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I have the following problem:

The time T until a new light bulb burns out is an exponential random variable with parameter A. Ariadne turns the light on, leaves the room, and when she returns, t time units later. finds that the light bulb is still on. which corresponds to the event A = {T > t}. Let X be the additional time until the light bulb burns out. What is the conditional CDF of X. given the event A?

(from Introduction to Probability,2nd edition, by Dimitri P. Bertsekas and John N. Tsitsiklis )

What I did to solve the problem is the following:

$Conditional$ $CDF$ $of$ $X$ $=\mathbb P(X<=x|A)=\mathbb P(T<=t+x|T>t)=$

$\mathbb P(T<=t+x$ $and$ $T>t|T>t)$ and after some calculations with $CDF$s I got the result:

$\cfrac {e^t-e^x}{e^t}$.

However, I looked in the book and they have a different solution to the problem. They found the following:

$\mathbb P(X>x|A)=\mathbb P(T>t+x|T>t)=...=e^{-\lambda x}$.

So, my question is this:

The problem clearly asks for the conditional cumulative distribution function of X, which I know to be $\mathbb P(X<=x|A)$, right? Because the definition of a $CDF$ is $\mathbb P(X<=x)$ and we use an integral to calculate it. But in their solution, which is obviously the correct one, they calculated $\mathbb P(X>x|A)$, not $\mathbb P(X<=x|A)$. But from what I know that is not the $CDF$. So why did they calculate that and not what I did?

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  • $\begingroup$ Your solution does not include $\lambda$. The solution in the book is $1-CDF$. If you show your calculation it would be clearer. $\endgroup$ – herb steinberg Mar 31 at 21:28
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Note that $$ \mathbb{P}[X \le x|A] = 1-\mathbb{P}[X>x|A], $$ so in terms of information, the two are equivalent, but computing one is very often much easier than the other one. They took a shortcut for quick computations.

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  • $\begingroup$ That would've been clear, but in their solution, they haven't used any 1-P(X>x|A), but just P(X>x|A), without the 1. So why did they do that, and how come that is correct? $\endgroup$ – Bogdan Vlad Mar 31 at 21:25
  • $\begingroup$ @BogdanVlad they are not claiming to be computing the cdf, what is written is correct. But to get the actual cdf, you need to subtract the result from 1... $\endgroup$ – gt6989b Mar 31 at 22:12

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