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x and y are joinly discrete with joint probability mass function (pmf)

$$ \begin{array}{|c|c|c|} \hline x/y & 0 & 1 \\ \hline 0& 0.2 & 0.1 \\ \hline 1 & 0.4 & 0.3 \\ \hline \end{array} $$

(a) Find $E(XY)$

Answer:

(a) $E(XY) = \sum xy P_X(x) = $

Not sure how to go about this

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The table gives you the probabilities $P(X=x, Y=y)$ for various values of $x$ and $y$. The formula of expectation you should be using here is $E[XY] = \sum_x \sum_y xy P(X=x, Y=y)$.

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There are four terms in $E(XY) = \sum_{x=0}^1\sum_{y=0}^1 xy\,f_{X,Y}(x.y),$ only one of the non-zero, so $E(XY) = 1(0.3) = 0.3.$

Alternatively, $V = XY$ has PDF $P(V=0)=0.7,\,P(V=1) = 0.3,$ so $E(V) = E(XY)= 0.3.$

Maybe the next step is to find $Cov(X,Y) = E(XY) - E(X)E(Y) = 0.3 - (0.7)(0.4) = 0.02.$

Suppose the joint PDF were given by the following table.

$$ \begin{array}{|c|c|c|} \hline x/y & 0 & 1 \\ \hline 0& 0.18 & 0.12 \\ \hline 1 & 0.42 & 0.28 \\ \hline \end{array} $$

Then $X$ and $Y$ would be independent and $Cov(X,Y) = 0.$

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