what is the mean of probability density function Suppose we have a PDF
$$p(x)=\frac{1}{2}\left[\frac{1}{\sqrt{2\pi}}\left(\exp \left(\frac {-(x-1)^2}{2}\right) + \exp \left(\frac {-(x+1)^2}{2}\right)\right)\right] \quad\text{for}\; -\infty  < x < +\infty$$
Can we say that the mean of the above is $0$ and variance is $1$?
Thank you.
 A: You can assign "mean" to be the expected value of the random variable $X$ whose distribution is your PDF, and you can assign "variance" to be the expected value of $(X - \mathrm{mean})^2$.  Specifically:
$$\mathrm{mean} = E(X) = \int_{-\infty}^{\infty} dx \: x \, p(x) $$
$$\mathrm{variance} = E[X-E(X))^2] = \int_{-\infty}^{\infty} dx \: (x-E(X))^2 p(x) $$
EDIT
Using these, I find the mean to be 0 and the variance to be 2.
I will say, however, that these concepts are a little sketchy if the distribution is truly bimodal.
A: Let $f_X$ denote the PDF of a random variable $X$, $(x_k)_k$ a collection of real numbers, and $(p_k)_k$ a collection of nonnegative real numbers summing to $1$. Then, 
$$
g:x\mapsto\sum\limits_kp_kf_X(x-x_k),
$$ 
defines a PDF and a random variable with this PDF is $Y=X+Z$, where $Z$ is independent of $X$ and $\mathbb P(Z=x_k)=p_k$ for every $k$. 
In particular, $\mathbb E(Y)=\mathbb E(X)+\mathbb E(Z)$ and $\mathrm{var}(Y)=\mathrm{var}(X)+\mathrm{var}(Z)$.
In your case $X$ is standard normal and $Z$ is $______$, hence $\mathrm{var}(Y)=$ $______$.
