I am given a joint distribution of two random variables, X and Y, which is as follows:

$$ f(x,y)= 8xy \\ 0≤y≤x≤1 $$

Now I need to find :

$$ P[X\leq 1/2] $$

Now I now I need to find marginal pdf first but I am confused between the limits. Will the limits be from $0$ to $1/2$, since we need to substitute the limits of y, if not, then what is the trick behind the selection of limits if we are given of this type?


1 Answer 1


That the region common to the support $\{(x,y):0\le y\le x\le 1\}$ of the joint distribution and the restriction $x<1/2$ is $\{(x,y):0\le y\le x\le 1/2\}$ should be clear if you draw a picture.

Then it follows that

\begin{align} P\left(X\le \frac{1}{2}\right)&=E\,[\mathbf1_{X\le1/2}] \\&=\iint \mathbf1_{x\le1/2}\,f(x,y)\,\mathrm{d}x\,\mathrm{d}y \\&=\iint \mathbf1_{x\le 1/2}\,8xy\,\mathbf1_{0\le y\le x\le 1}\,\mathrm{d}x\,\mathrm{d}y \\&=8\iint xy\,\mathbf1_{0\le y\le x\le 1/2}\,\mathrm{d}x\,\mathrm{d}y \end{align}

You are supposed to write the double integral as an iterated integral using Fubini's theorem, and depending on the order of integration, you have

$$\iint xy\,\mathbf1_{0\le y\le x\le 1/2}\,\mathrm{d}x\,\mathrm{d}y=\int_0^{1/2} y\left(\int_y^{1/2} x\,\mathrm{d}x\right)\mathrm{d}y=\int_0^{1/2} x\left(\int_0^x y\,\mathrm{d}y\right)\mathrm{d}x$$

Since the ranges of $x$ and $y$ are dependent, the inner integral is a function of the variable with respect to which you are evaluating the outer integral, while the outer integral runs free of the other variable, giving you a real value as the final answer.

  • $\begingroup$ then the limits will be $0$ to $1/2$ for outer integral and $y$ to $1/2$ for inner? $\endgroup$ Jan 11, 2019 at 17:09
  • $\begingroup$ @AhmadQayyum Yes. $\endgroup$ Jan 11, 2019 at 18:01

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