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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

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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.

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  • $\begingroup$ So the fact the $P(W>5)$ is a subset of $P(W>2)$ has no bearing here? $\endgroup$ – audiFanatic Nov 20 '13 at 20:11
  • $\begingroup$ And why are we subtracting $1$ twice from both the numerator and the denominator? $\endgroup$ – audiFanatic Nov 20 '13 at 20:17
  • $\begingroup$ @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 $1-P(W<2)=1-(1-e^{-2\lambda}$). $\endgroup$ – hejseb Nov 20 '13 at 20:21
  • $\begingroup$ Nevermind about the second comment, I see now that we're using the CDF $\endgroup$ – audiFanatic Nov 20 '13 at 20:23
  • $\begingroup$ And thank you very much for the clarification; it's been more than helpful! $\endgroup$ – audiFanatic Nov 20 '13 at 20:24

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