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Consider $N\sim\mathcal N(0,1)$ a standard normal random variable. Can we find a dependent random variable $D$, such that the random variable $M=N+D$ is still a standard normal distribution? I would also like to choose the variance of D that would be typically smaller than 1. $D$ would represent a kind of random noise that do not change the random distribution when added.

If $D$ is an independent distribution with average 0, the average is still $0$ but the variance of $D$ is added to that of $N$. This property is not true when the random variables are correlated. We might determine $D$ with its conditional distribution : $f(D | N)$

I have no idea how to solve this problem.

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    $\begingroup$ What about $D = -2N$? $\endgroup$ – Vasily Mitch Jun 10 at 20:01
  • $\begingroup$ Actually, as said, I would like to choose the variance of D. -2N has a variance of 4. (D=0 also works with a variance of 0) What if I want a variance of 0.1, for instance ? $\endgroup$ – Arnaud Mégret Jun 10 at 20:12
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    $\begingroup$ Let $D=-2N$ if $|N| \le k$ for some $k$ and $D=0$ if $|N| \gt k$. Make $k$ as close to $0$ as you want to get the variance of $D$ you want $\endgroup$ – Henry Jun 10 at 20:17
  • $\begingroup$ True. Well, that is definitely not the distribution I need as a random noise. But mathematically this perfectly solves the problem I described. $\endgroup$ – Arnaud Mégret Jun 10 at 20:25
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I finally found a simple and smooth solution. (sorry for having asked a question I did not spend enough time to try to solve myself before)

Consider a random variable $E \sim \mathcal{N}(0,\sigma^2)$ independent from N and $D = \frac{N+E}{\sqrt{1+\sigma^2}}-N$ then $M = \frac{N+E}{\sqrt{1+\sigma^2}} \sim \mathcal{N}(0,1)$

The variance of D is function of $\sigma^2$ so we can set $\sigma^2$ in order to have the choosen variance for $D$.

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