With $X_{i} \stackrel{i.i.d}{\sim} Expo(\lambda)$ find probability that all $X_{i} > 60$

Question

Given the following, $$X_{i} \stackrel{i.i.d}{\sim} Expo(\lambda)$$ Find the probability that every $X_{i} > 60$. My intuition is to do the following, $$P(X_{i} > 60) = 1 - F(60) = e^{-\lambda 60}$$ Then using the multiplication rule to find $(e^{-\lambda 60})^{n}$ where n is the number of r.v.s in $i$

Would this be correct?

Answer 1

My intuition for these are, if $X_1,\ldots,X_n \sim \mathrm{Expo}(\lambda)$ and they are iid, then we get

$P(\mathrm{min}(X_i)>60) = P(X_1>60 \cap \ldots \cap X_n>60)=P(X_1>60)\cdot\ldots\cdot P(X_n>60)=P(X_1>60)^n = (1-F(60))^n$

So your intuition is correct. I just wanted to show the general idea for these order statistics. Keep in mind this only works when independent and identical.