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Define a linear space $X=\Delta(Z)$. $Z=\mathbb R_+$ is the real valued outcome space and $\Delta$ is a probability simplex. Let $f$ and $g$ denote the elements in $X$. Suppose now I define a mixture $x=\lambda f+(1-\lambda) g$. The meaning of the mixture is as follows: Let R and G denote two independent real valued random variables that distributes as $f$ and $g$, respectively. Define a new random variable $H=\lambda R+(1-\lambda)G$. The mixture gives rise to $x$ which $H$ would distribute as.

I have two question:

  1. Is the definition of the mixture clear in mathematical sense? Or does it already have a specific name?
  2. How to write the mixture space formally? for example, I think the real value outcome space would not be Z anymore. So the mixture may result in elements does not exist in $X$.

Any advice would be much appreciated.

Emma
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    Convex combinations of densities are different than convex combinations of random variables. For example, suppose $X,Y \stackrel{\text{iid}}\sim f := \text{Unif}(0,1)$. Then a convex combination of the densities is $\lambda f + (1-\lambda) f = f$ while eg $\frac 12 X + \frac 12 Y$ has a density related to the convolution of $f$ with itself which here is a triangular distribution. So i think you'll have to be more precise about whether you want convex combos of RVs or densities/distributions – jld Jun 17 '21 at 18:20
  • Thanks for your comment. I mean the convex combinations of random variables. Therefore I write the mixture gives rise to a probability distribution describes the random variable =+(1−). Does this still look unclear? – Emma Jun 17 '21 at 21:14
  • Could you offer further suggestion on how to obtain the distribution for the convex combinations of random variables. In your example, I can see it is a triangular distribution. But in general, how could we obtain the new distribution based on the original distributions, especially when X and Y do not follow the same distribution. – Emma Jun 17 '21 at 21:17
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    Suppose we have two continuous RVs $X\sim f_X$ and $Y\sim f_Y$. I'll assume independence for simplicity. $\lambda X$ and $(1-\lambda )Y$ are both 1-1 transformations so $f_{\lambda X}(x) = f_X(x/\lambda)/\lambda$ and similarly for $(1-\lambda )Y$. Then the distribution of the sum of $\lambda X$ and $(1-\lambda)Y$ is given by the convolution of $f_{\lambda X}$ and $f_{(1-\lambda)Y}$. If we restrict to just Gaussians then these are closed under convolution so the result is still Gaussian, but in general the new distribution doesn't have to be of the same type as either $f_X$ or $f_Y$ – jld Jun 21 '21 at 13:51

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