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Sorry if the question is ill-phrased, but I'm pretty new to this and I have the following situation:

I know that samples X is drawn from a Gaussian Distribution G(u, v). Now that if u is substituted with another Gaussian random variable, what would this conjugate distribution be? Would it still be a Gaussian?

Edit:

Just to make this more clear, given the following experiment:

  1. sample a set of points Xs from a normal distribution with known mean and var,
  2. generate multiple normal distributions Ds where each D has one of Xs being mean and all Ds share the same known var,
  3. sample a set of Ys from each D, and combine the Ys together.

I'm basically trying to get a clue on what the distribution of Ys is now, is it still Gaussian, what would the mean and var be?

Edit:

Aaron's demo in his answer is pretty clear. Although in my study, I do tend to use pretty large n and m, per my own experiments, the shape of binning on the resulting points is still pretty Gaussian-like, it got me wondering again if there's theoretical proof behind that given large n and m, it is guaranteed to be Gaussian?

Here's a graph from one of my experiments (I ran it multiple times with different mean and variance, they all look similar): An example with n=1000, m=10000

The above graph is generated with n=1000, m=10000, mean=0.95, std_dev1=0.07, std_dev2=0.7

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Yes (assuming they are independent). Changing the mean of a variable from $0$ to $x$ is equivalent to adding $x$ to that variable by linearity of expectation. And if $x$ is normally distributed, the addition of two independent normal variables is still normal.

To be more precise, let $X \sim \mathcal{N}(\mu, \sigma_1^2)$ and $Y \sim \mathcal{N}(0, \sigma_2^2)$, and then the variable of interest is:

$$X + Y \sim \mathcal{N}(\mu, \sigma_1^2 + \sigma_2^2)$$

which is still normal.

If they are dependent, then I don't believe we can conclude very much about the resulting distribution.


Edit: To address the revised question, let $n$ be the number of times we sample $X$ in step 1, and then $m$ be the number of times we sample from $x_i + Y$ in step 3. My previous response addressed the case where $m = 1$.

For $m > 1$, each of the $m$ "subsamples" depends on the same realization $x_i$. You can think of this as $n$ separate normal distributions, each shifted by amounts that are normally distributed after being sampled $m$ times. There may not be any nicer description, but this depends on what you want to analyze and what questions you want to answer about the resulting distribution. Also note that depending on the relative sizes of $n$ and $m$ (e.g. if $m$ tends to infinity while $n$ stays constant, or $n$ tends to infinity while $m$ stays constant) or the relative sizes of $\sigma_1^2$ and $\sigma_2^2$, you may be able to come up with a suitable approximation.

For example, the following histogram depicts a random sample of $nm$ points according to your scheme where: $n = 2$, $m = 10000$, $\mu = 0$, $\sigma_1^2 = 50$, and $\sigma_2^2 = 1$.

                 example distribution

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    $\begingroup$ Well, you can still conclude a lot, depending on the nature of the dependency. If X and Y are Bivariate Normal with correlation coefficient $\rho$, and marginals as you stated, then $X + Y \sim \mathcal{N}(\mu, \sigma_1^2 + \sigma_2^2 + 2 \sigma_1 \sigma_2 \rho)$ $\endgroup$ – Mark L. Stone May 7 '16 at 0:03
  • $\begingroup$ Hi Aaron, thanks a lot! But I don't see how the result distribution is X + Y? Is it equivalent to the following experiment: 1. sample a set of points Xs from a normal distribution with known mean and var, 2. generate multiple normal distributions Ds where each D has one of Xs being mean and Ds share the same known var, 3. sample a set of Ys from each D, and combine the Ys together, now is the distribution of Ys the same as X + Y in your answer? Thanks! $\endgroup$ – weil0ng May 9 '16 at 16:55
  • $\begingroup$ Hi Mark, thanks for the additional info, but I'm assuming they are independent actually. $\endgroup$ – weil0ng May 9 '16 at 16:59
  • $\begingroup$ Based on these changes, what you are now effectively suggesting in steps 1-2 is that you repeat the experiment $n$ times independently. Then you get the same equation from my answer to give you each $D_i = X + Y$ for the $i^{th}$ experiment. Therefore, you are just sampling $nm$ times from $X + Y$, where $m$ is the number of times you sample from each $D_i$. However, for step 3, how do you intend to "combine" all of these samples? $\endgroup$ – Aaron Voelker May 9 '16 at 19:36
  • $\begingroup$ Correction: each of the $m$ subsamples needs to be conditioned on the same $X$. Once it's clear how you are combining all of these subsamples I will try to edit my question to address all of this. $\endgroup$ – Aaron Voelker May 9 '16 at 19:44

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