# Why doesn't integrating infinitesimally small likelihoods work in this sense?

We know that something is going to happen after $x$ amount of time, but the exact time at which the event occurs is random within $x$ time. (Like, say we did it a bunch of times where it happened in $x-y$ time, and all the $y$s were uniformly distributed reals $0$ to $x$.)

It is easy to see that you can compute the likelihood that the event has occurred by using $t/x$ with $t$ being the time that has elapsed and $x$ being the total amount of time. It's also easy to see the likelihood of it occurring in the next $n$ amount of time by computing $n/(x-t)$ (With $t$ again being the amount of time elapsed)

As $n$ gets very small, the likelihood of it happening in the next $n$ amount of time also clearly also becomes very small. This can be expressed by realizing that:

$$\lim_{n \to 0}\frac{n}{x-t}=0$$

But now, if you take that limit and integrate it over some period of $t$ s.t. $0<t<x$, you should get $t/x$ again, since we're essentially adding up all the likelihoods of it happening at each instant in the interval:

$$\int_{0}^{x-a} \lim_{n \to 0} \frac{n}{x-t} dt$$

Resolving the limit first clearly doesn't yield. Does that mean that the integral I have constructed does not represent what I think it does?

--

I considered using arclength in the integral but stopped myself in realizing that we are concerned with the values the function outputs, not the length of the curve.

-
Evaluate the limit first. That's essentially what the integral above means. As a word of advice, this sort of notation is not really... optimal, should I say? – ThisIsNotAnId Mar 24 '12 at 18:47
oh so you're saying that the integral I put does not well represent what I am asking? – enthdegree Mar 24 '12 at 18:53
Well, not quite. I'll just post an answer and explain what I mean. – ThisIsNotAnId Mar 24 '12 at 19:00

There are two problems:

• First, you didn't try to integrate the likelihood over an interval of infinitesimally small length -- you tried to integrate the likelihood over an interval of zero length

• Second, the probability you tried to compute was "The odds of it happening in $[t, t+n)$ conditioned on the assumption it did not happen in $[0, t)$", but what the integral actually needs is "The odds of it happening in $[t, t+n)$".

Setting it up via non-standard analysis, let $H$ be transfinite and $\epsilon = (x-a)/H$. The odds of the event happening in $[t, t + \epsilon)$ is $\epsilon / x$. The integral you were trying to set up is then approximated by the sum

$$\sum_{i=0}^{H-1} \frac{\epsilon}{x} \approx \int_{0}^{x-a} \frac{dt}{x} = \frac{x-a}{x}$$

In some cases, it's easier to just use standard means, though: if the probability of the event happening in $[0, t)$ is $t/x$, then the differential form is just $$d \left(\frac{t}{x} \right) = \frac{dt}{x}$$ (no $dx$ term because $x$ is a constant) and so the corresponding probability density is $\frac{1}{x}$.

-
Ok I see where I messed up except in your answer where it says "The odds of the event happening in [t,t+ϵ) is ϵ/x." This may be a total nitpick but did you mean "ϵ/(x-t)"? After t time has passed there is only (x-t) time left for the event to occur. – enthdegree Mar 24 '12 at 22:55
oops nevermind. – enthdegree Mar 24 '12 at 22:57
FWIW, In what situation would you choose to express it using transfinite numbers; initially this seems like it adds complexity without purpose. – enthdegree Apr 4 '12 at 20:24
Sometimes, problems are easier to set up or analyze when you use them. The use of nonstandard analysis doesn't preclude the use of familiar techniques: differentiation and differential forms are still useful! One particular simplification is that the notion of differential form can be made much closer to the notion of number than in standard analysis, so that textbooks feel comfortable actually introducing it and talking about it in an introductory calculus 101 class. – Hurkyl Jun 7 '12 at 9:04

The integral you have:

$$\int_{0}^{x-a}\lim_{n \to 0}\frac{n}{x-t} dt \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

is very similar to doing the following.

Let $f \text{ and } g$ be two integrable functions $\ni f,\ g: \mathbb{R} \rightarrow \mathbb{R},\ x \mapsto y$. Then:

$$\int_a^b f(x) + g(x)\ dx = \int_a^b f(x)\ dx + \int_a^b g(x)\ dx \ \ \ \ \ \ \ \ \ \ (2)$$

So, you're simply figuring out what to integrate before doing it.

-

First notice that $$\int_{0}^{x-a} \lim_{n \to 0} \frac{n}{(x-t)} dt$$ is not dimensionally correct. It has the dimensions of time. There is also a confusion about what is $t$.

You are confounding probability and the probability density function (pdf). If the event didn't happen up to time $t_0$, the pdf is $1/(x-t_0)$, not $n/(x-t_0)$. If we want to use this pdf to find the probability of the event occurring between time $0$ and some other time we must use the pdf $1/(x-t_0)$ with $t_0=0$, since $t_0$ here is the time before which the event had not occurred. Thus, the pdf is $1/x$ as expected and the probability is $t/x$.

-