# Suppose $X$ follows a normal mixture. In which cases is $X$ itself normally distributed?

Suppose $$X$$ follows a normal mixture with cdf:

$$F(x)=0.5\Phi(\frac{x-\mu_1}{\sigma_1})+0.5\Phi(\frac{x-\mu_2}{\sigma_2})$$,

where $$\Phi(\cdot)$$ denotes the normal cdf. Without further information, if I understand this correctly, $$X$$ does not necessarily follow a normal distribution. Is this correct?

• No. If $\mu_1$ and $\mu_2$ are different, but nearly equal, the density function may be unimodal and possibly mistaken for normal. See Wikipedia and search this site for additional discussions/examples. – BruceET Oct 3 at 17:22
• Cannot explain how did I miss that! Thanks! – econ86 Oct 3 at 17:38

When a finite mixture of Normal variables is Normal, then all the variables are identically distributed. Thus, if $$\mu_1\ne\mu_2$$ or $$\sigma_1^2\ne\sigma_2^2,$$ $$F$$ cannot be Normal.

A simple way to see why this assertion is true uses the characteristic function. Let the Normal components have variances $$\sigma_1^2 \le \sigma_2^2 \le \cdots \le \sigma_n^2,$$ corresponding means $$\mu_1, \ldots, \mu_n,$$ and nonzero proportions $$\pi_1, \ldots, \pi_n.$$ With no loss of generality, shift all means by a common amount to make the mixture mean zero. Suppose this mixture is Normal with the same zero mean and variance $$\sigma.$$ Its characteristic function must equal that of the mixture. Thus, for all real values $$t,$$

$$\exp(-\sigma^2 t^2/2) = \sum_{j=1}^n \pi_j \exp(i\mu_j t - \sigma_j^2 t^2/2).\tag{1}$$

Taking derivatives with respect to $$t$$ shows the left hand side is always real. The imaginary part of the derivative of the right side therefore must be constantly zero, especially in any neighborhood of $$0$$ where we may approximate the sine function by the linear term in its Taylor series:

$$0 = \frac{1}{2}\sum_{j=1}^n \pi_j \mu_j \sigma_j^2 \exp(- \sigma_j^2 t^2/2) \sin(\mu_j t) = \left(\frac{1}{2}\sum_{j=1}^n \pi_j \mu_j^2 \exp(-\sigma_j^2 t^2/2)\right) t + O(t^2).$$

This is impossible unless the coefficient of $$t$$ is identically zero, implying every $$\mu_j=0.$$ That is, all the component means must be equal if the mixture is Normal.

In light of this, $$(1)$$ asserts

$$\exp(-\sigma^2 t^2/2) = \sum_{j=1}^n \pi_j \exp(-\sigma_j^2 t^2/2) = \exp(-\sigma_1^2t^2/2 )\sum_{j=1}^n \pi_j \exp((\sigma_1^2-\sigma_j^2) t^2/2).$$

As $$t^2$$ grows large, the terms in the right hand sum either equal $$\pi_j$$ when $$\sigma_j=\sigma_1$$ or decrease rapidly to zero when $$\sigma_j \gt \sigma_1.$$ Taking logarithms, we find

$$\sigma^2t^2/2 = -\sigma_1^2 t^2/2 + \log\left(\sum_{j=1}^k \pi_j\right) + \epsilon,$$

where $$\epsilon$$ can be made arbitrarily small and $$\sigma_1^2=\sigma_2^2=\cdots=\sigma_k^2 \ne \sigma_{k+1}^2.$$ This is impossible unless the constant term on the right is zero, which means

$$1 = \sum_{j=1}^k \pi_j,$$

showing that $$k=n.$$ That is, all the $$\sigma_j$$ are equal, QED.

• This is an amazing answer. Thank you! – econ86 Oct 3 at 17:39