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 have a complicated thing I would like to find the distribution for.

Let's say I have a random variable $X\sim F_X(x)$ supported over $(0,1)$. I have two independent draws from $F_X$, which are $x_1$ and $x_2$.

Similarly, I have a random variable $Y\sim F_Y(y)$ supported over $(0,1)$. I have two independent draws from $F_Y$, which are $y_1$ and $y_2$.

I am not even sure what words to use to express this properly, but I want to find the distribution for (or any way to express) the "$x$" part of the $Minumum[x_1 + y_1, x_2 +y_2]$. That is probably not clear enough so let me try to explain more:

If $x_1 + y_1 < x_2 +y_2$ I want some way of expressing a distribution for $x_1$ that is more specific than $F_X$ since now we have more information about it. (And if $x_1 + y_1 > x_2 +y_2$ the of course I would want the way to describe $x_2$).

Does this question make sense? And if so, can anyone help me, even if it's just to have better terminology for describing what I'm looking for?

Thanks so much!

share|cite|improve this question
Certainly understandable. I would be more comfortable if capital letters were used. Have not thought it through, but it seems very reasonable to think that the distribution of the $Y_i$ is irrelevant. – André Nicolas May 21 '12 at 2:01
Thanks Andre! I could be wrong, but I think the distribution of Y is relevant. For example, if Y is always the same constant, then y1=y2 and the distribution I am looking for is just the minimum of two draws from $F_X$ , which is $1-(1-x)^2$. But, if Y can vary a lot, then it is possible that the $X_i$ I am looking for is not the minimum of $X_i$ and $X_j$ – Angada May 21 '12 at 3:14
I am not saying it is the minimum, only that it has the same distribution as the minimum. But need to verify, interesting question. – André Nicolas May 21 '12 at 4:30
Andre: what is the guideline for when to use capitalized vs. lowercase variables? I have seen both, but I don't know when to use which. Thanks for your help! – Angada May 21 '12 at 17:23
Habits differ. In teaching probability, I found that students are least confused if they always use caps for random variables. The issue comes up most commonly in random sampling. It is tempting to think of a sample of $10$ as $10$ numbers. But that is very unhelpful if we then ask for the probability that the sample mean differs from the true mean by less than $k$. – André Nicolas May 21 '12 at 17:30
up vote 1 down vote accepted

I might write it this way. Let $T = 1$ when $X_1 + Y_1 < X_2 + Y_2$, $2$ when $X_1 + Y_1 \ge X_2 + Y_2$. You want the distribution of $X_T$.

Now $P(X_T < x|Y_1, Y_2) = P(X_1 < x, X_1 +Y_1 < X_2 + Y_2 )) + P(X_2 < x, X_1 + Y_1 \le X_2 + Y_2)$. Suppose $X_1, X_2, Y_1, Y_2$ are all independent with continuous distributions (so I don't have to distinguish between $<$ and $\le$), and $|Y_2 - Y_1| = V $. There are several different cases, depending on the ordering of $x$, $V$ and $1-V$. For example, if $x < \min(V, 1-V)$, $P(X_T < x | Y_1, Y_2)$ is obtained by integrating $f_X(x_1) f_X(x_2)$ over a region that looks like this:

enter image description here

Then you'll have to integrate the result times $f_Y(y_1) f_Y(y_2)$ over the unit square to get unconditional probability.

share|cite|improve this answer

The distribution of $X_1$ conditional on the event $[X_1+Y_1\lt X_2+Y_2]$ has density $g$ defined by $$ g(x)=\frac1cf_X(x)u(x), $$ with $$ u(x)=\mathrm P(x+Y_1\lt X_2+Y_2), \qquad c=\int_{-\infty}^{+\infty} f_X(z)u(z)\mathrm dz. $$ Since $u$ is nonincreasing, $g$ puts more weight than $f_X$ on the small values.

Note that the distribution of $\min\{X_1,X_2\}$ has density $m$, where $$ m(x)=2f_X(x)v(x),\quad v(x)=\mathrm P(x\lt X_2)=1-F_X(x). $$

share|cite|improve this answer
Thank you. I am not sure how to interpret $c^{-1}$ though. Is it the inverse of the indefinite integral of $f_X(z)u(z)dz$? Then wouldn't I have $z$'s in the $g(x)$ equation? Or, in $g(x)$, do I sub in $f_x(x)u(x)$ as in $c^{-1}(f_x(x)u(x))$? I am sorry I am not understanding. – Angada May 21 '12 at 17:20
Also, is it easier or equivalent to stay in terms of CDFs rather than densities? I know I can change back to CDFs at the end, but since all I want are the CDFs, so I don't need the PDFs unless they are necessary to calculate what I'm looking for. – Angada May 21 '12 at 17:22

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.