In probability theory, calculation of the sum of normally distributed random variables is an instance of the arithmetic of random variables, which can be quite complex based on the probability distributions of the random variables involved and their relationships. This is not to be confused with the sum of normal distributions which forms a mixture distribution. 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 then 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 of the sum of two independent random variables X and Y is just the product of the two separate characteristic functions: of X and Y. The characteristic function of the normal distribution with expected value μ and variance σ2 is So This is the characteristic function of the normal distribution with expected value and variance Finally, recall that no two distinct distributions can both have the same characteristic function, so the distribution of X + Y must be just this normal distribution. Proof using convolutions For independent random variables X and Y, the distribution fZ of Z = X + Y equals the convolution of fX and fY: Given that fX and fY are normal densities, Substituting into the convolution: Defining , and completing the square: The expression in the integral is a normal density distribution on x, and so the integral evaluates to 1. The desired result follows: Using the convolution theorem It can be shown that the Fourier transform of a Gaussian, , is By the convolution theorem: Geometric proof First consider the normalized case when X, Y ~ N(0, 1), so that their PDFs are and Let Z = X