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.
 A: Solution using automated computer algebra systems:
Let $(X_1, ..., X_5)$ have joint pdf $f(x_1,..., x_5)$:

Then:

... where I am using the Prob function from the mathStatica package for Mathematica.
Always nice to check work and existence of exact solutions.
A: 
How can I calculate the probability that one random variable is bigger than a second one?


*

*You could integrate over the joint distribution in the area where the condition $X>Y$ is true.

*You could derive the distribution for the variable $X-Y$ and use it to compute $P(X-Y>0)$.

Below is an example of the first option

You variables $X = min(X_1,X_2,X_3)$ and $Y = max(X_4,X_5)$ are independent and follow the beta distribution with pdf's (without loss of generality I am scaling from [0,2] to [0,1])
$$\begin{array}{rcl} 
f_X(x) &=& 3(1-x)^2 \\ 
f_Y(y) &=& 2y
\end{array}$$
I've plotted a randomly generated sample of this in the image below. What you want to know is the probability that $X>Y$ and this corresponds to a point being on the bottom of the diagonal line $X=Y$.
You can find this probability by integrating the probability density of the points below that diagonal line.
$$\begin{array}{rcl}
P(X > Y) &=& \int_0^1 \int_0^x 6 (1-x)^2 y \, dy dx \\&=& \int_0^1 6(1-x)^2 \left[ \int_0^x   y \, dy \right] \, dx \\
&=& \int_0^1  3 (1-x)^2 x^2 \, dx \\
 &=& 3 \int_0^1 x^2 - 2x^3 + x^4 \, dx \\
 &=& 3 (\frac{1}{3} - \frac{2}{4} + \frac{1}{5}) \\
 & =& \frac{1}{10}
\end{array}$$


I believe that its is slightly more intuitive and easy than doing the quintuple integral. However, when your question would have been like $X = min(X_1,X_2,X_3)$ and $Y = max(X_3,X_4)$ then the variables are not independent and it may not be so easy to write down the joint pdf.
A: 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.
A: For all input values $0 \leq t \leq 2$ we have,
\begin{align*}
\mathbb P (\min(X_1,X_2,X_3) > t ) &= \mathbb P(X_1>t)^3 \\[12pt]
&= \Big (1 - \frac{t}{2} \Big )^3 \\[12pt]
\mathbb P (\max(X_4,X_5) \leq t ) &= \mathbb P(X_4\leq t)^2 \\[12pt]
&= \frac{t^2}{4}, \\[6pt]
\end{align*}
which also gives the corresponding density,
\begin{align*}
\quad f_{\max(X_4,X_5)}(t) &= \frac{t}{2}. \\[6pt]
\end{align*}
Thus, using the substitution $r = t/2$, we have,
\begin{align*}
\mathbb P (\min(X_1,X_2,X_3) > \max(X_4,X_5 )) 
&= \int \limits_0^2 \mathbb P (\min(X_1,X_2,X_3) > t ) \frac{t}{2} \ dt \\[6pt]
&=\int \limits_0^2 \Big (1 - \frac{t}{2} \Big )^3 \frac{t}{2} \ dt \\[6pt]
&= 2 \int \limits_0^1 (1-r)^3 r \ dr \\[6pt]
&= 2 \int \limits_0^1 (r - 3r^2 + 3r^3 - r^4) \ dr \\[6pt]
&= 2 \Bigg[ \frac{r^2}{2} - r^3 + \frac{3r^4}{4} - \frac{r^5}{5} \Bigg ]_{r=0}^{r=1} \\[12pt]
&= 2 \Bigg[ \frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5} \Bigg ] \\[12pt]
&= 0.1 \\[6pt]
\end{align*}
