Mean & Variance in Distribution

Question

Let $S_1, \ldots , S_n$ be the yet-to-be observed survey responses from $n$ individuals, where the event $S_i = 1$ corresponds to the $i$th individual subscribing to Spotify and the event $S_i = 0$ corresponds to the ith individual does not subscribe to Spotify $(i = 1,\ldots, n)$. Assume that $S_1, \ldots , S_n$ are i.i.d. $Ber(\pi)$.

You have a random variable $S = \sum\limits_{i=1}^n S_i$.

I think this is a binomial distribution so my mean is

$\theta =$ probability of success

$n \theta$

and variance is

$n (1- \theta)$

I am trying to figure out what the mean and variance would be if you multiply $S$ by $1/n$

so new random variable is $S = (1/n) \sum\limits_{i=1}^n S_i$.

Answer 1

Hints Note that $$ ES=\frac{1}{n}\sum_{i=1}^n ES_i $$ by linearity of expectation. Now use the fact that the $S_i$ are identically distributed. Moreover since the $S_i$ are independent $$ V(S)=\frac{1}{n^2}\sum_{i=1}^nV(S_i) $$ where $V(S)$ denotes variance of $S$.

Answer 2

The expected value $E[S]$ is linear hence $$ E(\alpha S) = \alpha E(S) $$ and the variance is $V[S] = E[(S-E[S])^2]$ hence $$ V[\alpha S] = E[(\alpha S-\alpha E[S])^2] = E[\alpha^2( S-E[S])^2] = \alpha^2 E[(S-E[S])^2] = \alpha^2 V[S]. $$

This extents to your case where $$ E\left[\frac{1}{n}S\right] = \frac{n\theta}{n} = \theta\\ V\left[\frac{1}{n}S\right] = \frac{n\theta(1-\theta)}{n^2} = \frac{\theta(1-\theta)}{n}. $$

Note that there is a typo where you state the variance of the binomial distribution.