Minimum of two iid exponential distributions

[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.
$P(X_1<X_2) = \frac{\mu_1}{\mu_1 + \mu_2}$

 

[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 $X_1$.

 

[Stephen Tashi]

$$
= \frac{1}{2} (E[X_1/X_1<X_2] + E[X_2/X_1>X_2] ) \\
= \frac{1}{2} (1/\mu + 1/\mu) \\
$$

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.

 

[ait.abd]

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.

Got it! Thanks!

 

[blue_raver22]

= \frac{\mu}{2\mu} = \frac{1}{2} and P(X_1>X_2) = \frac{\mu_2}{\mu_1 + \mu_2} = \frac{\mu}{2\mu} = \frac{1}{2}
Your method is actually correct and should work. However, there may be a small mistake in your calculation. When finding $E[Y/X_1>X_2]$, you should consider the case where $X_1>X_2$, which means $Y=X_2$. Therefore, $E[Y/X_1>X_2]=E[X_2]=1/\mu$.
Overall, your method should give the correct answer of $1/\mu$.