Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I just want to clarify where did the $1/4$ come from?

Let $X$ denote the diameter of an armored electric cable and $Y$ denote the diameter of the ceramic mold that makes the cable. Both $X$ and $Y$ are scaled so that they range between $0$ and $1$. Suppose that $X$ and $Y$ have the joint density

$$f(x, y) =\begin{cases} \frac1y,&0<x<y<1\\\\ 0,&\text{elsewhere} \end{cases}$$


$$\begin{align*} &P\left(X+Y>\frac12\right)=1-P\left(X+Y<\frac12\right)=1-\int_0^{1/4}\int_x^{1/2-x}\frac1y dy\,dx\\ &=\left. 1-\int_0^{1/4}\left[\ln\left(\frac12-x\right)-\ln x\right]dx=1+\left[\left(\frac12-x\right)\ln\left(\frac12-x\right)-x\ln x\right]\right\vert_0^{1/4}\\ &=1+\frac14\ln\left(\frac14\right)=0.6534. \end{align*}$$

share|cite|improve this question
up vote 1 down vote accepted

Since $0\le X,Y\le 1$, the region on which $X+Y<\frac12$ is the the triangle bounded by the axes and the line $x+y=\frac12$. To integrate over this region, you’d normally set up the integral like this:


However, in this case you know that $f(x,y)=0$ when $y\le x$, so you can ignore any part of the triangle lying below the diagonal $y=x$. Thus, the region over which you need to integrate is actually the triangle bounded by $x=0$, $y=x$, and $x+y=\frac12$, which is shaped roughly like this: $\triangleright$. Everywhere else, $f(x,y)=0$. The righthand vertex of that triangle is at the point $\left\langle\frac14,\frac14\right\rangle$, so you need only run $x$ from $0$ to $1/4$.

share|cite|improve this answer
@David: Yes; I was concentrating so much on the $1/4$ aspect that I forgot to change if. Thanks. – Brian M. Scott Feb 17 '12 at 4:41

As pointed out by Brian M. Scott, the region over which we need to integrate is the triangle with corners $(0,0)$, $(\frac{1}{4},\frac{1}{4} )$, and $(0,\frac{1}{2})$. I would like to point out a slightly different approach to the double integral that perhaps makes the integration easier.

You integrated first with respect to $y$, and then with respect to $x$. That has the disadvantage that you first get something that involves a couple of logarithms, which you then must integrate again.

Let us explore the alternative of integrating first with respect to $x$. Then the first integration will be trivial, since $1/y$ can be treated as a constant.

The downside (look at the picture!) is that for $0 \le y \le \frac{1}{4}$, $x$ goes from $0$ to $y$, while for $\frac{1}{4}\le y\le \frac{1}{2}$, $x$ goes from $0$ to $\frac{1}{2}-y$. So we will have to evaluate two integrals. We now do this, to show it is not hard.

First we evaluate $$\int_{y=0}^{\frac{1}{4}}\left(\int_{x=0}^y\frac{dx}{y}\right)dy.$$ Very easily, the inner integral is $1$, so our integral is $\dfrac{1}{4}$. Next we evaluate $$\int_{y=\frac{1}{4}}^{\frac{1}{2}}\left(\int_{x=0}^{\frac{1}{2}-y}\frac{dx}{y}\right)dy.$$ The inner integral is $\dfrac{1}{2y}-1$, so our integral is $(1/2)(\ln(1/2)-\ln(1/4)) -(1/2-1/4)$. This simplifies to $(1/2)\ln 2-1/4$. Add the two parts. We get $(1/2)\ln 2$. Finally, as in your calculation, the answer to the original question is $1-(1/2)\ln 2$.

Remark: That was easy, but in fact one can do better, by making the change of variable $w=x+y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.