# Using the weak law of large numbers

• Nov 24th 2007, 06:40 PM
johnj77
Using the weak law of large numbers
Hello,
To give a brief background
in my college math course on probability and statistics, the professor just finished going over Markov's inequality, Chebychev's inequality, and the Weak law of large numbers.

This is one of the examples given in the book, but I just can't seem to understand it. I especially do not understand the boxed step, and why the expected value of f(Ui) equals that integral.

If someone could provide a detiailed explanation, I would be very grateful.
• Nov 25th 2007, 01:23 AM
CaptainBlack
Quote:

Originally Posted by johnj77
Hello,
To give a brief background
in my college math course on probability and statistics, the professor just finished going over Markov's inequality, Chebychev's inequality, and the Weak law of large numbers.

This is one of the examples given in the book, but I just can't seem to understand it. I especially do not understand the boxed step, and why the expected value of f(Ui) equals that integral.

If someone could provide a detiailed explanation, I would be very grateful.

By definition of $\displaystyle f$ and $\displaystyle U_i$ :

$\displaystyle E(|X_i|) = E(|f(U_i)|)$

$\displaystyle E(|f(U_i)|) = \int_0^1 |f(x)| dx$

as again by given condition $\displaystyle U_i \sim U(0,1)$.

But by definition $\displaystyle f(x) \ge 0$ so we have:

$\displaystyle E(|X_i|) = E(|f(U_i)|)= \int_0^1 |f(x)| dx= \int_0^1 f(x) dx<\infty$

the last inequality because again by definition $\displaystyle f$ is integrable

RonL