Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am studying some basic probability and the lecture have defined the expectation of a $R.V$ ,$X$, in the cases that $X$ is discrete or continues.

In both definitions we first required some absolute convergence, for example, in the continues case we required that $\int|x|f(x)dx<\infty$ and defined the expectation to be $\int xf(x)dx$.

What is the reason for this requirement ? if $\int xf(x)dx$ then it seems that it should be called the expectation regardless of if $\int|x|f(x)dx<\infty$ is convergent.

share|improve this question
    
This answer of mine to the question Why does the Cauchy distribution have no mean? on stats.SE gives essentially the same illustration as @Marvis's answer below. Other answers to the question are also worth reading. –  Dilip Sarwate Nov 18 '12 at 18:50

2 Answers 2

up vote 1 down vote accepted

It is not just so that EX is finite. The problem is that we can have $E|X| = \infty$ but $EX$ is not well defined! If you open any good book on real analysis, say Rudin's mathematical analysis, you will find that if a series is not absolutely convergent but is "conditionally convergent",, then the series can be rearranged to have different limits!. But the expectation is one number and it shouldn't matter how the series is rearranged. $E|X| < \infty$ assures us of it's existence and well definedness.

Also EX can be $\infty$. Let $X= n \quad \mbox{w.p} \frac{6}{\pi^2n^2} \quad \forall n \geq 1$. The Expected value is $\infty$

share|improve this answer

Consider for instance, the Cauchy distribution i.e. $$f(x) = \dfrac1{\pi} \dfrac1{1+x^2}$$ Clearly, $f(x)$ is a valid probability density function since $$\int_{-\infty}^{\infty} f(x) dx = 1$$ However, $$\int_{-\infty}^{\infty} \dfrac1{\pi} \dfrac{x}{1+x^2} dx$$ doesn't exist. This is so since if you "interpret" the integral $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ as $\displaystyle \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{x}{x^2 + 1} dx$, then it is zero. (This is called the Cauchy principal value).

However, note that the integral $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ can also be interpreted as, for instance, $$\displaystyle \lim_{R \rightarrow \infty} \int_{-R}^{kR} \frac{x}{x^2 + 1} dx = \lim_{R \rightarrow \infty} \frac12 \log \left( \frac{1 + k^2R^2}{1 + R^2} \right) = \log(k) \text{ where }k > 0.$$ You could also take other functions of $R$ such that the lower limit tends to negative infinity and upper limit tends to infinity as $R \rightarrow \infty$ to get different answers.

Hence, $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ is not zero and in fact cannot be assigned any value unless you know how the lower limit and upper limit approach $\infty$.

This arises due to the fact the integral doesn't converge conditionally on $(-\infty, \infty)$ i.e. $$\int_{-\infty}^{\infty} \vert x \vert f(x) dx = \int_{-\infty}^{\infty} \dfrac1{\pi} \dfrac{\vert x \vert}{1+x^2} dx = \infty$$

Hence, $\displaystyle \int x f(x) dx$ is well-defined and exists only when $\displaystyle \int \vert x \vert f(x) dx < \infty$.

share|improve this answer
    
+1 Over on stats.SE, I used the same example to illustrate why the Cauchy principal value of $\int_{-\infty}^{\infty} xf(x) \, dx$ cannot be used to define the mean of a random variable. –  Dilip Sarwate Nov 18 '12 at 18:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.