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Let $Y_1, Y_2, Y_3$ be independent exponentially distributed random variables, with parameters $\lambda_1, \lambda_2, \lambda_3$ respectively. Why is it the case that:

$P(Y_1=min(Y_1,Y_2,Y_3))=\frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3}$?

I just came across this fact but I'm not sure where it comes from. I do know that if we define $Y=min(Y_1,Y_2),$ then $Y$ is an exponential random variable with parameter $\lambda_1+\lambda_2$. Is this fact related to the one above?

Thanks

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2 Answers 2

up vote 0 down vote accepted

Not sure what the intuition is, but it's a straightforward integral to work out. $$ \begin{aligned}\int_0^\infty \int_{0}^\infty& \int_{0}^\infty [x_1 \le x_2] [x_1 \le x_3] \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\ &= \int_0^\infty \int_{x_1}^\infty \int_{x_1}^\infty \lambda_1 \lambda_2 \lambda_3 e^{-\lambda_1 x_1 -\lambda_2 x_2-\lambda_3 x_3}dx_3 dx_2 dx_1\\ &=\int_{0}^\infty \lambda_1 e^{-(\lambda_1 +\lambda_2 +\lambda_3) x_1}dx_1 \\&=\frac{\lambda_1}{\lambda_1 +\lambda_2 +\lambda_3}\end{aligned}$$ (This is using the fact that expectation of the indicator function of an event equals the probability of that event.)

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This may have been what you meant by your statement in the (), but can you say one more word about how why that integral should work? That is, how did you come up with it? –  user66360 Nov 11 '13 at 23:37
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The event $x_1=\min(x_1,x_2,x_3)$ is equivalent to $x_1 \le x_2$ and $x_1 \le x_3$. So the probability of that event is given by integrating the pdf over that region. –  p.s. Nov 11 '13 at 23:46

This yields the same result as @p.s. provided, but this is how I'd think to solve it.

Let $f_{Y_1}(t)$ denote the density of $Y_1$, and note that $$\{ Y_1 = min(Y_1,Y_2,Y_3) \} \iff \{ Y_2 \geq Y_1, Y_3 \geq Y_1\}.$$ Then, by independence, \begin{align*} P(Y_1 = min(Y_1,Y_2,Y_3)) &= \int_0^{\infty} P(Y_2 \geq Y_1, Y_3 \geq Y_1| Y_1 = t)f_{Y_1}(t)dt \\ &= \int_0^{\infty} P(Y_2 \geq t) P(Y_3 \geq t)f_{Y_1}(t)dt \\ &= \int_0^{\infty} e^{-\lambda_2t}e^{-\lambda_3t}\lambda_1e^{-\lambda_1t}dt\\ &= \int_0^{\infty} \lambda_1 e^{-(\lambda_1+\lambda_2+\lambda_3)t}dt \\ &= \frac{\lambda_1}{\lambda_1+\lambda_2+\lambda_3} \end{align*}

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