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Find $\mathbb{E}[XY]$ given the joint probability density function:

$$ f_{X,Y}(x,y)= \begin{cases} 12y^2 & 0 \leq y\leq x\leq 1 \\ 0 & {\rm otherwise} \end{cases} $$

This is straightforward (I think):

$$ {\rm E}(XY) = \int_0^x \int_y^1 xy 12y^2 {\rm d}x{\rm d}y $$

The problem here is that I end up with a variable $(x)$ in the final answer($3x^4/2 - x^6$), rather than simply getting a result. How do I set up the limits of integration for this problem?

2 Answers 2

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Just to expand on Manuel's excellent answer a bit more.

How do I set up the limits of integration for this problem?

The support is $\{(x,y): 0\leq y\leq x\leq 1\}$.

The exterior integral, which we choose to take over the supported distribution of $Y$, must "run over the entire gauntlet": the interval, from the infimum to the supremum, for $Y$.   So we integrate respective to $y$ from $0$ to $1$.

$\def\E{\operatorname{\mathsf E}} \begin{align} \E(XY) & = \E(Y\E(X\mid Y)) \\ & = \int_0^1 y\,\mathsf E(X\mid Y=y)\,f_Y(y)\operatorname d y\end{align}$

The interior integral, which is over the support of $X$ when given a $Y$ value, must have its bounds set relative to that value.   So we integrate respective to $x$ from $y$ to $1$.

$\begin{align}\E(XY) & = \int_0^1\int_y^1 xy\,f_{X\mid Y}(x\mid y)\,f_Y(y)\operatorname d x\operatorname d y \\[1ex] &= \int_0^1\int_y^1 xy\,f_{X,Y}(x,y)\operatorname d x\operatorname d y \\[1ex] &= \int_0^1\int_y^1 12xy^3\operatorname d x\operatorname d y\end{align}$

Of course, there is nothing stopping us from doing it in a different order.   Fubini's Theorem states this should be the same, if the expectation of the absolute value is finite .

$\begin{align}\E(XY) & = \E(\E(XY\mid X)) \\[1ex] &= \int_0^1\int_0^x xy\,f_{X,Y}(x,y)\operatorname d y\operatorname d x \\[1ex] &= \int_0^1\int_0^x 12xy^3\operatorname d y\operatorname d x\end{align}$

Choose whichever path is easier to pass through.

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We do not say the variable has variable-defined support. The support is a fixed set in $\mathbb{R}^2$. It is the set: $$ R = \{(x,y) \in \mathbb{R}^2 \,\mid\, 0 \leq y \leq x \leq 1 \} $$

This random variable has probability density function defined over all points $(x,y) \in \mathbb{R}^2$, but we only care only about those in the support, the set $R$, visualized below:

enter image description here

The expectation of a continuous random variable is the integral of the product of the value of the random and the probability density function, taken over all elements of the sample space. It is the weighted average of the value of the random variable, weighted on how likely values are.

Therefore, the expectation of the random variable $XY$ is: \begin{align} \mathbb{E}[XY] &= \iint_{\mathbb{R}^2} xy f_{X,Y}(x,y) \, dA \\ &= \iint_R xy f_{X,Y}(x,y) \, dA \\ &= \int_0^1 \int_0^x xy (12y^2) \, dy \, dx \\ &= \int_0^1 \int_0^x 12xy^3 \, dy \, dx \\ &= \int_0^1 3x^5 \, dx \\ &= \frac{1}{2} \end{align}

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    This makes sense and aligns with what I did in Mathematica, but I'm not clear on why the limits of integration over $x$ are 0,1 rather than y=x,1. – 2017-02-26
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    I suggest looking over how to take double integrals over a region in $\mathbb{R}^2$. The reason is, we are integrating over this region from $y = 0$ to $y = x$ first, then integrating that function of $x$ from $x = 0$ to $x = 1$. If you wanted to integrate along the $x$-direction first, you would have to integrate from $x = y$ to $x = 1$, and then from $y = 0$ to $y = 1$. It is a matter of looking at the region of integration, and seeing how to integrate in one direction, and integrate that integral in the other direction. Again, I suggest reviewing multivariable calculus. – 2017-02-26