# Adding a dependent random variable to a standard normal variable without changing its distribution

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.

• What about $D = -2N$? – Vasily Mitch Jun 10 at 20:01
• 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 ? – Arnaud Mégret Jun 10 at 20:12
• 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 – Henry Jun 10 at 20:17
• True. Well, that is definitely not the distribution I need as a random noise. But mathematically this perfectly solves the problem I described. – Arnaud Mégret Jun 10 at 20:25

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 $$D$$.