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Let (X; Y ) have joint mass function $P(k, n) = \frac {C*2^-k}{n}$ , for k = 1, 2, and n = 1, 2, , k, and suitable constant C. Compute $E(X|Y = y)$.

Its easy to calculate the $P(X,Y)$ but i am getting confused in calculating $P(Y)$ also what should be the value(range) of X. Can someone please provide some direction?

Raveesh
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1 Answers1

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The conditional distribution of $X$ is $$p_{X|Y=y}(x) = \frac{p_{X,Y}(x,y)}{p_Y(y)}=\frac{P((X,y)=(x,y))}{P(Y=y)}.$$ You know the numerator already. The denominator can be found by partitioning the event $Y=y$ on the value of $X$: $$P(Y=y)=\sum_{k=1}^2 P(Y=y \text{ and } X=x).$$ Finally, $E(X \mid Y=y)= \sum_{k=1}^2 x \cdot p_{X|Y=y}(x)$. The sum is over $k=1,2$ because according to the formula you gave, $X$ only takes on the values $1$ or $2$.

kccu
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  • Thanks so much, i think i missed it in my question but K takes value from 1 to infinity but n takes value from 1 to K. I think the summation value will be from k=1 to infinity(denominator). Is that correct and how do i compute it in this case? – Raveesh Nov 10 '16 at 16:12
  • can you please check my comment? – Raveesh Nov 10 '16 at 16:24
  • Yes, both sums will be from $k=1$ to $\infty$. – kccu Nov 11 '16 at 19:51