The mean of the Probability Density Function

[p75213]

Homework Statement

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:
$$\lim \,b \to - \infty \,{1 \over { - b}}\int_b^0 {f(x)\,dx\,} + \,\,\lim \,a \to \infty {1 \over a}\int_0^a {f(x)\,dx = 0 + 0 = 0} $$

 

[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.

 

[Ray Vickson]

Homework Statement

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:
$$\lim \,b \to - \infty \,{1 \over { - b}}\int_b^0 {f(x)\,dx\,} + \,\,\lim \,a \to \infty {1 \over a}\int_0^a {f(x)\,dx = 0 + 0 = 0} $$

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

 

[p75213]

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