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Bob has $n$ coins, each of which falls heads with the probability $p$. In the first round Bob tosses all coins, in the second round Bob tosses only those coins which fell heads in the first round. Let $R_i$ the number of coins which fell heads in the round $i$.

  1. What is the distribution law for $R_2$?
  2. Find $Corr(R_1,R_2)$
  3. How does correlation behave when $p→0$ and $p→1$? Why?
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Looks like a nice homework problem, and if so, please add the homework tag. Also, what progress have you made on the problem? Where are you stuck? –  Dilip Sarwate Oct 31 '12 at 15:01
    
Ready to post an answer, as soon as @Dilip's suggestions are followed. –  Did Oct 31 '12 at 15:37
    
I have problems with understanding of how to build distribution law –  Xxx Oct 31 '12 at 21:32
    
Are you kidding? You added NOTHING to your post to comply with @Dilip's suggestion. –  Did Oct 31 '12 at 21:59
    
@did and now? it's a part of my answer –  Xxx Nov 5 '12 at 16:25

2 Answers 2

up vote 2 down vote accepted

Each coin falls head in the second round if and only it fell head in the first round (hence was tossed again) and fell head in the second round as well. Thus each coin falls head in the second round with probability $p^2$, and the number $R_2$ of these has binomial $(n,p^2)$ distribution.

Since $R_1$ is binomial $(n,p)$, $\mathbb E(R_1)=np$, $\mathrm{var}(R_1)=np(1-p)$ and $\mathbb E(R_1^2)=\mathrm{var}(R_1)+\mathbb E(R_1)^2=np(1-p)+n^2p^2$. Since $R_2$ is binomial $(n,p^2)$, $\mathbb E(R_2)=np^2$ and $\mathrm{var}(R_2)=np^2(1-p^2)$. For every $k$, conditionally on $R_1=k$, $R_2$ is binomial $(k,p)$, hence $\mathbb E(R_2\mid R_1)=R_1p$, and $$ \mathbb E(R_1R_2)=\mathbb E(R_1\mathbb E(R_2\mid R_1))=p\mathbb E(R_1^2). $$ Thus, $$ \text{Cov}(R_1,R_2)=\mathbb E(R_1R_2)-\mathbb E(R_1)\mathbb E(R_2)=np^2(1-p), $$ and $$ \text{Corr}(R_1,R_2)=\frac{\text{Cov}(R_1,R_2)}{\sqrt{\text{var}(R_1)\text{var}(R_2)}}=\sqrt{\frac{p}{1+p}}. $$

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you have sum with k, but k is the number of tails in 1st round. why is it so? –  Xxx Nov 5 '12 at 16:58
    
Sorry, I had mixed up heads and tails in the first and second rounds. The revised version should be OK. –  Did Nov 5 '12 at 18:31
    
can you please say how we get that formula for expectation of the product? –  Xxx Nov 5 '12 at 18:58
    
This exchange will become more and more difficult if you continue to say NOTHING about what it is exactly you do not understand, and why, and what you do understand and know about the subject. At present, I do not know the kind of answer you seek. –  Did Nov 5 '12 at 19:07
    
E(R1R2)=E(R1E(R2∣R1))=pE(R21). this thing I don't understand –  Xxx Nov 5 '12 at 19:34

$P(R_2=i)=\sum_{j=i}^{n}{C(n,j)}{C(j,i)}p^{j+i}(1-p)^{n-i}$

Is that right? I can't understand how to find an expectation of $R_2$.

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