# CDF of minimum of N random variables.

1. Nov 11, 2016

### ashwinnarayan

There's this problem that I've been trying to solve. I know the solution for it now but my initial attempt at a solution was wrong and I can't seem to figure out the mistake with my reasoning. I'd appreciate some help with figuring this one out.

1. The problem statement, all variables and given/known data
I have a set of random variables drawn independently from a distribution. And a new random variable.

$$Z = min\{X_1, X_2, ... X_N\}$$.

Each $X_i$ has the pdf $f_X(x)$ and CDF $F_X(x)$

What I want to do is to find the CDF (and then the PDF) of Z.

3. The attempt at a solution
So here's what I tried first.

$$P(Z<z) = P((\exists i\ s.t\ X_i < z) \cap (X_j > z\ \forall j \neq i))$$
$$P(Z<z) = \left(\sum_{i=1}^{N}P(X_i < z)\right) \left( \sum_{j=1, j\neq i}^{N}P(X_j < z) \right)$$
$$P(Z<z) = N(N-1)F_X(z)(1-F_X(z))$$

But I know this is wrong because I did some research and I know that the correct (and easier) way to do it is to find $P(Z > z)$. The actual answer is $1 - (1 - F_X(z))^N$.

Can someone help me find the flaw in my reasoning?

2. Nov 12, 2016

### Ray Vickson

You are claiming that $Z < z$ if and only if exactly one of the $X_i$ is $< z$ while all of the others are $> z$. This claim is false: $\min\{3,4,5 \} < 10$ but none of 3,4 or 5 is > 10. Also, $\min \{3,4,5 \} < 4.5$ but only one of the entries exceeds 4.5.

Also: be careful of inequalities. The usual definition of CDF is $P(Z \leq z)$, with a non-strict inequality. Some authors (very few) write the CDF as $P(Z < z)$, but in that case the complementary probability is NOT $P(Z > z)$, but rather, $P(Z \geq z)$. Of course, it makes no difference when you are dealing with continuous random variables having densities (as you seem to be), but if you want to deal with discrete, or mixed continuous-discrete random variables, then you must be very careful. The easiest way to be careful is to learn some rigid rules right from the start of your studies.

Last edited: Nov 12, 2016