I'm trying to follow the solution of this exercise: link.

It says: Let $X$ and $Y$ be two jointly continuous random variables with joint PDF

$ \begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} 2 & \quad y+x \leq 1, x>0, y>0 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation} $

To find $E[X]$, I would expect it to be:

$E[X] = \int_{0}^{1-y}{2dx}$

And have $E[X]$ be a function of $y$.

But the solution actualy integrates the function over $y$ to get a function of $x$ and then integrate again to get $E[X]$:

$ \begin{align}%\label{} \nonumber f_X(x)&=\int_{-\infty}^{\infty} f_{XY}(x,y)dy \\ \nonumber &=\int_{0}^{1-x}2dy\\ \nonumber &=2(1-x). \end{align} $

$ \begin{align}%\label{} \nonumber E[X]&=\int_{0}^{1}2x(1-x)dx\\ \nonumber &=\frac{1}{3}=EY, \end{align} $

What is the idea behind integrating a bivariate $PDF_{x,y}$ before integrating to calculate the $E[X]$.

  • $\begingroup$ Replacing an earlier, now deleted, comment ... $E[X] := \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\, x f_{XY}(x, y)\, dx \, dy.$ With the integration over $y$ first this reduces to $\int_{-\infty}^{\infty}\, x f_{X}(x)\, dx$ where $f_X(x) := \int_{-\infty}^{\infty}\, f_{XY}(x, y)\, dy.$ $\endgroup$ Jan 6 at 1:41

1 Answer 1


The expectation of $X$ is a number here, which we compute by first finding the marginal distribution of $X$ (hence the integration over the joint density for all possible values of $X$). From the probability distribution of a random variable...we can compute its expectation (if it exists).

I don't see any reason it would be a function of $Y$.


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