0
$\begingroup$

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 $N$?

$\endgroup$
  • $\begingroup$ 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 '16 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 '16 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 '16 at 8:42
1
$\begingroup$

$M$ 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.

$\endgroup$

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.