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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.

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Related: Variance of the weighted mixture of two gaussians –  leonbloy Jan 16 '13 at 13:41

2 Answers 2

up vote 1 down vote accepted

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.

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Sketchy? Really? –  Did Jan 16 '13 at 13:31
    
@did: Really. It is a term of rigor in Australia, where I lived for a time. –  Ron Gordon Jan 16 '13 at 13:35
    
There is nothing sketchy there, neither in the Australian meaning of the term nor in any other I can think of. –  Did Jan 16 '13 at 13:36
    
I was making a statement about bimodal (and multimodal) distributions, where concepts that fit nicely with a simple normal distribution don't work so well in general. It may be OK here, though. –  Ron Gordon Jan 16 '13 at 13:38
    
I know what you meant, and I reacted to it because this is deeply wrong: expectations and variances are perfectly well defined for (suitably integrable) multimodal distributions. (And I do not think your answer addresses the OP's question.) –  Did Jan 16 '13 at 13:41

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 $\mathrm{var}(Y)=$ $______$.

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