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Let $(X_i)_{i=1,\dots,n}$ be a finite sequence of random variables such that $X_i\sim\mathcal{E}(\lambda_i).$

We can prove that $Y:=\min_{1\le i\le n}X_i\sim\mathcal{E}(\lambda=\sum_{i=1}^n \lambda_i)$

Now I would like to compute $P(X_i=Y).$

We have $$P(\min_{j\ne i}X_j>x)=P(\cap_{j\ne i}\{X_j>x\})=\prod_{j\ne j}e^{-\lambda_j x}=e^{-(\lambda-\lambda_i)x}$$

Now not sure how I can continue.

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Manipulating the probabilities:

$$P(X_i = Y) = 1 - P(X_i \ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$

$$1 - P(X_i > \text{min}_{j = 1, \ldots, n}X_j) = 1 - P(X_i > \text{min}_{j \ne i}X_j)$$

Now, $X_i$ and $Y_i = \text{min}_{j \ne i}X_j$ are two independent exponential random variables with parameters $\lambda_i$ and $\tilde{\lambda}_i = \sum_{i \ne j} \lambda_j$. Then $P(X_i > Y_i) = P(\mathcal{E}(\lambda_i) > \mathcal{E}(\tilde{\lambda}_i)) = \tilde{\lambda_i} / (\lambda_i + \tilde{\lambda}_i) = \tilde{\lambda_i} / \lambda$.

So $P(X_i = Y) = 1 - \tilde{\lambda_i} / \lambda = (\lambda - \tilde{\lambda_i}) / \lambda = \lambda_i / \lambda$.

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  • $\begingroup$ @Julien it was just a fast way to denote $P(Z>W)$, where $Z = \mathcal{E}(z), W = \mathcal{E}(w)$ and are independent (you can find the calculation here) $\endgroup$ – dcolazin Mar 17 at 7:56

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