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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.

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  • $\begingroup$ 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

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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 $\mathbb P\{u_1\lt U_k\}$.

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  • $\begingroup$ 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

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