How can I calculate the probability that one random variable is bigger than a second one?

I have five random variables which are independent and each one of them has a continuous uniform distribution on the interval $ [0,2]:$
$$X_i = \operatorname{Uniform}[0,2].$$

I want to calculate the probability $$\Pr(\min(X_1, X_2, X_3)\gt \max(X_4, X_5)).$$

I’m aware there is combinatorial solution, but I’m trying to solve this problem using coordinates with $X$ as the minimum and $Y$ as the maximum, but I don’t know how to sketch the function and calculate the function space in order to know the probability.

Answer

To be brutally mindless about it, we may begin with the full five-dimensional integral and then proceed to evaluate it. Because this is carried out over a region in $\mathbb{R}^5,$ I will not attempt to sketch it :-).

As a simplification of the notation (and to reveal the ideas), let the joint density of $(X_1,X_2,X_3)$ be $f_{123} $ and the joint density of $(X_4,X_5)$ be $f_{45}.$ Then, with $P = \Pr(\min(X_1,X_2,X_3) \gt \max(X_4,X_5)),$

$$P = \iint f_{45}(x_4,x_5) \iiint_{\max(x_4,x_5)} f_{123}(x_1,x_2,x_3)\,\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3\ \mathrm{d}x_4\mathrm{d}x_5.$$

The first (double) integral extends over all $\mathbb{R}^2$ while the second (triple) integral extends only over those points $(x_1,x_2,x_3)$ in $\mathbb{R}^3$ where all three coordinates exceed both $x_4$ and $x_5.$

It is usually easiest to deal with a maximum in an integral’s endpoint by breaking the integral into parts: almost surely either $X_4$ or $X_5$ will be the larger of those two and these two events (namely, $\mathcal{E}_4:X_4=\max(X_4,X_5)$ and $\mathcal{E}_4:X_5=\max(X_4,X_5)$) are mutually exclusive. Therefore we may compute the probabilities of these two events and add them.

Because $X_4$ and $X_5$ are iid, they are exchangeable, implying $\mathcal{E}_4$ and $\mathcal{E}_5$ have the same probability. Consequently, taking the case $X_4\gt X_5$ (event $\mathcal{E}_4$), we obtain

$$P = 2\int\int_{x_5} f_{45}(x_4,x_5) \iiint_{x_4} f_{123}(x_1,x_2,x_3)\,\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3\ \mathrm{d}x_4\mathrm{d}x_5.$$

Specializing now to iid uniform$[0,1]$ variables we may compute this integral using the most elementary techniques as

$$\begin{aligned}
P &= 2\int_0^1\int_{x_5}^1\int_{x_4}^1\int_{x_4}^1\int_{x_4}^1\,\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3\mathrm{d}x_4\mathrm{d}x_5 \\
&= 2\int_0^1\int_{x_5}^1 \left(\int_{x_4}^1\mathrm{d}x_1\right)\left(\int_{x_4}^1\mathrm{d}x_2\right)\left(\int_{x_4}^1\mathrm{d}x_3\right)\,\mathrm{d}x_4\mathrm{d}x_5 \\
&= 2\int_0^1\int_{x_5}^1(1-x_4)^3\mathrm{d}x_4\mathrm{d}x_5 \\
&= 2\int_0^1 \frac{1}{4}(1-x_5)^4\,\mathrm{d}x_5 \\
&= 2\left(\frac{1}{4}\right)\left(\frac{1}{5}\right) = \frac{1}{10}.
\end{aligned}$$

This gives the answer for any continuous iid variables with common density $f$ because the Probability Integral Transform

$$u(x) = \int^x f(t)\,\mathrm{d}t,$$

converts the variables $(X_1,\ldots, X_5)$ into variables $U_i = u(X_i)$ that are iid with a Uniform$[0,1]$ distribution without changing the order statistics, thereby leading to the calculation of $P$ that was just performed.

Attribution
Source : Link , Question Author : Ben , Answer Author : whuber

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