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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?

econ86
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    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](https://en.wikipedia.org/wiki/Mixture_distribution) and search this site for additional discussions/examples. – BruceET Oct 03 '19 at 17:22
  • Cannot explain how did I miss that! Thanks! – econ86 Oct 03 '19 at 17:38
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    I found a more elementary proof of a generalization asked at https://stats.stackexchange.com/questions/451850. The present question concerns the case where a Normal mixture with $n=2$ equally weighted components is compared to a mixture with $k=1$ components. The generalization allows for any number of components and arbitrary mixture coefficients. The answer is the same: a mixture uniquely determines its components and their weights. – whuber Feb 28 '20 at 18:51

1 Answers1

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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.

whuber
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