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Let $X_1 \sim \text{exp} \left( {\lambda}_1 \right)$ & $X_2 \sim \text{exp} \left( {\lambda}_2 \right)$, and they are independent.

Now consider the random variable $Y = \min \left[X_1, X_2 \right]$.

I need to estimate the $\mathbb{P} \left[Y = X_1 \right]$.

I can calculate the random variable $Y \sim \text{exp} \left( {\lambda}_1 + {\lambda}_2 \right)$

But how can I calculate above probability? Initially I thought that probability is zero given that we are having continuous distribution, but that assertion does not seem to be correct.

I am looking for a general framework for this type of problem.

Glen_b
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Brian Smith
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1 Answers1

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One way to consider this problem is as the exponential distribution being the distance between events in a Poisson point process.

The $\lambda_1$ and $\lambda_2$ relate to the rates at which two processes are occuring, and the distance untill the next event is related to the sum of those processes. That's how you can see the reason that you get $\lambda_1 + \lambda_2$ for the distance untill the next event from either process $1$ or process $2$.

The probability for $Y = \min(X_1,X_2) = X_1$ is the probability that an event is from process $1$. That probability is related to the frequency of the different processes. Given that there is an event, the probability that it is from process 1 is $\frac{\lambda_1}{\lambda_1+\lambda_2}$.

Sextus Empiricus
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