Limited Expected Value Problem

[RockOn]

Hello,

I'm working on a Limited Expected value problem [x ^ 10,000] for the Single Parameter Pareto distribution with alpha = 1 and theta = 1000.

Since the tables are not helpful, the solution works it using first principles by integrating the Survival function. The solution splits the integral into two pieces - a value of 1000 + the integration of the survival function from 1000 to 10000.

The solution states that the integral from zero to 1000 equals 1000.

What principle is being used to determine that the first piece of the integral equals 1000?

Sorry if I'm missing something obvious.
Thank you,

[RockOn]

I think I just figured it out - are they using the expected mean formula?

[Gandalf]

The survival function is a constant 1 on that interval. the integral of 1 on an integral is equal to the length of the interval. It is not a difficult integration.

[RockOn]

Gandalf - thank you for the quick reply!

Correct - it is not a difficult integration. However, ln(0) is undefined.

[Gandalf]

Ln(0) is irrelevant. The solution throughout is integrating the survival function. The survival function is a constant 1 below 1000, so that’s what you are integrating there.

[RockOn]

Gandalf - Thank you.
OK. I'm dense. How is it 1 below 1000?

[Gandalf]

The published tables give you the survival function S or cumulative distribution function F for x>=theta, I forget which. Below that, you need the (universal) definition of a survival function: S(x) = Pr(X>x). From the tables, you can conclude S(theta)=1 (directly, or by 1-F(theta), if that´s what they give you.)

If S(theta) is 1, then S must be a constant 1 below theta, since S is a non increasing function and probabilities can’t exceed 1.

[RockOn]

AAAaaahhhhh. Right. I forgot about that stipulation of x>theta - which makes sense.

Thank you bunch!!!