# Find $E(\text{min}(X_1,X_2,X_3))$ where each $X_i$ is exponential with parameter $i$

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?

• Why the wrong title?
– Did
Commented Jan 15, 2019 at 23:03
• @Did - I'm sorry I'm a bit tired now. I don't see what you're refering to? Commented Jan 15, 2019 at 23:05
• What was your title and what is it now that I have corrected it? Tired or not...
– Did
Commented Jan 15, 2019 at 23:07
• @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. Commented Jan 15, 2019 at 23:31

## 1 Answer

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 $$\min(X_1,X_2,X_3) > t \iff X_1, X_2, X_3 > t$$.