I learned that the standard normal distribution is unique because the mean and variance are fixed at 0 and 1 respectively. By this fact, I wonder if any two standard random variables must be independent.

  • 13
    $\begingroup$ Why should they be..? Independence has nothing to do with distribution. $\endgroup$
    – Tim
    Commented Dec 7, 2016 at 15:03
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    $\begingroup$ Consider $X$ and $X$. They are not independent. $\endgroup$
    – djechlin
    Commented Dec 7, 2016 at 18:57
  • $\begingroup$ You may find this helpful from a practical standpoint. stats.stackexchange.com/questions/15011/… $\endgroup$ Commented Dec 7, 2016 at 19:34
  • $\begingroup$ In addition to the nice examples given consider generally a bivariate normal distribution with N(0,!) marginal distributions. It is possible to have any correlation between -1 and 1. The examples below are all special cases. As an aside it is possible for two standard normal variables to be dependent but not have a bivariate distribution. $\endgroup$ Commented Dec 7, 2016 at 22:03
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    $\begingroup$ I notice Batman gives a general result that may be the same as what I am suggesting. The Y = -X case has correlation -1 and so is a degenerate form of a bivariate normal. I haven't seen an example here(on this post) that illustrate a case that isn't bivariate normal. $\endgroup$ Commented Dec 7, 2016 at 22:10

4 Answers 4


The answer is no. For example, if $X$ is a standard random variable, then $Y=-X$ follows the same statistics, but $X$ and $Y$ are clearly dependent.


No, there is no reason to believe that any two standard gaussians are independent.

Here's a simple mathematical construction. Suppose that $X$ and $Y$ are two independent standard normal variables. Then the pair

$$ X, \frac{X + Y}{\sqrt{2}}$$

are two dependent standard normal variables. So, as long as their are two independent normal variables, there must be two dependent ones.

The second variable is normal because any linear combination of independent normal variables is again normal. The $\sqrt{2}$ is there to make the variance equal to $1$.

$$ V \left(\frac{X + Y}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}^2} (V(X) + V(Y)) = 1 $$

Intuitively, these are dependent because knowing the value of $X$ gives you additional information you can use to predict the value of the second variable. For example, if you know that $X = x$, then the conditional expectation of the second variable is

$$ E \left[\frac{X + Y}{\sqrt{2}} \mid X = x \right] = \frac{x}{\sqrt{2}} $$


Here's a fairly wide answer:

Let $X,Y$ be jointly Gaussian random variables (i.e. for any $a,b$ real numbers, $a X + bY$ has a Gaussian distribution). Then, $X$ and $Y$ are independent if and only if $E[(X-E[X])(Y-E[Y])]=0$ (i.e. they are uncorrelated). See these notes, for example, for details.

How can you generate standard normal random variables which are not independent? Pick your favorite matrix of the form $\Sigma=\begin{bmatrix} 1 & p \\ p & 1 \end{bmatrix}$ such that $(\lambda-1)^2 - p^2$ has positive roots in $\lambda$. Then, apply the Cholesky decompositon to $\Sigma= R R^T$. Then, take two independent standard normal random variables $U,V$ and then the vector $R \begin{bmatrix} U \\ V \end{bmatrix}$ has standard normal components, but the components are independent if and only if $p=0$.


A non-bivariate normal example (as Michael Chernick suggests in the comments):

Let $f_{X,Y}(x,y) = \begin{cases} \frac{1}{\pi} e^{-\frac{x^2+y^2}{2}} & xy\geq 0 \\ 0 & o.w. \end{cases}$.

This is not a bivariate normal distribution, but a simple integral shows that both marginals are standard normal. They're obviously not independent since $f_{X,Y}(x,y) \neq f_X(x) f_Y(y)$.


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