A coin (with probability of getting head equal to $p$) is tossed twice. Let $X$ be the total number of heads and $Y$ be the difference between the total number of heads and the total number of tails. Find $\rho(X, Y)$.
Let $X$ be the total number of heads and $Y$ be the difference between the total number of heads and the total number of tails. Find $\rho(X, Y )$
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0Where are you stuck? – 2012-12-15
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0This cannot be: please indicate what you tried. (Or are you flatly using MSE as a site to quickly get fully written solutions to your homework? I hope not.) – 2012-12-15
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0Interesting: you flout spectacularly the rules of the site, you delete a previous comment of yours to which mine answered (better to erase your footsteps?), and you libel as *disturbance* my reminder... Let me note that **none** of your questions, so far, indicates **anything** about what you tried or what you know. Why is that? – 2012-12-15
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0Three comments by the OP now deleted by the OP. – 2012-12-16
1 Answers
If the coin is tossed twice, there are four possible outcomes, corresponding to the following values of $X$ and $Y$. (I take the definition of $Y$ to mean "number of heads (H) minus number of tails (T)")
event probability X Y HH pp 2 2 HT p(1-p) 1 0 TH (1-p)p 1 0 HH (1-p)(1-p) 0 -2
To compute the correlation between $X$ and $Y$ (I assume that's what you mean by $\rho(X,Y)$), we observe that $E X = 2p^2+2p(1-p)=2p$, $E X^2 = 4p^2+2p(1-p)=2p(p+1)$ and thus $$ \operatorname{Var} X = E X^2 - (E X)^2 = 2p(1-p). $$ Similarly, one computes $E Y=2p^2-2(1-p)^2=2(2p-1)$, $E Y^2=4p^2+4(1-p)^2=8p^2-8p+4$, and thus $$ \operatorname{Var} Y = E Y^2 - (E Y)^2 = -8p^2+8p=8p(1-p). $$ One also computes that $EXY=4p^2$, and thus the covariance between $X$ and $Y$ is $$ \operatorname{Cov}(X,Y) =E XY-E X EY = 4p^2 - 16p^2(1-p)^2. $$
Finally, the covariance between $X$ and $Y$ becomes $$ \rho(X,Y)=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var} X \operatorname{Var} Y}} = \frac{p \left(4p(2-p)-3\right)}{1-p}. $$
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0I would have liked to comment to ask Bs11 what he has tried himself (herself), but my reputation doesn't seem to allow this. Hence the full answer. I agree that MSE should not be used to get full answers to homework questions. – 2012-12-15
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0Eckhard This is a friendly advice: when the OP forgets (or refuses) to show what they tried or what they know, one may prefer to provide hints rather than a full solution (I know, this is not always easy). To get the idea, you might wish to consult answers to other questions by same OP. – 2012-12-16