In a computer program I am writing, I have several independent variables, each of which is uniformly distributed between two real values. I seek the probability of each variable being the maximum of the group.

That is, let $X_1, X_2, X_3,\ldots, X_n$ be my random variables. I seek a probability function $P$ such that $P(X_m)$ is the probability that $X_m = \max\{X_1, X_2, X_3,\ldots, X_n\}$. How would I find such a function given the ranges of each random variable? I don't need the actual maximum value - rather, I only need the probabilities that each variable produces that maximum value.

Since this is for a computer program, I would like to have a solution which can be easily algebraically calculated for an arbitrary number of variables, each with arbitrary ranges. However, if this is not feasible, I would naturally accept a calculus based program.

  • $\begingroup$ For n=2 you can have this approach, perhaps extensible. Scale the range with higher upper bound to $1$, let the other range be $a$ with overlap $x$. The probability of the first rv is greater than the second can be written as $1-x + \frac{x^2}{2a}$ $\endgroup$ – karakfa Mar 25 '13 at 21:04

As I understand it, the interval of each $X_i$ is a parameter of your algorithm.

Let $X_1,X_2...,X_n$ be the random variables that survived step 1. Let $F_1,F_2...,F_n$ be their cdf(cumulative distribution function) and let $f_1,f_2...,f_n$ be their pdf(probability density function). Let $[r_i,\:s_i]$ be the interval of $X_i$. Let $r_k$ be the biggest of all $r_i$.

I suggest a reduction phase. Remove all random variables $X_i$ that have $s_i<r_k$. Those cannot be the maximum.

After this step of reducing the problem to a subset of the random variables, we can start looking at the probability aspect of this.

Now the probability that the following holds $X_m = \max\{X_1, X_2, X_3,\ldots, X_n\}$ amounts to:

$$\int_{r_k}^{s_m}\: f_m(x)\prod_{i=1,i\neq m}^n F_i(x) \; dx$$

$$=\frac{1}{s_m-r_m}\int_{r_k}^{s_m}\: \prod_{i=1,i\neq m}^n F_i(x) \; dx$$

For uniform distributed random variables we know their cdf, namely:

$$F_i(x)=\left\{ \begin{array}{ll} 0 & x<r_i \\ \frac{x-r_i}{s_i-r_i} & r_i \leq x < s_i \\ 1 & x \geq s_i \end{array} \right.$$

Or equivalently: $$F_i(x)=\left\{ \begin{array}{ll} 0 & x<r_i \\ \frac{1}{s_i-r_i}(\min(x,s_i)-r_i) & r_i \leq x \end{array} \right.$$

Using the fact that we choose $r_k$ to be the biggest of all lower bounds of the intervals, we only have case 2 for our cdf. This is the equation you get :

$$\prod_{i=1}^n{\frac{1}{s_i-r_i}}\int_{r_k}^{s_m}\: \prod_{i=1,i\neq m}^n (\min(x,s_i)-r_i) \; dx$$

Now some the knowledge of the interval can be used. Let the upper bounds of the intervals $s_i$ be increasing or equal if $i$ increases, this can always be done by changing the index of each random variable. Now cutting the integration area in parts we get:

$$\int_{r_k}^{s_m}\: \prod_{i=1,i\neq m}^n (\min(x,s_i)-r_i) \; dx=$$ $$\int_{r_k}^{s_1}\: \prod_{i=1,i\neq m}^n (x-r_i) \; dx + \sum_{j=1}^{m-1}\int_{s_j}^{s_{j+1}}\: \prod_{i=j+1,i\neq m}^n (x-r_i) \; dx=$$

Now you have all integrals of polynomials of degree at most n, this can be computed.

I reckon this won't have a lovely runtime in the number of variables left after the reduction, because a random variable makes the degree of all polynomials one higher in the worst case.

I hope this helps.


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