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

Let $X$ and $Y$ be two independent random variables, who's supports are $[0,\infty]$. We can express $\mathbb{P}[X<Y]$ as:

$$\mathbb{P}[X < Y] = \int_{y=0}^{\infty}\int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy.$$

Can we find a similar expression for:

$$\mathbb{P}[X < Y|Y<k],\; \mathrm{given}\;k\in[0,\infty)?$$

share|cite|improve this question
Without any further information about random variables $X$ and $Y$ this question can not be answered. – Sasha May 19 '13 at 14:47
@Sasha , can the solution not be given in terms of pdf's / cdf's and integrals? I'm assuming X and Y are independent and belong to the same support. – David Simmons May 19 '13 at 14:53
No, you need much more information about the distributions involved. What does $k$ have to do with anything? – dfeuer May 19 '13 at 15:07
k belongs to the support – David Simmons May 19 '13 at 15:08
What extra information do you need? $X$ and $Y$ are independent, the support of $X$ and $Y$ are the same, $k$ is a constant that also belongs to the support. – David Simmons May 19 '13 at 15:13
up vote 1 down vote accepted

You can find an expression for the desired conditional probability. Assume for simplicity that $X$ and $Y$ have respectively density functions $f_X(x)$ and $f_Y(y)$. Then $$\Pr(X\lt Y|Y\lt k)=\frac{\Pr((X\lt Y)\cap (Y\lt k)}{\Pr(Y\lt k))}.$$ Both numerator and denominator can be expressed as integrals. For the numerator, we want $\int_{y=0}^k\int_{x=0}^y f_X(x)f_Y(y)\,dx\,dy$. For the denominator, it is much the same, except that $x$ goes from $0$ to $\infty$.

Remarks: $1.$ The independence does not play a large role here, apart from (in concrete cases) making the integrations easier. For joint density functions $f_{X,Y}(x,y)$ the expression for the conditional probability is $$\frac{\int_{-\infty}^k\int_{-\infty}^y f_{X,Y}\,dx\,dy}{\int_{-\infty}^k\int_{-\infty}^\infty f_{X,Y}\,dx\,dy}.$$

$2.$ Note that as pointed out by @Jon Claus, the denominator is just the probability that $Y\lt k$, so it can be expressed in the simple form $\int_0^k f_Y(y)\,dy$.

share|cite|improve this answer
I assume the result is a consequence of Bayes' theorem, I wasn't sure whether we could apply the theorem in this case. Thank you very much. – David Simmons May 19 '13 at 16:44
I guess we can call it Bayes' Theorem. I think of it as the definition of conditional probability. – André Nicolas May 19 '13 at 16:46
For the denominator, you need not include an integral for $X$. After all, you can factor the double integral using Fubini's theorem and use the fact that the integral across the reals of $f_X(x)$ is one. – Jon Claus May 19 '13 at 17:05
@JonClaus: True. I wanted a generic setup, where $f_X(x)f_Y(y)$ can be replaced by $f_{X,Y}(x,y)$. But thanks, I will add a remark about the "general" case. – André Nicolas May 19 '13 at 17:09
Thank you again, this is all very useful. – David Simmons May 19 '13 at 17:26

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.