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.


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

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}.

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.

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

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