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.
It is possible. Suppose
$$(X,Y)'\sim\mathcal{N}\bigg(\begin{pmatrix}0\\0\end{pmatrix},\begin{pmatrix}1&amp;-\frac{1}{2}\\-\frac{1}{2}&amp;1\end{pmatrix}\bigg)$$
And consider the vector $(1,1)$. Then $(1,1)\cdot (X,Y)'=X+Y$ is distributed Normal with mean
$$(1,1)\cdot\begin{pmatrix}0\\0\end{pmatrix}=0$$
and variance
$$(1,1)\cdot \begin{pmatrix}1&amp;-\frac{1}{2}\\-\frac{1}{2}&amp;1\end{pmatrix}\cdot\begin{pmatrix}1\\1\end{pmatrix}=1$$
Which is what you want.
Note: By a similar argument, using the vectors $(1,0)$ and $(0,1)$ you have that $X$ and $Y$ are standard normals on their own.
Note: One should be careful and make sure that the variance-covariance matrix is a valid one. One way of doing it is by directly checking it is possitive semi-definite, which is true in this case as the first principal minor is $1$ and the determinant is $5/4$, both strictly positive.
There is an easy second way to verify that a $2$ by $2$ matrix is a valid variance-covariance matrix. Call this matrix $\Sigma$. If you can verify that the diagonal entries are strictly positive, the matrix is symmetric and that there exists $\rho\in[-1,1]$, a valid correlation coefficient, such that $\Sigma_{21}=\Sigma_{12}=\rho\cdot(\Sigma_{11}\cdot\Sigma_{22})^{1/2}$, you are done. In this case $\rho=-\frac{1}{2}$.
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. :)