Finding the conditional distribution.

Asked Oct 22 '16 at 19:30

Active Jun 18 '17 at 15:27

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A penny and dime are tossed. Let $X$ denotes the number of heads up. Then the penny is tossed again. Let $Y$ denotes the number of heads up on the dime(from the first toss) and the penny from the second one. We need to find the conditional distribution of $Y$ given $X=1$ , i.e , $Y|X=1$;

In order to somehow simplify the situation I treated this as a coin tossed three times. Thus the sample space becomes : $\{HHH , HHT,HTH,THH,TTT,TTH,THT,HTT\}$

So the sample space for the event $Y|X=1$ becomes $\{HTH,THH,THT,HTT\}$

$$(Y|X=1) = \begin{cases}\dfrac{1}{4},& y=0\\ \dfrac{1}{2},& y=1\\ \dfrac{1}{4},&y=1\end{cases}$$

Is this correct?

Also $cov[X,Y]$ is also required , according to me it should be zero since the number of heads on the first toss doesn't depend on the number of heads on the second one.

Is the interpretation correct?

edited Jun 18 '17 at 15:27

jvdhooft

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asked Oct 22 '16 at 19:30

User9523

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The covariance is not $0$ since $X=D+P$ and $Y=D+P'$ for independent $(D,P,P')$ Bernoulli random variables, hence Cov$(X,Y)=$Var$(D)=\frac14$. – Did Oct 22 '16 at 20:29
Is the remaining interpretation correct ? @Did – User9523 Oct 23 '16 at 16:52
Dunno, what do you call "the remaining interpretation"? – Did Oct 23 '16 at 16:55
The function $(Y|X=1)$ ? @Did – User9523 Oct 23 '16 at 18:22
This one is simple: the object $$(Y\mid X=1)$$ does not exist hence every solution requiring it is invalid. – Did Oct 23 '16 at 18:31

Finding the conditional distribution.

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