Assume μ1 and μ2 are the two means and the standard deviation is σ for two normal distribution.
The probability of x in the mixture of these two normal distributions is $$ p(x)=0.5N(x;μ1,σ)+0.5N(x;μ2,σ) $$ How can I prove these two things?
- The first derivative can be shown to be 0 at $\frac{μ1+μ2}{2}$
- The second derivative is strictly less than zero for |μ1−μ2|<2σ
Really appreciate any thoughts.
Assuming that your function $N(x;\mu, \sigma)$ is the cumulative distribution function for the normal distribution with mean $\mu$ and variance $\sigma^2$, i.e. $$N(x;\mu, \sigma) = \frac{1}{\sqrt{2\pi}\ \sigma} \int_{-\infty}^{x}e^{-\frac{(z-\mu)^2}{2\sigma^2}}dz$$
Then it's derivative is simply given by the probability density function for the same distribution, i.e.
$$f(x;\mu,\sigma)=\frac{d}{dx} N(x;\mu, \sigma) = \frac{1}{\sqrt{2\pi}\ \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
Now, since derivation is linear, the derivative of interest is:
$$\frac{d}{dx}p(x;\mu_{1}, \mu_{2}, \sigma) = \frac{1}{2}f(x;\mu,\sigma) + \frac{1}{2}f(x;\mu_{2},\sigma)$$
Which, when evaluated at $x=\frac{\mu_{1}+\mu_{2}}{2}$ will be 0.