# Homework Help: The mean of the Probability Density Function

1. Nov 14, 2012

### p75213

1. The problem statement, all variables and given/known data
The mean of a function is as follows:
$${1 \over {a - b}}\int_b^a {f(x)\,dx}$$

So why is the mean of the PDF as follows:
$$\int_{ - \infty }^\infty {xf(x)\,dx}$$

I thought it would have been this way:
$$\lim \,b \to - \infty \,{1 \over { - b}}\int_b^0 {f(x)\,dx\,} + \,\,\lim \,a \to \infty {1 \over a}\int_0^a {f(x)\,dx = 0 + 0 = 0}$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 14, 2012

### I like Serena

Hi p75213!

The difference is that the mean of the function is an average of y-values.
The mean of the PDF is a weighed average of x-values.

3. Nov 14, 2012

### Ray Vickson

The integral $$\int_{ - \infty }^\infty {xf(x)\,dx}$$ is not the "mean of the pdf"; it is the mean of the random variable X for which f is the density function. Sometimes you will see authors use rather sloppy language and say something like "let μ be the mean of the pdf", but when they say that they do not mean it literally: they mean that μ is the mean corresponding to the pdf.

RGV

4. Nov 14, 2012

### p75213

Thanks guys. I've done some more reading and investigation. Rather than the mean I prefer to think of it as "expected value".