Let [itex] X_1, X_2 \sim Exp(\mu) [/itex] and [itex]Y = min(X_1, X_2) [/itex] then find [itex]E[Y].[/itex](adsbygoogle = window.adsbygoogle || []).push({});

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 [itex]1/(2\mu)[/itex]. I don't understand why the above method doesn't work ?

I used the following property of exponential distribution.

[itex] P(X_1<X_2) = \frac{\mu_1}{\mu_1 + \mu_2}[/itex]

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# Minimum of two iid exponential distributions

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