Just began studying statistical inference and I'm hoping someone could help clarify an example in the literature.
If we perform $10$ trials and note that the event $A$ happens in x = 4 of these, then the relative frequency $p* = x/10 = 0.4$. Since $x$ is an observation of $X \sim \mathrm{Bin}(10, p)$ with $\mathrm{Var}(X) = 10p(1-p)$ then $\mathrm{Var}(p*) = p(1-p)/10.$
I don't quite understand the last equality. Why is it divided by $10$? Is it because the variance for the binomial distribution is calculated as $np(1-p)$ where $np$ is the mean, but here we have estimated $x/10$ as the mean? As you can probably tell, I'm a bit confused, so any clarification would be very helpful.
You know that $Var(X)=n\cdot p \cdot (1-p)$.
Now you can calculate $Var\left(\frac{X}{n} \right)=\frac{1}{n^2}\cdot Var\left(X \right)$
$\frac{X}{n} $ is the random variable of the proportion of successful $n$ trials. This estimator is unbiased, because $E\left(\frac{X}{n}\right)=p$
$n$ is a constant. Therefore it can be factored out.
$\frac{1}{n^2}\cdot Var\left(X \right)=\frac{1}{n^2}\cdot n\cdot p \cdot (1-p)=\frac{p \cdot (1-p)}{n} $