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Let $X$ be the standard uniform random variable. That is, $X$ has the density $f_X(x) = 1$ for $0 < x < 1$ and $0$ elsewhere. Suppose we toss a fair coin (independently of the value of $X$) and set

$Y = \left\{\begin{matrix} X\text{ if the coin shows "heads",}\\1\text{ if the coin shows "tails".} \end{matrix}\right.$

(a) Calculate Var$(Y^p)$ for any $p > 0$.
(b) Calculate the limit of Var$(Y^p)$ as $p\to\infty$. Can you think of a way to figure out that limit without having to do first the calculation in part (a) ?

Schnokeyboy
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1 Answers1

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Let $W=Y^p$. A reasonable approach to calculating the variance of $W$ is to calculate $E(W^2)-(E(W))^2$.

So first let us calculate $E(W)$. We use a conditional expectation approach. With probability $\frac{1}{2}$, we have $W=X^p$, and with probability $\frac{1}{2}$ we have $W=1^p=1$.

If $W=X^p$, then $E(W)=\int_0^1 x^p \,dx=\frac{1}{p+1}$. And if $W=1$ then $E(W)=1$. It follows that $$E(W)=\frac{1}{2}\cdot\frac{1}{p+1}+\frac{1}{2}\cdot 1=\frac{p+2}{2(p+1)}.$$ More formally, let random variable $H$ be equal to $1$ if the coin toss results in a head, and $0$ otherwise. Then $$E(W)=E(W|H=1)\Pr(H=1)+E(W|H=0)\Pr(H=0).$$

For $E(W^2)$, the calculation is very similar. Now calculating $E(W^2)-(E(W))^2$ is routine.

For the intuition about the limit, note that if $p$ is very large, and we tossed a head, then $Y^p$ is very likely to be close to $0$.

André Nicolas
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