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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.

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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 $$\Pr(B|A)=\frac{\int_{3/4}^1 6x(1-x)\,dx}{\int_{1/2}^1 6x(1-x)\,dx}.$$

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