# Minimum of two exponential variates: What's wrong with this derivation?

Suppose we have $\newcommand{\E}{\mathrm{Exp}} X \sim \E(\lambda)$, $Y \sim \E(\mu)$, and $W = \min(X,Y)$.

I know that $W \sim \E(\lambda+\mu)$. I know how to derive it. But, I tried this alternate derivation that gave me a different distribution for $W$, and I still can't figure out what's wrong with it.

I started with


Now I need $\Pr(X < Y)$. Seems straightforward:

\begin{align} \Pr(X < Y) &= \int_0^\infty [1-F_X(t)] f_Y(t) \rd t \\ &= \int_0^\infty e^{-\lambda t} \mu e^{-\mu t} \rd t \\ &= \frac{\mu}{-(\lambda + \mu)} \left[ e^{-(\lambda + \mu)t} \right]^{\infty}_0\\ &= \frac{\mu}{\mu+\lambda} \>. \end{align}

And, if I do the same thing for $\Pr(Y < X)$, I get $\Pr(Y < X) = \frac{\lambda}{\mu + \lambda}$.

So $\Pr(X < Y)$ and $\Pr(Y < X)$ sum up to 1 as expected. Encouraging.

And now I substitute that into my original equation:

\begin{align} f_W(t) &= \frac{\mu}{\mu + \lambda} \lambda e^{- \lambda t} + \frac{\lambda}{\mu + \lambda} \mu e^{- \mu t} \\ &= \frac{\lambda\mu}{\lambda + \mu}\left(e^{-\lambda t} + e^{-\mu t}\right) \>. \end{align}

That... That's no exponential.

Where did I go wrong?

• The distribution of $X$ conditional on the fact that it is less than $Y$ is not the same as the distribution of $X$. Apr 7 '12 at 15:41

The distribution of $X$ conditional on the fact that it is less than $Y$ is not the same as the distribution of $X$. This is probably easiest understood if $X$ and $Y$ are identical distributions. Then your mixture $$f_W(t)=f_X(t)P(X<Y)+f_Y(t)P(Y\le X)=f_X(t),$$ and this is obviously not true.