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I am faced with an approximation that replaces a probability density function with the indicator function and I am at a loss as to why this is valid.

We want to model the lifetime $T$ of a website using the density function $Pr[X > t] \approx e^{-\rho t}$. Then, according to the math I am reading,

$\begin{eqnarray} P[X > k] & = & \int_0^t Pr[X > k\,|\,T = u] \frac{d Pr[T \leq u]}{du} du \\ & \approx & \rho \int_0^t e^{-\rho u} Pr[X > k\,|\,T = u] du \\ & \approx & \rho \int_0^t e^{-\rho u} 1_{\{l(u) > k\}} du \end{eqnarray}$

if $l(u) \sim e^{\beta u}$ for some $\beta < \rho$ (Here, $\sim$ means that $l(u)$ grows as fast as $e^{\beta u}$ - some would write $l(u) = \Theta(e^{\beta u})$). $1_{\{l(u) > k\}}$ is the indicator function that is equal to $1$ if $l(u) > k$, otherwise it is $0$.

I want to understand the intermediate steps required to jump from the PDF to the indicator function.

As a bit of context, this expression comes from mathematics that tries to model the lifetime $t$ of a website. This is used to explain why the world-wide web network is (roughly) scale-free.

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    Also, if you know of a good book that deals with these kinds of approximations/identities as well as intricate Lebesgue-Stieltjes integral theory (since so much advanced probability theory uses such integrals), please add that to your answer!2011-03-06
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    What do you mean by $Pr[X > k] \approx e^{-\rho t}$? The two sides have no free variables in common. Did you mean $e^{-\rho k}$?2011-03-06
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    Sorry, that was a typo. Should be fixed now.2011-03-06
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    Also, what we need to understand the approximation is the conditional probability that appears in the approximated integral; I don't see how the unconditional probability helps in understanding the approximation.2011-03-06
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    Sorry joriki, I was trying too hastily to reduce the math content of the question to a minimum and I guess I completely destroyed the meaning. Thanks for your patience.2011-03-06
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    No problem. You can let someone be notified that you've replied to their comment by preceding it with a ping like this: "@joriki:". (I'm not doing that since you get notified of comments to your question anyway.) I still don't understand the equations. $t$ appears as a free variable on the right-hand sides but not on the left-hand side -- what is $t$?2011-03-06
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    @joriki $t$ should no appear as a free variable on both sides. In the full equation, $k$ is used instead of $t$ (I copied it directly from the source).2011-03-06
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    I don't understand what that means. In the equations as you wrote them, there's a free $k$ on the left-hand side and both a free $k$ and a free $t$ on the right-hand sides, so I can't make sense of "$k$ is used instead of $t$".2011-03-06
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    Arg! I forgot something essential. $T$ is the lifetime of a website.2011-03-06
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    It's getting more confusing by the minute. I think you'll need to make more of an effort to pose the question correctly if you want to get an answer -- I for one am starting to lose interest. You've now added the contradictory statements that both $T$ and $t$ are the lifetime of a website. Both $T$ and $t$ appear in the same formula, $T$ apparently as the name of a random variable and $t$ as a limit of an integral. $k$ and $X$ are nowhere explained. I suggest that you read through the question from top to bottom and make sure that you're no longer using any unexplained symbols.2011-03-06
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    Could you provide a link to the paper or whatever you are reading?2011-03-06

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