Let's say I have $N$ independent uniformly distributed random variables $U(a_i, b_i)$.

I'm wondering how I'd calculate: $$P(U_1 < U_2 \text{ & } U_1 < U_3 \cdots \text{ & } U_1 < U_N)$$

I'm assuming it is the same as $$1 - P(U_1 > U_2 \text{ or } U_1 > U_3 \cdots \text{ or }U_1 > U_N)$$ but I have no idea how to calculate that too.

When I do the simulations I can easily write these expressions and see that they are correct but I can't figure out the algorithm to calculate the numbers explicitly.

I can easily calculate $P(U_1 < U_i)$ but the joint probability is out of reach.

I've been thinking of just doing all possible permutations of $U_1$ being smaller than every other sequence of variables but it would be too slow.

  • $\begingroup$ As your another question, it is pretty hopeless for general $a_i,b_i$. The best you can get is $ \int_{a_1}^{\min_{i\ge 2} b_i} \prod_{i=2}^n (b_i-x) dx /\left(\prod_{i=1}^n (b_i-a_i)\right).$ $\endgroup$
    – zhoraster
    Apr 11, 2020 at 12:11
  • $\begingroup$ @zhoraster I've been trying to do it by hand for 3, 4 variables but can't seem to figure out the efficient way ($N = 400$). $P(U_1 < U_2)P(U_2 < U_3) + P(U_1 < U_3)P(U_3 < U_2)$ would be for three, but as the number of variable grows there's just too much (exponential) calculation needed. $\endgroup$ Apr 11, 2020 at 12:25
  • $\begingroup$ @zhoraster What also occured to me is that maybe the full range $min(a_i), max(b_i)$ can be partitioned into equal ranges and then calculate the overlapping parts of individual variables as all having equal uniform distribution, that maybe has a closed form efficient solution. $\endgroup$ Apr 11, 2020 at 12:46
  • $\begingroup$ Are the involved random variables independent? $\endgroup$ Apr 14, 2020 at 22:29
  • $\begingroup$ @DavideGiraudo Yep. $\endgroup$ Apr 15, 2020 at 8:32

1 Answer 1


We use the fact that if $(U_1,\dots,U_n)$ is a vector of independent random variables, then for all measurable bounded function $f\colon\mathbb R^n$, $$ \mathbb E\left[f(U_1,\dots,U_n)\right]=\int_{\mathbb R}f(u,U_2,\dots,U_n)d\mathbb P_{U_1}(u) $$ (this is a consequence of the fact that the law of $(U_1,\dots,U_n)$ is the product of the law of $U_1$ with that of $(U_2,\dots,U_n)$ and Fubini's theorem.

Applying this fact to $f\colon (u_1,\dots,u_n)\mapsto \prod_{k=2}^n\mathbf 1\{u_1\lt u_k\}$, we get that the wanted probability $p$ is $$ p=\frac 1{b_1-a_1}\int_{a_1}^{b_1}\mathbb E\left[\prod_{k=2}^n\mathbf 1\{u_1\lt U_k\}\right]du_1 $$ and using independence of the $U_i$, we derive that $$ p=\frac 1{b_1-a_1}\int_{a_1}^{b_1} \prod_{k=2}^n\mathbb P\{u_1\lt U_k\} du_1, $$ which can be further simplified by computing $\mathbb P\{u_1\lt U_k\}$.

  • $\begingroup$ What would you get calculating this with $U_1(1, 5)$, $U_2(1.2, 8)$, $U_3(0, 22)$, $U_4(0, 7)$? I assumed $\mathbb P\{u_1\lt U_k\}$ is 1 or 0 depending on the interval. I've tried splitting the integral into simple sections but can't get the ~0.398. Also, do we need to integrate the product or can it be applied separately? $\endgroup$ Apr 15, 2020 at 13:12
  • $\begingroup$ " I assumed $\mathbb P\{u_1\lt U_k\}$ is 1 or 0 depending on the interval: actually no, it could be any value between 0 and 1 a priori. The point is that we have to look how the intervals $(a_1,b_1)$ and $(a_k,b_k)$ overlap. $\endgroup$ Apr 15, 2020 at 13:48
  • $\begingroup$ I still haven't found the time to compute with the above distributions, it's on my todo list, if everything works, I'll accept the answer. I did try some computation by separating the intervals (calculating inside overlaps, so that I can work with calculating areas of triangles and rectangles) but couldn't get the correct answer, so I'll do it again soon. $\endgroup$ Apr 20, 2020 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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