# Distribution of $min_{i\in[1,2,3,…,n}X_i$ if $X_i$'s are exponentially distributed and independent

Doing some recreational probability problems. Came across this set and got stuck:

If $X$ is exponentially distributed with rate $r$, then the p.d.f. of $X$ is given by: $$p(x) = re^{−rx}, x ∈ [0, ∞)$$ (a) Compute the characteristic function of $X$.

Suppose $X_i ∼ Exp(r_i)$ for $i = 1, ..., n$ and that the $X_i$ are independent.

(b) Find the distribution of $min_{i∈[1,...,n]} X_i$

the first part is trivially $\dfrac{r}{r+ix}$. However, I don't even know where to start on the second part.

The question itself seems a little vague. I don't think that it means to just list the exponential distribution with the smallest rate parameter $r$.

If anyone has any idea about how to go about answering this question, please let me know. It seems like an interesting problem!

EDIT: doing some simulations, the distribution is (as you might expect) definitely exponentially distributed with a smaller rate parameter. Here is some code and a graph showing that:

#! /usr/bin/python
from scipy.stats import expon
from numpy.random import exponential

rs = np.logspace(0,1,10)

def exponential_sim(rs):
return min([exponential(_) for _ in rs])

x = np.linspace(0,10,100)
simulated_data = [exponential_sim(rs) for _ in range(1000)]
loc,scale = expon.fit(simulated_data,floc=0)
print "fit scale ={0}".format(scale)
plt.figure()
for r in rs:
rv = expon(loc=0,scale=r)
plt.plot(x, rv.pdf(x), lw=2, label='{0}'.format(r))

rv = expon(loc=0,scale=scale)
plt.plot(x, rv.pdf(x), 'o-',lw=2, label='min_scale = {0}'.format(scale),color='r')

plt.legend()
plt.show()


• Note that if you have $n$ independent random variables $X_i$, $\min_i X_i > x$ is equivalent to $X_1 > x \cap X_2 > x \cap \dots \cap X_n > x$, which is easy to handle if you know the CDF of each $X_i$, using the independence. (Really you are using the "complementary CDF" i.e .$1-F$, but whatever.) And the CDF for the exponential distribution is quite simple. – Ian Mar 15 '17 at 22:11
• Ah, okay, that makes things easy. Thanks Ian – lstbl Mar 15 '17 at 22:13