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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!

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  • $\begingroup$ 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. $\endgroup$ – whuber Sep 29 '20 at 14:18
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    $\begingroup$ @whuber Yes, you are right. I want to show the mixture is symmetric. $\endgroup$ – 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.

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  • $\begingroup$ Thanks! This is helpful! $\endgroup$ – T34driver Sep 29 '20 at 17:06

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