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Regarding the following problem:

A fair coin is tossed 5 times. Let $X$ be the number of heads in all $5$ tosses, and $Y$ be the number of heads in the first $4$ tosses. What is $\operatorname{Cov}(X,Y)$?

My attempt:

I know that I should calculate the following: $$ \operatorname{Cov}(X,Y) = E[XY]-E[X]E[Y]$$

Well, the right flank is pretty easy: $E[X]=2.5, E[Y]=2 $

But what about $E[XY]$?

I searched online and saw that:

$$ E[XY]=\sum_{x}\sum_{y}xyP(X=x, Y=y) $$

But any attempt to imply that in the problem only led me for further more confusion.

Thanks in advance.

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    $\begingroup$ Use the fact that $X=Y + Z$ where $Z$ is independent of $Y$ and $\mathbb{P}(Z=1)=\mathbb{P}(Z=0)=1/2$, this is because $X$ and $Y$ can differ only due to the last toss $\endgroup$ – Hayk Jan 22 at 16:37
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Here's a way to do it without messy arithmetic. Let $T_i$ be the result of the $i$'th toss ($0$ if tails, $1$ if heads). These are independent, with $X = T_1 + \ldots + T_5$ and $Y = T_1 + \ldots + T_4$. Then $$\text{Cov}(X,Y) = \text{Cov}(Y+T_5, Y) = \text{Cov}(Y,Y) + \text{Cov}(T_5,Y) = \text{Cov}(Y,Y) = \text{Var}(Y)$$ Now since $T_1, \ldots, T_4$ are independent, $$\text{Var}(Y) = \text{Var}(T_1) + \ldots + \text{Var}(T_4) = 4 \text{Var}(T_1)$$ and $\text{Var}(T_1)$ is easy to find...

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  • $\begingroup$ why does $Cov(T5,Y)=0$ ? thanks! $\endgroup$ – superuser123 Jan 22 at 17:00
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    $\begingroup$ Because they are independent. $\endgroup$ – Robert Israel Jan 22 at 17:10
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$X$ can assume the values $0,1,2,3,4,5$ and $Y$ can assume the values $0,1,2,3,4$. One of the terms in the sum, for example, is $$5\cdot 4\cdot P(X=5,Y=4)=5\cdot 4\cdot \frac{1}{2^{5}} $$

The rest of the terms are computed similarly. Note that you'll want to use $$P(X=x,Y=y)=P(Y=y)\cdot P(X=x|Y=y)$$ A helpful observation: $P(X=x,Y=y)=0$ if $x<y$ (since we cannot have more heads in the first $4$ tosses than in all $5$), so many of the terms in the sum are $0$.

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  • $\begingroup$ If I sum it up it turns out to be $ \frac{70}{32}$ which is incorrect, what did I miss here? $\endgroup$ – superuser123 Jan 22 at 17:27
  • $\begingroup$ That is also what I'm getting for the sum. What makes you say it's incorrect? $\endgroup$ – pwerth Jan 22 at 17:30
  • $\begingroup$ $Cov(X,Y)=1$ according to the answers, and it doesn't come up like that, unless I'm missing something $\endgroup$ – superuser123 Jan 22 at 17:44
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    $\begingroup$ My apologies, the sum is not equal to $\frac{70}{32}$. I believe you assumed each outcome to have the same probability of $\frac{1}{2^{5}}$ which is not correct. This is true for the example I gave in my answer because there is only $1$ outcome of the possible $2^{5}=32$ in which all $5$ flips are heads. But there are multiple outcomes where, say, we have $X=2,Y=1$, since the first "heads" can appear at any of the first $4$ flips. $\endgroup$ – pwerth Jan 22 at 17:52
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    $\begingroup$ The sum should work out to $\frac{192}{32}=6$, which means the covariance is indeed $6-(2)(2.5)=1$ $\endgroup$ – pwerth Jan 22 at 17:54

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