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&lt;X_2) = \frac{\mu_1}{\mu_1 + \mu_2}

 

[Gullik]

On the first glance E(X_1|X_1&lt;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&lt; X_2] = E[X_1] = E[X_1| X_1 &gt; 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&lt; X_2] = E[X_1] = E[X_1| X_1 &gt; 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!