# How do I find the mean of this?

• May 11th 2009, 06:14 PM
MMath09
How do I find the mean of this?
I have $\displaystyle \mathbb{E}[X]=\mu$

$\displaystyle f(x) = \left\{ \begin{array}{lr} 2x^2 + x, & x \leq \mu\\ x^2 + 3x + 1, & x \geq \mu \end{array} \right.$

How do I find $\displaystyle \mathbb{E}[f(X)]$?
• May 11th 2009, 08:50 PM
CaptainBlack
Quote:

Originally Posted by MMath09
I have $\displaystyle \mathbb{E}[X]=\mu$

$\displaystyle f(x) = \left\{ \begin{array}{lr} 2x^2 + x, & x \leq \mu\\ x^2 + 3x + 1, & x \geq \mu \end{array} \right.$

How do I find $\displaystyle \mathbb{E}[f(X)]$?

You don't have enough information to do any thing but write the expectation as an integral:

$\displaystyle E(f(X))=\int_{-\infty}^{\infty} f(x) p(x)~dx=\int_{-\infty}^{\mu}(2x^2 + x) p(x)~dx + \int_{\mu}^{\infty}(x^2 + 3x + 1) p(x)~dx$

where $\displaystyle p(x)$ the density of $\displaystyle X$.

CB
• May 12th 2009, 03:08 PM
MMath09
Hmmm... what if I also have the variance below the mean and the variance above the mean? Any better?
• May 12th 2009, 07:47 PM
CaptainBlack
Quote:

Originally Posted by MMath09
Hmmm... what if I also have the variance below the mean and the variance above the mean? Any better?

Depending on what exactly you mean by those may-be.

CB