Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Time $t \ge 0$ after which a new lightbulb burns out is determined by a distribution that has density $f(t) = \lambda e^{-\lambda t}$..... λ is a positive constant. How do you do conditional probability when you have a function like this? If you can answer both in a way that explains how to do it in general using the title as an example, would be great.

share|cite|improve this question
which question do you want answered? the one in the title or the one in the body of your post? ;) – Nana Oct 3 '12 at 0:49
up vote 4 down vote accepted

I will call the times $t_1$ and $t_2$, since in probability it is useful to reserve caps for events and random variables.

There is ambiguity in the question. Maybe it means what is the probability that it lasts a total of at least $t_2$ hours. Or maybe it means what is the probability it lasts at least $t_2$ additional hours beyond $t_1$. I will take the second interpretation. It is an easy matter to adapt the answer to the first interpretation.

First, let $T$ be a random variable that has the distribution given by density function $\lambda e^{-\lambda t}$ if $t\ge 0$, and $0$ otherwise. Then $$\Pr(T\gt a)=\int_a^\infty \lambda e^{-\lambda t}\,dt=e^{-\lambda a}.\tag{$1$}$$

A very important property of the exponential distribution above is memorylessness. If an object has a lifetime governed by an exponential distribution, then the probability it lives at least $t_2$ additional years, given it has lived $t_1$ years, is just the same as the probability that an object fresh out of the box lives at least $t_2$ years. By $(1)$, this is $e^{-\lambda t_2}$.

This sounds implausible, and in fact is not a terrific model for the lifetime of lightbulbs. But it works very nicely for the decay of a radioactive substance.

Now we do a formal calculation, without using memorylessness. Actually, the calculation will prove the fact of memorylessness of the exponential distribution. Similar calculations can be done for other distributions, it is just a matter of using the basic conditional probability formula.

Let $A$ be the event the lightbulb lasts at least $t_1$ years. Let $B$ be the event the lightbulb lasts at least $t_1+t_2$ years. We want the conditional probability $\Pr(B|A)$.

By the usual formula for conditional probability, we have $$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$ Note that $\Pr(A\cap B)=\Pr(B)$. By formula $(1)$ (that is, by integration) we have $\Pr(A\cap B)=e^{-\lambda(t_1+t_2)}$.

Similarly, $\Pr(A)=e^{-\lambda t_1}$. It follows that $$\Pr(B|A)=\frac{e^{-\lambda(t_1+t_2)}}{e^{-\lambda t_1}}=e^{-\lambda t_2}.$$

Remark: The exponential distribution is pretty special. So let $X$ have density function say $6x(1-x)$ on the interval $(0,1)$ and $0$ elsewhere. Let $A$ be the event $X\gt 1/2)$, and let $B$ be the event $X\gt 3/4$. We want $\Pr(B|A)$. This is given by the same basic conditional probability we gave earlier. Note that $\Pr(A\cap B)=\Pr(B)$, so $$\Pr(B|A)=\frac{\int_{3/4}^1 6x(1-x)\,dx}{\int_{1/2}^1 6x(1-x)\,dx}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.