Standard deviation for weighted sum of normal distributions I have 2 normally distributed random variable $H_0$ and $H_1$, which are combined to give the weighted distribution $H$ as follows:
$H_0 \sim N(\mu_0, \sigma_0)$
$H_1 \sim N(\mu_1, \sigma_1)$
$$f_H = p * f_1(x) + (1-p) * f_0(x),$$
where $H$ has pdf $f_H$ and $H_1$ and $H_0$ have pdfs $f_1$ and $f_0$ respectively.
The mean value $\mu$ of the combined distribution $H$ is:
$\mu = p * \mu_1 + (1-p) * \mu_0$
Now, what is the standard deviation $\sigma$ of $H$?
Is there a simple formula for $\sigma$, in terms of $p, \mu_0, \sigma_0, \mu_1, \sigma_1$?
 A: (I am assuming $H_1$ and $H_0$ are independent)
Let $f_H = pf_1(x) + (1-p)f_0(x)$, where $f_1$ and $f_0$ are pdfs of $H_1$ and $H_0$. Then the random variable $H$ is the mixture of two normal distributions. For the mean of $H$
$$E(H) = \int x\left(pf_1(x) + (1-p)f_0(x) \right) dx = p\mu_1 + (1-p)\mu_0. $$
Similarly for the second moment of H
\begin{align*}
E(H^2)& = \int x^2 \left(pf_1(x) + (1-p)f_0(x) \right) dx\\
& = pE(H_1^2) + (1-p)E(H_0)^2\\
& = p(\sigma_1^2 + \mu_1^2) + (1-p)(\sigma_0^2 + \mu_0^2)
\end{align*}
Finally,
\begin{align*}
Var(H) & = E(H^2) - [E(H)]^2\\
& = p(\sigma_1^2 + \mu_1^2) + (1-p)(\sigma_0^2 + \mu_0^2) - \left[p\mu_1 + (1-p)\mu_0 \right]^2\\
& = \left[p\sigma_1^2 + (1-p)\sigma_0^2\right] + [p\mu_1^2 + (1-p)\mu_0^2]- \left[p\mu_1 + (1-p)\mu_0 \right]^2\\
& = p\sigma^2_1+(1−p)\sigma^2_0+p(1−p)(\mu_1−\mu_0)^2
\end{align*}
Taking the square root of $Var(H)$, you get the standard deviation.
A: Let $Z$ denote a Bernoulli random variable with parameter $p$. Then, the random variable $H$ can be thought of as having conditional density
$N(\mu_i, \sigma_i^2)$ according as $Z$ equals $i$, $i=0,1$, and thus
unconditional density
$$f_H(x) = pf_{H_1}(x) + (1-p)f_{H_0}(x).$$
The unconditional mean is thus the weighted sum of conditional means, viz.
$$\mu_H = p\mu_1 + (1-p)\mu_0$$
while the unconditional (or total) variance is given by the total variance formula:

mean of the conditional variance plus the variance of the conditional mean

The first quantity is clearly $p\sigma_1^2 + (1-p)\sigma_0^2$. On the
other hand, the conditional mean takes on value $\mu_1$ and $\mu_0$with 
probabilities $p$ and $1-p$ respectively  and thus has variance
\begin{align}p\mu_1^2 + (1-p)\mu_0^2 - \left(p\mu_1 + (1-p)\mu_0\right)^2
&= (p-p^2)\mu_1^2 + (1-p-(1-p)^2)\mu_0^2 -2p(1-p)\mu_0\mu_1\\
&= p(1-p)(\mu_1-\mu_0)^2
\end{align}
Adding these two quantities together, we have that
$$\sigma_H^2 = p\sigma_1^2 + (1-p)\sigma_0^2
+ p(1-p)(\mu_1-\mu_0)^2$$
as in Greenparker's answer.
