A coin $C_1$ has probability $p$ of turning up head and a coin $C_2$ has probability $2p$ of turning up head. All we know is that $0 < p < 1/2$. You are given $20$ tosses. You can choose all tosses from $C_1$ or all tosses from $C_2$ or some tosses from each (the total is $20$). If the objective is to estimate $p$, what do you do?
The following is the given solution.
If we choose to have all $n = 20$ tosses from $C_1$, then we get $X_1, \ldots , X_n$ that are i.i.d. $\mathrm{Ber}(p)$. An estimate for $p$ is $\bar{X_n}$ which is unbiased. Hence, MSE over $\bar{X_n}(p)$ is $V[\bar{X_n}] = p(1 − p)/n$.
On the other hand, if we choose to have all $20$ tosses from $C_2$, then we get $Y_1 , \ldots , Y_n$ that are i.i.d. $\mathrm{Ber}(2p)$. The estimate for $p$ is now $\bar{Y_n}/2$ which is also unbiased and has MSE over $\bar{Y_n}/2(p)$ that is $V[\bar{Y_n}] = 2p(1 − 2p)/4 = p(1 − 2p)/2$.
It is not hard to see that for all $p < 1/2$, MSE of $Y_n(p)/2$ is smaller than MSE of $X_n(p)$ and hence choosing $C_2$ is better, at least by mean-squared criterion.
I understood how MSE for $C_1$ is calculated but I couldn't understand how variance of $C_2$ is calculated. Why is it $2p(1-2p)/4$? where does that division by $4$ come from? Can someone help me understand this? Please elaborate the answer as I'm very new to statistics.
With some corrections