14
$\begingroup$

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.

$\endgroup$
1
15
$\begingroup$

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.

$\endgroup$
12
$\begingroup$

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*}

$\endgroup$
3
  • $\begingroup$ thank you but as i said i know the combinatorical solution. I'm trying to solve this using coordinates system $\endgroup$
    – Ben
    Sep 16 at 13:45
  • 1
    $\begingroup$ @Ben Ok sorry, I edited my answer $\endgroup$
    – periwinkle
    Sep 16 at 14:16
  • $\begingroup$ @periwinkle: Nice answer (+1). I have taken the liberty of editing to make your answer a bit prettier (lining up some equations, moving density into equations, adding more paragraph space between equations, etc.) and I have added a couple more steps in the final integral to assist readers. Please feel free to revert or edit if you do not like what I've changed. $\endgroup$
    – Ben
    Oct 1 at 22:28
4
$\begingroup$

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

enter image description here

Then:

enter image description here

... where I am using the Prob function from the mathStatica package for Mathematica.

Always nice to check work and existence of exact solutions.

$\endgroup$
3
  • 2
    $\begingroup$ Because the question refers to the combinatorial argument (which is simple and clear, yielding the answer $1/\binom{5}{2}$), a good confirmation of the answer is already in place. Indeed, had your software obtained any other answer I would blame it on a bug rather than trusting the software! $\endgroup$
    – whuber
    Sep 16 at 18:39
  • 2
    $\begingroup$ Agreed - the attraction for me is how the parsing of the CAS solution can so closely and naturally match the question. $\endgroup$
    – wolfies
    Sep 17 at 17:07
  • 1
    $\begingroup$ Yes, that is a really nice feature of the software. $\endgroup$
    – whuber
    Sep 17 at 17:25
4
$\begingroup$

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.

$$\begin{array}{rcl} P(X > Y) &=& \int_0^1 \int_0^x 6 (1-x)^2 y \, dy dx \\&=& \int_0^1 6(1-x)^2 \left[ \int_0^x y \, dy \right] \, dx \\ &=& \int_0^1 3 (1-x)^2 x^2 \, dx \\ &=& 3 \int_0^1 x^2 - 2x^3 + x^4 \, dx \\ &=& 3 (\frac{1}{3} - \frac{2}{4} + \frac{1}{5}) \\ & =& \frac{1}{10} \end{array}$$

example


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.