Quick question here. I'm trying to make sense of this problem...

$$f(x_1,x_2) =\begin{cases} 8x_1x_2 & \text{for } 0 < x_1 < x_2 < 1\\ 0&\text{otherwise} \end{cases}$$

If $Y = X_1(X_2)^3$ , what is the expected value of $Y$?

My thoughts:

I'm not entirely sure where to start, but here is where I would start: \begin{align} E(Y) &= E\left(X_1(X_2)^3\right)\\ & = E(X_1) E(X_2^2) \end{align}

Then after this step I'm not sure of what to do. There must be a way to use the pdf to solve for the expected value but I'm not sure. I'm not looking for answers but guidance would be greatly appreciated!!

  • 13
    $\begingroup$ Joint probability density functions do not have expected values; random variables do. A very useful result called the law of the unconscious statistician says that if $Y = g(X)$, then the expected value of $Y$ can be found from the distribution of $X$ via $$E[Y]=\int_{-\infty}^\infty g(x)f_X(x)\,\mathrm dx,$$ that is, it is not necessary to find the distribution of $Y$ first. This applies more generally too. If $X$ and $Y$ are random variables and $Z = g(X,Y)$, then $$E[Z]=\int_{-\infty}^\infty\int_{-\infty}^\infty g(x,y))f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy.$$ See? That was $EZ$. $\endgroup$ Mar 27, 2013 at 21:18
  • $\begingroup$ Thanks! This is excellent! A question however arose. The limits of integration would not be (- infinity to infinity) every time right? So in the particular case, the limits for the integral for both x and y would be from 0 to 1 since it is in the pdf right? Thanks again! $\endgroup$
    – nicefella
    Mar 27, 2013 at 21:46
  • $\begingroup$ The limits are always $\infty$ to $\infty$ but (in the one variable case) if $f_X(x) = 0$ for $x < 0$ and $x>3$, say, then $$\begin{align}E[X]&=\int_{-\infty}^\infty xf_X(x)\,\mathrm dx\\&=\int_{-\infty}^0 xf_X(x)\,\mathrm dx+\int_{0}^3 xf_X(x)\,\mathrm dx+\int_{3}^\infty xf_X(x)\,\mathrm dx\\&=\int_{-\infty}^0 0\,\mathrm dx+\int_{0}^3 xf_X(x)\,\mathrm dx+\int_{3}^\infty 0\,\mathrm dx\\&=\int_{0}^3 xf_X(x)\,\mathrm dx\end{align}$$ $\endgroup$ Mar 27, 2013 at 21:51
  • $\begingroup$ Excellent! Thank you so much! :) $\endgroup$
    – nicefella
    Mar 27, 2013 at 22:32
  • $\begingroup$ Nice one Dilip Sarwate $\endgroup$ Feb 28, 2017 at 17:48

1 Answer 1


The proposed start will not work: $X_1$ and $X_2^3$ are not independent.

I would suggest first making a name change, $X$ for $X_1$, $Y$ for $X_2$, and $W$ for $XY^3$. You need to calculate the expectation $E(W)$ of the random variable $W$.

Call the joint density $8xy$ over the region with $0\lt x\lt y\lt 1$.

Now draw a picture (this was the whole purpose of the name changes). The region where the density function is $8xy$ is the part of the square with corners $(0,0)$, $(0,1)$, $(1,1)$, and $(0,1)$ which is above the line $y=x$. (The density is $0$ everywhere else.)

The region where the density is $8xy$ is a triangle. Call it $T$. Then $$E(W)=E(XY^3)=\iint_{T} (xy^3)(8xy)\,dx\,dy.$$

It remains to calculate the integral. This should not be hard. Express as an iterated integral. Things will be a little simpler if you first integrate with respect to $x$.


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