Welch's procedure proof

Question

In Welch's procedure, how does $E(\bar{Y}_i)=E(Y_i)$ and $V(\bar{Y}_i)=V(Y_i)/n$. I do not understand how it works?

Answer 1

Based on your simulation tag, it seems you're interested in the so-called Welch's procedure that uses a variance-reduction method applied to $n$ independent "replicates" of a time series $Y_1,...,Y_m$.   (Each replicate is supposed to have the same distribution as the time series; e.g., if $Y_{ji}$ is the $i$th observation in the $j$th replicate, then for all $j$, $Y_{ji}$ has the same distribution as $Y_i$.)   Letting $\overline{Y_i}$ be the arithmetic mean of the $i$th observation across the $n$ replicates:

$$\begin{align}\mathsf E(\overline{Y_i}) ~=~& \mathsf E(\frac{1}{n}~\sum_{j=1}^n Y_{ji}) \\[1ex] =~& \frac{1}{n}~\mathsf E(\sum_{j=1}^n Y_{ji}) \\[1ex] =~&\frac{1}{n}~\sum_{j=1}^n \mathsf E(Y_{ji}) \\[1ex] =~&\frac{1}{n}~\sum_{j=1}^n \mathsf E(Y_{i}) \\[1ex] =~&\frac{1}{n}~n~\mathsf E(Y_{i}) \\[1ex] =~& \mathsf E(Y_{i}) \\[3ex]\mathsf V(\overline{Y_i}) ~=~& \mathsf V(\frac{1}{n}~\sum_{j=1}^n Y_{ji}) \\[1ex] =~&\frac{1}{n^2}~\mathsf V(\sum_{j=1}^n Y_{ji}) \\[1ex] =~&\frac{1}{n^2}~\sum_{j=1}^n \mathsf V(Y_{ji}) \\[1ex] =~&\frac{1}{n^2}~\sum_{j=1}^n \mathsf V(Y_{i}) \\[1ex] =~&\frac{1}{n^2}~n~\mathsf V(Y_{i}) \\[1ex] =~&\frac{1}{n}~\mathsf V(Y_{i}) \end{align}$$