Are there random variables $X,Y$ such that $X,Y,X+Y \sim N(0,1)$? Let's suppose we have two, not necessarily independent, random variables $X,Y \sim N(0,1)$.
Then, is it possible that $X+Y \sim N(0,1)$?
What I have managed to show is that because $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ it would be necessary that $Cov(X,Y) = -1/2$.
But I'm still looking for an example or else a proof that it is not possible.

The pdf of the sum of two absolutely continuous random variables is simply given by the convolution of the pdf's of the variables. In our case, $f_X(x) = f_Y(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, so $f_{X + Y}(x) = \int_{- \infty}^\infty f_X(t)f_Y(x - t) dt = \frac{1}{4} \int_{- \infty}^\infty \exp(- \frac{t^2 + (x - t)^2}{2}) dt = \frac{1}{2 \pi} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2})^2 - 2 t \sqrt{2} \frac{x}{\sqrt{2}} + \frac{x^2}{2} + \frac{ x^2}{2}}{2}) dt = \frac{1}{2 \pi} e^{-\frac{ x^2}{4}} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2} - \frac{x}{\sqrt{2}})^2}{2}) dt = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \int_{- \infty}^\infty e^{- \frac{u^2}{2}} du = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \sqrt{2 \pi} = \frac{1}{2 \sqrt{\pi}} e^{- \frac{x^2}{4}}.$ In particular, the sum of two normals of mean $0$ and variance $1$ is not a normal of mean $0$ and variance $1$. I hope this helps. :)
