Sum of normally distributed random variables

Property

1. If $X\sim N(\mu _{X},\sigma _{X}^{2})$ and $Y\sim N(\mu _{Y},\sigma _{Y}^{2})$ are independent normal random variables, then:
   • Their sum is normally distributed with $Z=X+Y\sim N(\mu _{X}+\mu _{Y},\sigma _{X}^{2}+\sigma _{Y}^{2})$.

Proof

$p(Z)=\iint _{X\,Y}p(X,Y,Z)\,dX\,dY$

$p(Z)=\iint _{X\,Y}p(X)p(Y)p(Z)\,dX\,dY$

$p(Z)=\iint _{X\,Y}p(X)p(Y)\delta (Z-(X+Y))\,dX\,dY$

$p(Z)=\int _{X}p_{x}(X)p_{y}(Z-X)\,dX$

$p(Z)=\int _{X}{\frac {1}{{\sqrt {2\pi }}\sigma _{x}}}e^{-{\frac {(X-\mu _{x})^{2}}{2\sigma _{x}^{2}}}}{\frac {1}{{\sqrt {2\pi }}\sigma _{y}}}e^{-{\frac {((Z-X)-\mu _{y})^{2}}{2\sigma _{y}^{2}}}}\,dX$

$p(Z)={\frac {1}{2\pi \sigma _{x}\sigma _{y}}}\int _{X}e^{-{\frac {1}{2}}[{\frac {(X-\mu _{x})^{2}}{\sigma _{x}^{2}}}+{\frac {((Z-X)-\mu _{y})^{2}}{\sigma _{y}^{2}}}]}\,dX$