=1+\frac{n}{n+1}-(2-\lambda)$$
thus the estimator is biased.
For consistency we need $\lim_{n \to \infty}\mathbb{E}(\hat \lambda)=\lambda$ and $\lim_{n \to \infty}\mathbb{Var}(\hat \lambda)=0$, so:
$$\lim_{n \to \infty}\mathbb{E}(\hat \lambda)=\lim_{n \to \infty}\left(1+\frac{n}{n+1}-(2-\lambda)\right)=\lambda$$
How can I find $\mathbb{Var}(\hat \lambda)$?
First edit
Based on Michael's suggestion:
$\mathbb{Var}(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2=\left(0\cdot\frac{5}{12}(1-\lambda)^2+1^2\cdot\lambda+2^2\cdot\lambda(1-\lambda)+3^2\cdot\frac{1}{3}(1-\lambda)^2+4^2\cdot\frac{1}{4}(1-\lambda)^2\right)-(2-\lambda)^2=3-5\lambda+2\lambda^2$.
$\mathbb{Var}(\widehat\lambda)=\dfrac{3-5\lambda+2\lambda^2}{n}$ so $\lim_{n \to \infty}\mathbb{Var}(\hat \lambda)=0$