0
$\begingroup$

We flip a coin n times and calculate the number of heads. for the obtained number of heads of the previous n tosses we will toss the coin again. what is the expected value of heads in this tossing process?

$\endgroup$
2
  • 1
    $\begingroup$ What have you tried? Where have you gotten stuck? $\endgroup$ – WaveX Oct 6 '18 at 6:16
  • $\begingroup$ How we can calculate this expected value after tossing for the obtained number of heads of the previous n tosses? $\endgroup$ – lighting Oct 6 '18 at 6:21
1
$\begingroup$

Let $H_1$ be the number of head in the first $n$ tosses, and $H_2$ be the number of head in the second. We are looking to compute $E(T)$, where $T=H_1 + H_2$. Note that $E (T)=E(H_1)+E(H_2)$. $E(H_1)$ is simply $n*0.5 = n/2$. To compute $H_2$, we will use the law of total expectation, \begin{align} E(H_2) = E(E(H_2|H_1)) &=\sum_{i=0}^n E(H_2|H_1=i)P(H_1=i) \\ &= \sum_{i=0}^n i/2 \binom{n}{i}(0.5)^n \\ &= (0.5)^{n+1}n2^{n-1}\\ &= n/4 \end{align} Thus, $E(T)=n/2+n/4 = 3n/4$

$\endgroup$
1
  • 1
    $\begingroup$ +1 More concise: $\mathbb E[\mathbb E[H_2\mid H_1]]=\mathbb E\frac12H_1=\frac14n$ $\endgroup$ – drhab Oct 6 '18 at 8:03

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.