We have $n$ independent variables, $X_1, X_2, \ldots , X_n$ uniformly distributed over the interval $(0,1)$. We then define two new variables, $M = \min(X_1, X_2, \ldots , X_n)$ and $N = \max(X_1, X_2, \ldots , X_n)$.

I want to find the joint distribution of a pair $(M,N)$. I also want to find the CDF for $M$ and the CDF for $N$. What I'm confused about is how these even have distributions. Isn't there one unique value for M and one unique value for $N$?

  • $\begingroup$ Your question is related to mainly related to the Order statistics. See en.wikipedia.org/wiki/Order_statistic For the sample maximum (and minimum), you will need to use the identity $\max\{X_1, X_2, \ldots, X_n\} \leq x \iff X_1 \leq x \text{ and } X_2 \leq x \text{ and } \ldots X_n \leq x$ $\endgroup$
    – BGM
    Jan 25, 2016 at 4:38
  • $\begingroup$ Do I know from the wording of the question that $X_1 = min{X_1, X_2, ... X_n}?$ I'm really struggling with the notation here. $\endgroup$
    – Taylor
    Jan 25, 2016 at 7:04
  • $\begingroup$ Not sure about your question. One usual convention is that $X_1, X_2, \ldots, X_n$ represent a sequence of random variables (random sample) which is i.i.d. and has no specific ordering - the index is just arbitrary. And for ordered statistic, we sort the realization of this random sample accordingly, and denote the sorted result as $X_{(1)} \leq X_{(2)} \leq \ldots \leq X_{(n)}$. $\endgroup$
    – BGM
    Jan 25, 2016 at 8:42

1 Answer 1


$M$ and $N$ are random variables. They take values when the $X_{i}$ are determined but when the $X_{i}$ are unknown then so are $M$ and $N$ and we can find their distributions. You seem happy that $X_{i}$ have distributions, however in any realisation they will take a unique value. What we can do is look at what probability they have of taking the different values, exactly the same for $M$ and $N$.

Using the identity in BGM's comment for $0 < x < 1$, \begin{align} \mathbb{P}[M\le x] &= \mathbb{P}[X_{1} \le x, \dots, X_{n} \le x] \\ &= \mathbb{P}[X_{1} \le x] \cdots \mathbb{P}[X_{n} \le x] \quad\text{[Independence]} \\ &= x^{n}. \quad \text{[As $X_{i} \sim U(0,1)]$} \end{align}

This is the CDF for $M$. In any realisation $M$ will take a particular value, however we can look at the distribution of this value. Using a similar argument you can produce the CDF for $N$ and combine them to form the joint distribution.


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