# Probability minimum is “reached”

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.

$$P(X_i = Y) = 1 - P(X_i \ne Y) = 1 - (P(X_i < Y) + P(X_i > Y)) = 1 - P(X_i > Y) =$$
$$1 - P(X_i > \text{min}_{j = 1, \ldots, n}X_j) = 1 - P(X_i > \text{min}_{j \ne i}X_j)$$
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 $$P(X_i = Y) = 1 - \tilde{\lambda_i} / \lambda = (\lambda - \tilde{\lambda_i}) / \lambda = \lambda_i / \lambda$$.
• @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) – dcolazin Mar 17 at 7:56