In probability theory, calculation of the sum of normally distributed random variables is an instance of the arithmetic of random variables. This is not to be confused with the sum of normal distributions which forms a mixture distribution. Addition of random variables, on the other hand, are the convolution of their probability distributions.
Independent random variables Let X and Y be independent random variables that are normally distributed (and therefore also jointly so), then their sum is also normally distributed. i.e., if
X ∼ N ( μ X , σ X 2 ) {\displaystyle X\sim N(\mu _{X},\sigma _{X}^{2})}
Y ∼ N ( μ Y , σ Y 2 ) {\displaystyle Y\sim N(\mu _{Y},\sigma _{Y}^{2})}
Z = X + Y , {\displaystyle Z=X+Y,}
then
Z ∼ N ( μ X + μ Y , σ X 2 + σ Y 2 ) . {\displaystyle Z\sim N(\mu _{X}+\mu _{Y},\sigma _{X}^{2}+\sigma _{Y}^{2}).}
This means that the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances (i.e., the square of the standard deviation is the sum of the squares of the standard deviations). In order for this result to hold, the assumption that X and Y are independent cannot be dropped, although it can be weakened to the assumption that X and Y are jointly, rather than separately, normally distributed. (See here for an example.) The result about the mean holds in all cases, while the result for the variance requires uncorrelatedness, but not independence.
Proofs
Proof using characteristic functions The characteristic function
φ X + Y ( t ) = E ( e i t ( X + Y ) ) {\displaystyle \varphi _{X+Y}(t)=\operatorname {E} \left(e^{it(X+Y)}\right)}
of the sum of two independent random variables X and Y is just the product of the two separate characteristic functions:
φ X ( t ) = E ( e i t X ) , φ Y ( t ) = E ( e i t Y ) {\displaystyle \varphi _{X}(t)=\operatorname {E} \left(e^{itX}\right),\qquad \varphi _{Y}(t)=\operatorname {E} \left(e^{itY}\right)}
of X and Y. The characteristic function of the normal distribution with expected value μ and variance σ2 is
φ ( t ) = exp ( i t μ − σ 2 t 2 2 ) . {\displaystyle \varphi (t)=\exp \left(it\mu -{\sigma ^{2}t^{2} \over 2}\right).}
So
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