1
$\begingroup$

Suppose we have a (possibly infinite) collection k-variate gaussian distributions $\{(\mathcal{N}(\mu_{\lambda}, \Sigma_{\lambda}))\}$ ($\lambda$ is just a label), and for each distribution $\mu \in [-1,1]^k$, the variance of each coordinate $X_i$ is $1$, and all other covariances $Cov[X_i ,X_j] \in [-1,1]$.

Suppose also we have a distribution $\Lambda$ over labels $\lambda$. Sample a vector $X \in \mathbb{R}^d$ by first sampling $\lambda \sim \Lambda$, and then sampling $X \sim \mathcal{N}(\mu_{\lambda}, \Sigma_{\lambda})$.

Now, sample $Y$ from $\mathcal{N}(\mathbb{E}_{\lambda \sim \Lambda}[(\mu_\lambda, \Sigma_\lambda)])$.

Are $X$ and $Y$ ever identically distributed?

$\endgroup$
1
  • $\begingroup$ I think the answer is "no" in nontrivial cases because the variance of the mixture is greater than the variances of the things being mixed on account of the variance of $\mu_\lambda$. But this is not a proof. $\endgroup$ Aug 6, 2015 at 0:00

2 Answers 2

1
$\begingroup$

No, a mixture of Gaussians is not Gaussian: $\exp[-(x-1)^2]+\exp[-(x+1)^2]$ is very different from a constant times $\exp[-x^2]$.

$\endgroup$
2
  • $\begingroup$ I think you missed the word "ever". Mixtures of gaussians are gaussian sometimes. For example, if u is gaussian then a mixture of gaussians with the same variance and mean u is gaussian. $\endgroup$ Aug 5, 2015 at 13:37
  • $\begingroup$ You're right, it's my bad understanding of "ever". I was about to give the same example. But with the conditions given here ($\mu\in[-1,1]$, unit variance) I bet the answer is no for $k=d=1$. $\endgroup$ Aug 5, 2015 at 14:32
1
$\begingroup$

A sum of independent random variables is Gaussian if and only if each of them is Gaussian (Cramér, 1939). Apply this to $X_1$, which is the sum of a centered Gaussian with variance $1$ and an independent random mean $\mu_1$: $X_1$ would be Gaussian only if $\mu_1$ was...

Except of course in the trivial case of a non random label $\lambda=\lambda_0\ a.s.$

In view of Brendan's comment to the question, in nontrivial cases the answer is no for two reasons: $X_i$'s variance is greater than $Y_i$'s (even if both were Gaussian, which could happen if the condition $\mu_{\lambda,i}\in[-1,1]$ was suppressed), and $X_i$ cannot be Gaussian (of any variance) if $\mu_{\lambda,i}$ is not Gaussian.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.