1
$\begingroup$

I've been stuck on this problem for a while; I can get the expected number of heads, but the solution set I saw gives $\frac{11n}{16}$ for the variance, whereas I get $\frac{9n}{16}$:

Each coin flip, $X_i$ for $0,\ldots,n$, can be represented by a Bernoulli random variable with $E(X_i) = p = 0.5$. The total number of heads for the initial round of throws is then a binomial random variable, $N$, with $E(N) = np = \frac{1}{2}n$.

The expected total number of heads for the second set of throws, $N'$ is conditional on $N$, $$ E(N') = E(E(N' \vert N)) = E(N)E(X_i) = \frac{1}{4}n $$

To get the expected total number of heads we sum these two expectations, $$E(N + N') = E(N) + E(N') = \frac{1}{4}n$$

So the expected total number of heads is a function of a random variable, $h(N) = N + E(X_i)N$. Using the fact that $Var(Y) = Var[E(Y \vert X)] + E[Var(Y \vert X)] $

$$ Var(N + E(X_i)N) = Var[E(N + E(X_i)N \vert N)] + E[Var(N + E(X_i)N \vert N)] \\ = Var\left[E\left(\frac{3}{2}N \mid N \right)\right] + E\left[Var\left(\frac{3}{2}N \mid N \right) \right] \\ = \left(\frac{3}{2}\right)^2Var[E(N \mid N)] + \left( \frac{3}{2}\right)^2E[Var(N\mid N)] $$

Factoring out the $\left(\frac{3}{2}\right)^2$, we're left with an expression that is equivalent to $Var(N) = \frac{n}{4}$ (per the equality above), hence $Var(N + N') = \frac{9n}{16}$.

What am I missing?

$\endgroup$
1
$\begingroup$

For $i\in\{1,\dots,n\}$ define $$H_i=\begin{cases} 0\text{ if tails on the }i^{\text{th}}\text{ toss,}\\ 1\text{ if heads on the }i^{\text{th}}\text{ toss, tails on the corresponding second round toss,}\\ 2\text{ if heads on the }i^{\text{th}}\text{ toss, heads on the corresponding second round toss}. \end{cases}$$

Then $$\operatorname{Var}(H_i)=E(H_i^2)-E(H_i)^2=\frac54-\left(\frac34\right)^2=\frac{11}{16}.$$

The total number of heads is $$H=H_1+\cdots+H_n.$$ Since the variables $H_1,\dots,H_n$ are mutually independent, we have $$\operatorname{Var}(H)=\operatorname{Var}(H_1)+\cdots+\operatorname{Var}(H_n)=\frac{11}{16}+\cdots+\frac{11}{16}=\frac{11n}{16}.$$

$\endgroup$
  • $\begingroup$ I like this a lot, but how do you compute the expectations of the $H_i$s? $\endgroup$ – HoHo Sep 6 '17 at 8:45
  • $\begingroup$ $$E(H_i)=0\cdot P(H_i=0)+1\cdot P(H_i=1)+2\cdot P(H_i=2)=0\cdot\frac12+1\cdot\frac14+2\cdot\frac14=\frac34$$ $\endgroup$ – bof Sep 6 '17 at 21:19
  • $\begingroup$ $$E(H_i^2)=0^2\cdot P(H_i=0)+1^2\cdot P(H_i=1)+2^2\cdot P(H_i=2)=0\cdot\frac12+1\cdot\frac14+4\cdot\frac14=\frac54$$ $\endgroup$ – bof Sep 6 '17 at 21:22
0
$\begingroup$

HINT: Note that $N + \mathbb{E}(X_i)N$ is not equal in distribution to $N + N'$. Apply the law of total variance to $N + N'$ directly.

$\endgroup$
0
$\begingroup$

Alright, I found the problem: I was effectively ignoring the effect of the conditionals.

Since $E(N + N' \vert N = x) = E(x + \sum_{i=0}^{x}X_i) = x + xE(X_i)$, $$ E(N + N' \vert N) = N + NE(X_i)$$

Similarly, since $Var(N + N' \vert N = x) = Var(x + \sum_{i=0}^{x}X_i) = xVar(X_i)$ (being independent) $$Var(N + N' \vert N) = NVar(X_i)$$

Putting these two together, \begin{equation*} \begin{split} Var(N + N') &= Var[N + E(X_i)N] + E[N Var(X_i)] \\ &= Var\left(\frac{3}{2}N\right) + E\left(\frac{1}{4}N\right) \\ &= \left(\frac{3}{2}\right)^2\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{1}{2}\right) = \frac{11}{16} \end{split} \end{equation*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.