# Joint distribution proof

I am trying to study for an exam and I am kind of lost on how my professor came to a particular result on his practice exam.

Let $W$ be an exponentially distributed random variable with $\lambda = 2$

Prove that $P(W > 5 | W > 2) = P( W > 3)$

I made it as far as re-writing the problem as $$\frac {P(W>5,W>2)}{P(W>2)}$$

However, he managed to cancel out the $P(W > 2)$ in the numerator and I'm lost as to why he can do that.

The entire problem is here (problem 4,b) And the solutions are here

In the numerator you have the probability that $W$ is greater than 5 and greater than 2. This only happens when $W$ is greater than 5. Thus $$\frac{P(W>5, W>2)}{P(W>2)}=\frac{P(W>5)}{P(W>2)}$$
What is this then? Well, $$\frac{P(W>5)}{P(W>2)}=\frac{1-(1-e^{-\lambda 5})}{1-(1-e^{-\lambda 2})}=\frac{e^{-5\lambda}}{e^{-2\lambda}}=e^{-(5-2)\lambda}=e^{-3\lambda}=1-(1-e^{-3\lambda})=1-P(W<3)=P(W>3)$$ which is the answer you wanted.
• So the fact the $P(W>5)$ is a subset of $P(W>2)$ has no bearing here? – audiFanatic Nov 20 '13 at 20:11
• And why are we subtracting $1$ twice from both the numerator and the denominator? – audiFanatic Nov 20 '13 at 20:17
• @audiFanatic That is why $P(W>5, W>2)=P(W>5)$. Which is why I wrote: "In the numerator you have the probability that $W$ is greater than 5 and greater than 2. This only happens when $W$ is greater than 5." As for your second question, $1-e^{-\lambda w}$ is the cumulative density function, i.e. $F(w)=P(W<w)$. So then since $P(W>w)=1-P(W<w)$, $P(W>2)$ for example is equal to $1-P(W<2)=1-(1-e^{-2\lambda}$). – hejseb Nov 20 '13 at 20:21