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Suppose $f_{-a}(x)$ is the pdf for $N(-a,\sigma^2)$ and $f_{a}(x)$ is the pdf for $N(a,\sigma^2)$.

Let $f(x)=0.5f_{-a}(x)+0.5f_{a}(x)$ be the mixture density.

Is $c=0$ the unique center for $f(x)$ in the sense that $f(x)=f(2c-x)$ for any $x$?

My guess is that it is the unique center, but I don't know how to rigorously show it.

Thanks!

T34driver
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  • https://stats.stackexchange.com/questions/46843 rigorously shows $0$ is *a* center (in the sense of being a mean or median). That it is unique follows immediately from observing the mixture density is not zero there. The sense given in your penultimate statement is that the mixture is *symmetric.* That's a much stronger result and is rigorously shown in the duplicate. – whuber Sep 29 '20 at 14:18
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    @whuber Yes, you are right. I want to show the mixture is symmetric. – T34driver Sep 29 '20 at 17:06

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Suppose $f_{-a}(x)$ and $f_{a}(x)$ are independent, $f(x)\sim N(0,\frac{1}{2}\sigma^2)$, $f(2c-x) \sim N(2c,\frac{1}{2}\sigma^2)$. let $\sigma_1=\sqrt{\frac{\sigma^2}{2}}$, therefore, $f(x) \sim N(0,\sigma_1^2) $ and $f(2c-x) \sim N(2c,\sigma_1^2).$ Write down their corresponding pdfs, we have:
$f(x) = \frac{e^{-(x - 0)^{2}/(2\sigma_1^{2}) }} {\sigma_1\sqrt{2\pi}}=\frac{e^{-(x - 2c)^{2}/(2\sigma_1^{2}) }} {\sigma_1\sqrt{2\pi}}=f(2c-x)$ for any x. Take the log on both sides and cancel out common terms, the equation becomes as follows:
$x^2=(x-2c)^2$ for any x. Therefore, $c=0$ is the unique center for this mixture of distribution.
Hope this answer helps you.

Linxing Yao
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