4
$\begingroup$

I have random variables $X_0,X_1,…,X_n$ that are IID, and are normally distributed with mean $\mu$ and variance $\sigma$. I want to find the distribution of the minimum value, in the set $X_i$, let's call it Y.

I'm aware of the answer as listed here : How is the minimum of a set of random variables distributed?

My problem is that for a normally distributed variable $X$, its cdf is given by:

$\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^{2}/2}dt $

which has no analytical solution. Furthermore, even if I did have the answer to an approximation to the integral, I'm not sure how to calculate the distribution of $Y$.

$\endgroup$
4
  • $\begingroup$ (1) Your expression for $\Phi(x)$ is just a complicated way to write the number $1$. You probably meant something different. (2) You calculate $y$ by scanning through the $X_i$, comparing the current minimum to the current value, and always retaining the smallest. When the scan is over you have found the smallest. If that's not what you were looking for, please edit your question to clarify it. $\endgroup$
    – whuber
    Aug 1 '16 at 14:39
  • 1
    $\begingroup$ @AnthonyW i've made some edits to the question that i hope clarify what you're trying to ask. If they do not reflect your actual questions, feel free to roll them back $\endgroup$
    – jld
    Aug 1 '16 at 14:52
  • $\begingroup$ @Chaconne. Thank you. This makes sense to me now. I have a large set of RVs that are dependent on an underlying RV that is normally distributed. Hence I can find the minimum by as suggested above, but it takes a long time. Therefore, I wanted to determine the minimum through recourse to the moments of an underlying distribution. $\endgroup$
    – Anthony W
    Aug 1 '16 at 15:37
  • $\begingroup$ you might find the paper Moments of Order Statistics from a Normal Population to be of interest: jstor.org/stable/2333540 $\endgroup$
    – jld
    Aug 1 '16 at 18:17
6
$\begingroup$

I'll present the standard normal case. Let $Z_1, ..., Z_n \sim \ \text{iid} \ \mathcal N(0,1)$. Then $P(Z_{(1)} \leq z) = 1 - P(Z_{(1)} \geq z) = 1 - P(Z_1 \geq z, ..., Z_n \geq z) = 1 - (1 - \Phi(z))^n$.

We can differentiate with respect to $z$ to find the pdf: $$ f_{(1)}(z) = \frac{d}{dz} \left(1 - (1 - \Phi(z))^n\right) = n(1-\Phi(z))^{n-1}\varphi(z) $$

This is the distribution of the minimum value of your set of $n$ iid standard normal random variables. No integration required (unless you want to actually evaluate it).

Update: limiting case

Suppose we let $n \rightarrow \infty$. $$ \lim_{n \rightarrow \infty} P(Z_{(1)} \leq z) = \lim_{n \rightarrow \infty} 1 - (1 - \Phi(z))^n. $$

Because $\forall z \in \mathbb R$ $\Phi(z) \in (0,1)$, we have that $$ \lim_{n \rightarrow \infty} P(Z_{(1)} \leq z) = 1 - 0 = 1 $$ for an arbitrary $z \in \mathbb R$. This means that if we have a large enough collection of iid standard normal RVs, eventually we'll get one that's less than any particular real number $z$. This makes sense since the support of $\mathcal N(0,1)$ is all of $\mathbb R$.

$\endgroup$
2
  • $\begingroup$ Thanks. So if I take the expectation of this, for a given value of Z I will effectively obtain a confidence level that the Pr(min Z <z) = 0.95 (say)? Also, what happens for large n - does the answer tend to a limit? $\endgroup$
    – Anthony W
    Aug 1 '16 at 14:36
  • 1
    $\begingroup$ @AnthonyW i've added a section about the $n \rightarrow \infty$ case $\endgroup$
    – jld
    Aug 1 '16 at 14:46

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