Calculating Variance to the power of a variable Let X be  standard uniform random variable. That is, X has the density $f_x(x) = 1$ for 0 < x < 1 and 0 elsewhere. Suppose that we toss a fair coin (independently of the value of X) and set $Y = X$ if the coin shows heads, and $Y = 1$ if the coin shows tails. 
Calculate $Var(Y^p)$ for any p>0
Calculate the limit of $Var(Y^p)$ as p $\to \infty$ Can you think of a way to figure out the limit without having to do the calculation in the first part?
So I'm not exactly sure how to start this problem, as far as I know there is no real rule for variances to take out the power of p factor. I imagine I would need to try something akin to
$Var(Y^p) = E(Y^{2p}) - (E(Y^p))^2 $. However I'm not sure how to calculate $E(Y^p)$. I know $E(Y)$ would simply be $E(Y) = (\frac12)(\frac12) + (\frac12)(1) $ As there is a 1/2 chance of heads and if it is heads the distribution $f_x$ has an expected value 1/2, similarly if it is tails it has 1/2 chance of being 1. Would $E(Y^p)=(\frac12)(\frac12)^p + (\frac12)(1)$ make any sense?
For part (b) the limit is simply 0 or infinity depending whether the variance is a fraction or greater than 1. I'm not sure what another method would be.
 A: Your approach is fine but you have to modify $\mathbb E(X^p)$ which is calculated as follows:
$$
\mathbb E(X^p)=\int_0^1 x^pdx=\frac{1}{p+1}
$$
Therefore:
$$
\mathbb E(Y^p)=\frac{1}{2}+\frac{1}{2}\frac{1}{p+1}\\
\mathbb E(Y^{2p})=\frac{1}{2}+\frac{1}{2}\frac{1}{2p+1}
$$
and finally
$$
\mathbb{Var}(Y^p)=\frac{1}{2}+\frac{1}{2}\frac{1}{2p+1}-\left(\frac{1}{2}+\frac{1}{2}\frac{1}{p+1}\right)^2
$$
and we can say:
$$
\lim_{p\to\infty}\mathbb{Var}(Y^p)=\frac{1}{4}.
$$
A: This is called a mixture distribution. Since you have a weighted average of two or more distributions. Namely: $Y=\frac{1}{2}X+\frac{1}{2}1$. The second distribution is called a degenerate one since it is always equal to the same value. Your calculation of $E(Y^p)$ is incorrect because $E(X^p)$ is wrong.  But the idea as a weighted average is right. Write out the full equation for $Var(Y^p)$ after getting $E(Y^{2p})$ and $E(Y^p)$ and take the limit. I think the answer will not be limited to $0$ or infinity as you stated. 
A: For the second part $X^p$ converges towards $0$ as $p$ increases without limit. Meanwhile $1^p=1$.
So $Y^p$ converges towards a Bernoulli random variable taking values $0$ or $1$ with equal probability, and this has mean $\frac12$ and variance $\frac14$. 
