# Give correct answer when have two random variables, X and Y. X is the

Give correct answer when have two random variables, X and Y. X is the value of a fair die, Y is the result of a coin flip, with heads being 1 and tails being 0.
$E\left[X\right]=\sum _{k=1}^{6}\frac{k}{6}=\frac{7}{2},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}E\left[Y\right]=\frac{1}{2}$. Thus
$E\left[X\right]E\left[Y\right]=\frac{7}{4}$.
That the expectation of XY is not multiplicative, i.e.: E[X]E[Y] is not necessarily equal to E[XY]. But I'm confused about what E[XY] means in the first place. That is, is E[XY] each possible value of the two events combined, multiplied by the probability that the two events occur?
That is, is $E\left[XY\right]=\frac{1}{2}\left(1\right)\sum _{k=1}^{6}\frac{k}{6}+\frac{1}{2}\left(0\right)\sum _{k=1}^{6}\frac{k}{6}=\frac{7}{4}?$ If not, what is it?
Due to a typo the last equation $\left[XY\right]=\frac{1}{2}\left(1\right)\sum _{k=1}^{6}\frac{k}{6}+\frac{1}{2}\left(0\right)\sum _{k=1}^{6}\frac{k}{6}$ was evaluated as . I'd appreciate it if responders told me whether this is a correct value for E[XY].
That concerned with how to calculate E[XY] in the quickest way possible, but how to interpret what E[XY] means. E[X]E[Y]=E[XY] for independent events doesn't concern me as much as why that is the case, and how to manually evaluate E[XY] in order to prove that indeed E[XY]=E[X]E[Y].

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accimaroyalde
We should separate the meaning of E(XY) from whatever devices we use to compute the expectation.
What does XY mean? It is a random variable. In our particular example, imagine the following game. We roll a fair die and a fair coin. If the die shows k and the coin shows Head, we get (k)(1) dollars. If the die shows k and the coin shows Tail, we get (k)(0) dollars, that is, 0.
The random variable XY is the amount of money we get. It takes on values 0,1,2,3,…,6, with various probabilities, it has a certain distribution. Then E(XY) is the expected value (mean) of XY, in the usual sense.
In this case, $Pr\left(XY=0\right)=\frac{1}{2},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Pr\left(XY=n\right)=\frac{1}{12}f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}n=1,2,\dots ,6$. Now we can compute E(XY). We have
$E\left(XY\right)=\frac{1}{2}\left(0\right)+\frac{1}{12}\left(1\right)+\frac{1}{12}\left(2\right)+\cdots +\frac{1}{12}\left(6\right)$.
But in this case, there is a shortcut. Since X and Y are independent, we have E(XY)=E(X)E(Y), and therefore there is no need to find the distribution of XY.

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Janet Young
XY denotes a new random variable. You could call it Z if you want, and work out its distribution. Then E[XY] just means E[Z].
So in your case, since X takes the discrete uniform distribution on {1,2,3,4,5,6} and Y takes the discrete uniform distribution on {0,1}, the product XY takes values in {0,1,2,3,4,5,6}. XY does NOT have a uniform distribution, though. See if you can work out its distribution directly.
Hint: $P\left(XY=0\right)=P\left(X=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}Y=0\right)=P\left(Y=0\right)=\frac{1}{2}$
And $P\left(XY=1\right)=P\left(X=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y=1\right)=P\left(X=1\right)\cdot P\left(Y=1\right)=\frac{1}{6}\cdot \frac{1}{2}=\frac{1}{12}$
And so on. (I've assumed X and Y are independent.)

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