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$x$ and $y$ have joint pmf $p(X=x,Y=y)=\frac{1}{3x};\ \ y=1,2,\dots,x ;\ \ x=1,2,3$. Then what is the conditional expectation $E(Y \vert X=3)$?

I have found $E(X=3)$.

Then I am confused on what I should do,

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    $\begingroup$ Your expression for the pmf is ambiguous. Do you mean $\frac{1}{3x}$ or $\frac13 x$? Can you clarify what you mean by your second paragraph? $\endgroup$ – Glen_b -Reinstate Monica Jan 1 '16 at 10:10
  • $\begingroup$ @Glen_b Presumably $\frac1{3x}$, since that pmf is normalized and $\frac13 x$ is not. Tesla, though, you should clarify what you mean by $E(X=3)$ – do you mean $P(X=3)$? Have you tried plugging in the definition of conditional expectation and going from there? $\endgroup$ – Dougal Jan 1 '16 at 10:19
  • $\begingroup$ @Glen_b I have found answer . question was right and Adam is wonderful . $\endgroup$ – Tesla Jan 1 '16 at 11:00
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    $\begingroup$ The conditions, in words, assert that whatever value $X$ might have, $Y$ has equal probabilities of being any of the values $1$ through $X$. The question focuses on $X=3$. How, then, would you go about finding the expectation of a random variable that has equal chances of being $1$, $2$, or $3$? $\endgroup$ – whuber Jan 1 '16 at 15:21
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    $\begingroup$ The incomprehensible $\mathbb{E}(X=3$ is still in place. $\endgroup$ – Xi'an Jan 1 '16 at 20:58
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Assuming the probability you stated $P(X = x, Y = y) = \frac{1}{3x}$ for $y = 1, \dots, x$ and $x=1, 2, 3$, we can compute this probabilities and marginal distributions as follows:

\begin{array}{|c||c|c|c||c|} \hline \frac{x_i}{y_i} & 1 & 2 & 3 & P_Y(y_i) \\ \hline \hline 1 & \frac{1}{3} & \frac{1}{6} & \frac{1}{9} & \frac{11}{18} \\ \hline 2 & 0 & \frac{1}{6} & \frac{1}{9} & \frac{5}{18} \\ \hline 3 & 0 & 0 & \frac{1}{9} & \frac{1}{9} \\ \hline \hline P_X(x_i) & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \\ \hline \end{array}

Rows numbers regards to $y_i$ and columns numbers denotes $x_i$.

The marginal distributions are calculated as $$ P_X(x_i) = \sum\limits_{y_j}P(x_i, y_j)$$

$$ P_Y(y_i) = \sum\limits_{x_j}P(x_j, y_i)$$

We can express conditional probabilities $$ P(Y = y_i | X = x_j) = \frac{P(x_j, y_i)}{P_X(x_j)}$$

$$ P(Y = 1 | X = 3) = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}$$

$$ P(Y = 2 | X = 3) = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}$$

$$ P(Y = 3 | X = 3) = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3}$$

$$ \mathbb{E}(Y | X = 3) = 1 \cdot \frac{1}{3} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{1}{3} = 2 $$

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  • $\begingroup$ I have done my job . $\endgroup$ – Tesla Jan 1 '16 at 11:10
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    $\begingroup$ Given that this is a self-study question, providing the complete answer is not the best approach to help the OP understand the issues behind the question and his difficulties. $\endgroup$ – Xi'an Jan 1 '16 at 11:56
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    $\begingroup$ @Xi'an When it comes to the educational value, you're in 100% right, but I assumed he knew well all the theory behind conditional expectation and he was only looking for working example to make sure. Like as in Andrew's Lang quote, he was looking for support rather than for illumination. $\endgroup$ – Adam Przedniczek Jan 1 '16 at 12:29
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    $\begingroup$ @Xi'an I promise that next time I'll follow your guidance and pay more attention to the tags. By the way, from my own experience, if I post question on the forum, that means I have exhausted all other options and I'll appreciate the most accurate explanantion. $\endgroup$ – Adam Przedniczek Jan 1 '16 at 12:37

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