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I want to calculate $E(\text{min}(X_1,X_2,X_3))$ where $X_1\sim\text{exp}(1), \ X_2\sim\text{exp}(2)$ and $X_3\sim\text{exp}(3).$

Denote $M=\text{min}(X_1,X_2,X_3)$. By independence of the $X_i$ we have that

\begin{align} F_M(t)&=\mathbb{P}(M\leq t)=1-\mathbb{P}(M>t)=1-\mathbb{P}(X_1>t)\mathbb{P}(X_2>t)\mathbb{P}(X_3>t)=1-e^{-6t}, \end{align}

thus $f_M(t)=F_M'(t)=6e^{-6t}\implies E(M)=1/6$.

Question:

Why isn't it the case that

\begin{align} F_M(t)&=\mathbb{P}(M\leq t)=\mathbb{P}(X_1\leq t)\mathbb{P}(X_2\leq t)\mathbb{P}(X_3\leq t)=(1-e^{-x})(1-e^{-2x})(1-e^{-3x})\\ &=... ? \end{align}

Why do I have to use the complement rule?

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  • $\begingroup$ Why the wrong title? $\endgroup$
    – Did
    Commented Jan 15, 2019 at 23:03
  • $\begingroup$ @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to? $\endgroup$
    – Parseval
    Commented Jan 15, 2019 at 23:05
  • $\begingroup$ What was your title and what is it now that I have corrected it? Tired or not... $\endgroup$
    – Did
    Commented Jan 15, 2019 at 23:07
  • $\begingroup$ @Did - Ah yes, the reason for that was because this is just a part of an assignment regarding Poisson distributed arrivals and I made a typo. I apologize that my lack of focus at this hour is causing you discontent and wasting your time. I'll better myself and triple-read my future posts before I submit them. $\endgroup$
    – Parseval
    Commented Jan 15, 2019 at 23:31

1 Answer 1

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Because $\Bbb{P}(M\leq t) \neq \Bbb{P}(X_1 \leq t)\Bbb{P}(X_2 \leq t)\Bbb{P}(X_3 \leq t)$. Indeed - if $X_1 \leq t$ but $X_2 > t$ and $X_3 > t$ then we still have $M \leq t$.

However, it is not too hard to see that $\min(X_1,X_2,X_3) > t \iff X_1, X_2, X_3 > t$.

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