Let's say I 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.
$E[X] = \sum_{k=1}^{6}{\frac{k}{6}} = \frac{7}{2}$, and $E[Y] = \frac{1}{2}$. Thus $E[X]E[Y] = \frac{7}{4}$.
I'm aware 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 $E[XY] = \frac{1}{2}(1)\sum_{k=1}^{6}{\frac{k}{6}} + \frac{1}{2}(0)\sum_{k=1}^{6}{\frac{k}{6}} = \frac{7}{4}$? If not, what is it?
Edit 1: Typo
Due to a typo the last equation $E[XY] = \frac{1}{2}(1)\sum_{k=1}^{6}{\frac{k}{6}} + \frac{1}{2}(0)\sum_{k=1}^{6}{\frac{k}{6}}$ was evaluated as $\frac{7}{2}$, when I believe it should be $\frac{7}{4}$. I'd appreciate it if responders told me whether this is a correct value for $E[XY]$.
Edit 2: Clarification
I'm not 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 $E[XY] = E[X]E[Y]$.
$\endgroup$ 52 Answers
$\begingroup$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 $0,1,2,3,\dots,6$, 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, $\Pr(XY=0)=\frac{1}{2}$, and $\Pr(XY=n)=\frac{1}{12}$ for $n=1,2,\dots, 6$. Now we can compute $E(XY)$. We have $$E(XY)=\frac{1}{2}(0)+\frac{1}{12}(1)+\frac{1}{12}(2)+\cdots+\frac{1}{12}(6).$$
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 $XY$.
$\endgroup$ 6 $\begingroup$$XY$ denotes a new random variable. You could call it $Z$ if you want, and work out its distribution. Then $E[XY]$ just means $E[Z]$.
So in your case, since $X$ takes the discrete uniform distribution on $\left\{1, 2, 3, 4, 5, 6\right\}$ and $Y$ takes the discrete uniform distribution on $\left\{0, 1\right\}$, the product $XY$ takes values in $\left\{0, 1, 2, 3, 4, 5, 6\right\}$. $XY$ does NOT have a uniform distribution, though. See if you can work out its distribution directly.
Hint: $$P(XY=0)=P(X=0\,\text{or}\,Y=0)=P(Y=0)=\frac{1}{2}$$ And $$P(XY=1)=P(X=1\,\text{and}\,Y=1)=P(X=1)\cdot P(Y=1)=\frac{1}{6}\cdot\frac{1}{2}=\frac{1}{12}$$ And so on. (I've assumed $X$ and $Y$ are independent.)
$\endgroup$
