# Minimum of two iid exponential distributions

1. Dec 9, 2012

### ait.abd

Let $X_1, X_2 \sim Exp(\mu)$ and $Y = min(X_1, X_2)$ then find $E[Y].$
My attempt is as follows:
$$E[Y] = E[Y/X_1<X_2]P(X_1<X_2) + E[Y/X_1>X_2]P(X_1>X_2) \\ = \frac{1}{2} (E[X_1/X_1<X_2] + E[X_2/X_1>X_2] ) \\ = \frac{1}{2} (1/\mu + 1/\mu) \\ = 1/\mu$$
But, we know that minimum of two exponentially distributed RVs is another exponentially distributed RV with mean $1/(2\mu)$. I don't understand why the above method doesn't work ?
I used the following property of exponential distribution.
$P(X_1<X_2) = \frac{\mu_1}{\mu_1 + \mu_2}$

Last edited: Dec 9, 2012
2. Dec 9, 2012

### Gullik

On the first glance $E(X_1|X_1<X_2)≠E(X_1)=1/\mu$

The instances of $X_1$ where it is smaller than $X_2$ should average lower than an unconditional $X_1$.

3. Dec 9, 2012

### Stephen Tashi

Your method assumes $E[X_1|X_1< X_2] = E[X_1] = E[X_1| X_1 > X_2]$
I don't think that's true. Could you prove it by writing out the integrals involved? The probability densities would be distributions of the "order statistics" of the exponential distribution.

4. Dec 9, 2012

### ait.abd

Got it! Thanks!