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Suppose that there're two independent random variables X and Y, both of which are normal. X has parameters N(-1, 1) while Y has parameters N(1, 1).

And for the third random variable Z, its value is equal to either X or Y, with both probabilities as 50%. Is Z normally distributed? If it is what is its variances?

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  • $\begingroup$ Related: stats.stackexchange.com/questions/16608/…. For more similar questions, please search our site: stats.stackexchange.com/search?tab=votes&q=mixture%20moment. $\endgroup$ – whuber Jan 8 '18 at 23:29
  • $\begingroup$ You could plot the density function of $Z$ at several points, e.g., $-3, -2.9, -2.8, ..., 3$ to get an idea of what it looks like. This would provide you with a lot of information! $\endgroup$ – jbowman Jan 8 '18 at 23:44
  • $\begingroup$ @whuber Thank you! For my question, can RV Z be expressed as Z = 0.5X + 0.5Y? For all "weighted mixtures" Z, can they be expressed as Z = XPx + YPy? $\endgroup$ – Yujian Jan 9 '18 at 0:05
  • $\begingroup$ @jbowman Thank you jbowman! I did plot that with R and it was normal. But I want a mathematical prove. The link above is helpful. $\endgroup$ – Yujian Jan 9 '18 at 0:07
  • $\begingroup$ @Glen_b The wikipedia page solved my last confusion. Thank you Glen_b! $\endgroup$ – Yujian Jan 9 '18 at 0:13
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The random variable $Z$ is NOT $0.5X + 0.5Y$.

The expression $0.5X + 0.5Y$ describes averaging the values of $X$ and $Y$, not choosing one of them with probability $\frac12$.

With mixtures, it is instead the density (and cdf) that are averaged:

$$f_Z(z) = 0.5 f_X(z) + 0.5 f_Y(z)\,.$$

See Wikipedia's page on mixture distributions for basic information on this. (For means $\mu_i$ and sd's $\sigma_i$ with probability weights $w_i$, it gives the variance of the mixture as $\sum _{i=1}^n w_i((\mu _i-\mu )^2+\sigma_i^2)$ (where $\mu$ is the mean of the mixture). The results for mean and variance there are pretty easy to show.

If you plot the density of Z correctly you will see that it is not normal.

mixture density, and normal density with same mean and variance; we see the mixture has a "flat top" with bigger shoulders- higher density near 1 standard deviation from the mean

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    $\begingroup$ This one only has one mode but a larger separation in means (ceteris paribus) will be bimodal. $\endgroup$ – Glen_b Jan 9 '18 at 1:38
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    $\begingroup$ @MichaelChernick I simulated this, and it looks normal (granted, I did it once, but with n = 1000): gist.github.com/markhwhiteii/201c6bafbec5d54e5081535508bc9c8e If it looks normal, is it still normal, even if we know it comes from a mixture of two similar distributions? Is that too theoretical? If it looks normal, I would say, "It is a normal distribution", but strictly speaking, it is not... since, by definition, it is a mixture. $\endgroup$ – Mark White Jan 9 '18 at 2:29
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    $\begingroup$ @MarkWhite I will draw the density so that you can see how it compares to a normal. $\endgroup$ – Glen_b Jan 9 '18 at 4:45
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    $\begingroup$ @Mark It's there now; you can see it's noticeably different. $\endgroup$ – Glen_b Jan 9 '18 at 4:56
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    $\begingroup$ @Weiwen This distribution is unimodal, not bimodal. Modes refer to properties of the distribution, not to how it is expressed. So although this one is constructed from two distributions that have two separate modes, it nevertheless is unimodal. You can verify this by finding the zeros of the derivative of its density, which is a positive multiple of $x+1+e^{2x}(x-1)$: it has a unique zero at $x=0$. Whether approximating it with a Normal distribution is "acceptable" will depend on the application. $\endgroup$ – whuber Jan 9 '18 at 15:29

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