Find a real number a < 1 so that the events are independent.

Question

Let $X \sim \mathrm{Exp}(2)$. Find a real number $a < 1$ so that the events $\{X \in [0, 1]\}$ and $\{X \in [a, 2]\}$ are independent.

I'm stuck on this one. So far I found $P(X \in [a, 2])$ and $P(X \in [0, 1])$. Where do I go from here?

Answer 1

Let $E=\{X\in[0,1]\}$ and $F=\{X\in[a,2]\}$. For $E$ and $F$ to be independent, we must have $$ \mathbb P(E\cap F) = \mathbb P(E)\mathbb P(F). $$ Now, $$ \mathbb P(E) = \int_0^1 2e^{-2t}\ \mathsf dt = 1-e^{-2} $$ and $$ \mathbb P(F) = \int_a^2 2e^{-2t}\ \mathsf dt = e^{-2a}-e^{-4}. $$ The product is then given by $$ \mathbb P(E)\mathbb P(F) = \left(1-e^{-2}\right)\left(e^{-2a}-e^{-4}\right).\tag1 $$ Since this quantity is never zero, $E\cap F$ must be nonempty. For $a\in(0,1)$ we have $\mathbb E\cap F = \{X\in[a,1]\}$, and so $$ \mathbb P(E\cap F) = \int_a^1 2e^{-2t} = e^{-2a} - e^{-2}.\tag2 $$ Equating $(1)$ and $(2)$, we have $$ \left(1-e^{-2}\right)\left(e^{-2a}-e^{-4}\right) = e^{-2a} - e^{-2}, $$ and hence $$ a = 2-\frac{1}{2} \log \left(1-e^2+e^4\right)\approx 0.06223. $$