$$u (\text{mean}) = \frac{X_1} 5 + \frac 4 {(5N-1)} \cdot (X_2 +X_3 + \cdots + X_N)$$
I'm supposed to find the bias and variance of this estimator, but not sure how to do this. The sample is independent and normally distributed. This is a review problem set and we didn't cover this in class, so I'm a bit rusty. Would appreciate guidance.
What I have so far:
$$\frac{5X_1}{5N} – \frac{NX_1}{5N} = \frac{X_1(5-N)}{5N}$$
$$ \frac{X_2 +X_3 + \cdots + X_N} N – \frac 4 {5N-1} \cdot (X_2 +X_3 + \cdots + X_N) = \frac 1 5 \cdot \frac{X_2 +X_3 + \cdots + X_N} N$$
So bias: $ (X_1(5-N)/5N) + (1/5) (X_2 +X_3 + \cdots + X_N)/N$ right?
I totally forgot how to find variance, would appreciate guidance on this. What I have so far on variance:
$$\text{Var} = \frac 1 N \left(\sum_i X_i^2 – \left[N \cdot \frac 4{5N-1} \cdot (X_2 +X_3 + \cdots + X_N)\right]^2\right)$$
that's all I have so far. Do I need to simplify further? Also that's all I need to find MSE right?
Best Answer
For a computation of the bias you should to tell us the parametric distribution of the (iid?) random variables $X_i$. (I post this as an answer because my reputation is not sufficient to post it as a comment).
EDIT: Ok, then let $\mu$ and $\sigma$ be the parameters of the normal distributed iid random variables $X_i$. According to the comments above the corrected estimate is:
$$U(X_1,...,X_N)=\frac{X_1}{5}+\frac{4}{5(N-1)}(X_2+...+X_N),$$
with a slight modification of the denominator $(5N-1)\to5(N-1)$ (which make sense). Then, the expectation value is
$$\text{E}[U(X_1,...,X_N)]=\frac{\mu}{5}+\frac{4}{5(N-1)}(N-1)\,\mu=\mu.$$
That is, the estimator is unbiased since $\text{E}[U-\mu]=0$.
Similar to the Variance:
$$\text{Var}[U]=\left(\frac{1}{5}\right)^2\sigma^2+\left(\frac{4}{5(N-1)}\right)^2(N-1)\,\sigma^2=\frac{N+15}{25(N-1)}\,\sigma^2.$$
For large $N\to\infty$ the variance approaches to
$$\text{Var}[U]\to\left(\frac{\sigma}{5}\right)^2$$