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