<s>[INST] 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.
 [/INST] 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. :)
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