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Suppose $X_1 \ldots X_n$ be dependent zero mean Gaussian random variables, and the dependence rule is that for any $1<i \leq n$, $X_i$ only depends on $X_1 + \ldots + X_{i-1}$. You can imagine that $X_i \sim \mathcal{N}(0, \sigma (X_1+\ldots+X_{i-1}))$, for some function $\sigma$.

Questions:

Is $X_1+\ldots+X_n$ also a Gaussian? If so, what are its parameters? If not, is it at least sub-Gaussian? If so, what are its parameters? If not, what is the best concentration bound known for such random variable?

  • https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables#Correlated_random_variables – SmallChess Mar 07 '17 at 02:53
  • Wikipedia has exactly what you're looking for. Please take a look. – SmallChess Mar 07 '17 at 02:53
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    @StudentT It's not clear that this answers the question actually being asked here. For example, I don't think it is at all obvious that the conditions stated here necessarily rule out the $X_i$ following a distribution with Gaussian margins that is not multivariate Gaussian (maybe they do, but I think an answer would need to make that argument) – Glen_b Mar 07 '17 at 05:58
  • The first several questions have a nice answer (in the negative) at http://stats.stackexchange.com/questions/30159. Since we can't appreciably alter tail behavior through addition, the answer to the sub-Gaussian question is evident. It's unclear what you mean by "best" concentration bound. – whuber Mar 07 '17 at 15:44

1 Answers1

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Assume that $X_1$ is Gaussian, say $X_1 \sim N(\mu_1,\sigma_1^2)$. Then, the assumption is that

  • The conditional density of $X_2$ given $X_1=x_1$ is a Gaussian density. Note that we are not specifying what the mean and variance of this conditional density are and and how they depend on $x_1$ or $\mu_1$ or $\sigma_1^2$; just that the conditional density is a Gaussian density.
  • The conditional density of $X_3$ given that $X_1+X_2=y$ is a Gaussian density (once again, no specification about how this density depends on $y$ or on any of the parameters).
  • More generally, the conditional density of $X_i$ given that $X_1+X_2+\cdots + X_{i-1} =z$ is a Gaussian density (once again, no specification about how this density depends on $z$ or on any of the parameters).

The question asked is "Is the unconditional density of $X_n$ a Gaussian density?"

Let's start with $X_2$. We have that $$f_{X_2\mid X_1}(x_2 \mid X_1=x_1) = \frac{1}{h(x_1,\mu_1,\sigma_1)\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{x_2-g(x_1,\mu_1,\sigma_1)}{h(x_1,\mu_1,\sigma_1)}\right)^2\right)$$ where $g(x_1,\mu_1,\sigma_1)$ and $h(x_1,\mu_1,\sigma_1)$ are functions that describe how the mean and the standard deviation of this conditionally Gaussian density depend on $x_1$ and the mean and standard deviation of $X_1$. Since $g$ and $h$ are essentially arbitrary (we do need to insist that $h$ be nonnegative, of course, but that is a minor restriction), it is going to be the exception rather than the rule that $X_1$ and $X_2$ are jointly Gaussian. Thus, in general, the answer to the question asked is NO for the case $n=2$ and since the same argument can be extended to larger values of $n$, the answer to the question "Is the unconditional density of $X_n$ a Gaussian density?" is No, not in general.

What about the exceptional case? If $g$ and $h$ can be expressed in the form $$g(x_1,\mu_1,\sigma_1) = \mu_2 + {\rho}{\sigma_2}\left(\frac{x_1-\mu_1}{\sigma_1}\right), \quad h(x_1,\mu_1,\sigma_1) = \sigma_2\sqrt{1-\rho^2},\tag{1}$$ where $\sigma_2 > 0$ and $|\rho|\leq 1$, then $X_1$ and $X_2$ are indeed jointly Gaussian and $X_2 \sim N(\mu_2,\sigma_2^2)$. It follows that $X_1+X_2$ is also a Gaussian random variable. So, $X_3$ is conditionally Gaussian given $X_1+X_2 = y$, and if the amazing relationships described in $(1)$ apply mutatis mutandis in this case, then we get that $X_3$ and $X_1+X_2$ are jointly Gaussian, that $X_3$ is unconditionally Gaussian, and that $X_1+X_2+X_3$ is Gaussian. Continuing to believe that that such fortuitous coincidences will continue to occur exactly when we need them, we arrive at the conclusion that $X_n$ is unconditionally a Gaussian random variable, and indeed that $X_1, X_2, \ldots, X_n$ are jointly Gaussian random variables. But when $n > 8$ early in the morning, you need to be better than the Red Queen in Lewis Carroll's (Through the Looking Glass) who believed as many as eight impossible things before breakfast.

Dilip Sarwate
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