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Portion of integral below a certain number
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The value of this integrals seems to be a finite number, and not infinity as you said. Maybe you have a logical error somewhere in your thinking. Would you share what were your though process?
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(Edited because I re-watched the video a bit more closely.) I noticed that Tony wrote "∝n^(-1.3)", i.e. proportional to and not necessarily equal. So let's assume that the probability function f(n)=an^(-1.3) for some constant a. The anti-derivative of that is  is F(n)=K-(10/3)an^(-3/10). Because F(n) is the cumulative distribution, the limit as n->infinity of F(n) needs to be 1, so K=1, and we have F(n)=1-(10/3)an^(-3/10).

The other requirement for f(n) to be a probability function is that the area under the curve has to be 1. Let’s find m such that the area from (m,inf) is equal to 1. The area should just be equal to:

F(infinity) - F(m) = 1

1-(10/3)a(infinity)^(-3/10) - (1-(10/3)am^(-3/10)) = 1

1 - 0 - 1 + (10/3)am^(-3/10) = 1

(10/3)am^(-3/10) = 1

(10/3)a = m^(3/10)

a = (3/10)m^(3/10)

Now let's say that we want this distribution to be defined for n=1. Then we want to set m=1 above and solve for a:

a = (3/10)(1)^(3/10)

a = 3/10

And now we can go back to our original probability function definition and say that f(n)=(3/10)n^(-3/10). If you integrate that from 1 to infinity, you'll find that the area is indeed equal to 1. (Never mind that this is a discrete distribution we're talking about, but it's close enough.) That's the part that wasn't covered in the video. The cumulative distribution for that is in fact F(n)=1-n^(-3/10), as shown in the video.

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