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I have two exponentially distributed numbers (with rate 3). I need to calculate E(min) given that one number is ≥ x and one number is ≤ x.

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Let $X$ and $Y$ be independent exponential random variables each with rate $\mu$. $$\begin{multline} \mathbb{E}(\min(X,Y) | (\min(X,Y) \leqslant x, \max(X,Y) \geqslant x)) = \\ \mathbb{E}(X | (X \leqslant x, Y \geqslant x)) \mathbb{P}(X \leqslant Y)+ \mathbb{E}(Y | (Y \leqslant x, X \geqslant x)) \mathbb{P}(Y \leqslant X) \end{multline} $$ Due to symmetry $\mathbb{P}(Y \leqslant X) = \mathbb{P}(X \leqslant Y) = \frac{1}{2}$, and $\mathbb{E}(X | (X \leqslant x, Y \geqslant x)) = \mathbb{E}(Y | (Y \leqslant x, X \geqslant x))$, hence we need to compute $$ \mathbb{E}(X | (X \leqslant x, Y \geqslant x)) \stackrel{\text{independence}}{=} \mathbb{E}(X | X \leqslant x) = \frac{\mathbb{E}(X \cdot [ X \leqslant x])}{\mathbb{P}(X \leqslant x)} =\\ \frac{\int_0^x \mathrm{e}^{-\mu z} z \mathrm{d} z}{1-\mathrm{e}^{-\mu x}} = \frac{1}{\mu} \cdot \frac{1- \mathrm{e}^{-\mu x} (1 + x \mu)}{1-\mathrm{e}^{-\mu x}} $$

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