Inprobability theory, calculation of thesum of normally distributed random variables is an instance of the arithmetic ofrandom variables.

This is not to be confused with thesum of normal distributions which forms amixture distribution.

LetX andY beindependentrandom variables that arenormally distributed (and therefore also jointly so), then their sum is also normally distributed. i.e., if

${\displaystyle X\sim N({\mu }_X,{\sigma }_X^2)}$

${\displaystyle Y\sim N({\mu }_Y,{\sigma }_Y^2)}$

${\displaystyle Z=X+Y,}$

then

${\displaystyle Z\sim N({\mu }_X+{\mu }_Y,{\sigma }_X^2+{\sigma }_Y^2).}$

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).^[1]

In order for this result to hold, the assumption thatX andY are independent cannot be dropped, although it can be weakened to the assumption thatX andY arejointly, rather than separately, normally distributed.^[2] (Seehere for an example.)

The result about the mean holds in all cases, while the result for the variance requires uncorrelatedness, but not independence.

Thecharacteristic function

${\displaystyle {\phi }_{X+Y}(t)=\mathrm{E}\left(e^{it(X+Y)}\right)}$

of the sum of two independent random variablesX andY is just the product of the two separate characteristic functions:

${\displaystyle {\phi }_X(t)=\mathrm{E}\left(e^{itX}\right),{\phi }_Y(t)=\mathrm{E}\left(e^{itY}\right)}$

ofX andY.

The characteristic function of the normal distribution with expected value μ and variance σ² is

${\displaystyle \phi (t)=\exp \left(it\mu -\frac{{\sigma }^2{t}^2}{2}\right).}$

So

This is the characteristic function of the normal distribution with expected value${\displaystyle {\mu }_X+{\mu }_Y}$ and variance${\displaystyle {\sigma }_X^2+{\sigma }_Y^2}$

Finally, recall that no two distinct distributions can both have the same characteristic function, so the distribution ofX + Y must be just this normal distribution.

For independent random variablesX andY, the distributionf_Z ofZ =X + Y equals the convolution off_X andf_Y:

${\displaystyle f_Z(z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{f}_Y(z-x)f_X(x)dx}$

Given thatf_X andf_Y are normal densities,

${\displaystyle \begin{array}{r}{f}_X(x)=\mathcal{N}(x;{\mu }_X,{\sigma }_X^2)=\frac{1}{\sqrt{2\pi }{\sigma }_X}{e}^{-(x-{\mu }_X{)}^2/(2{\sigma }_X^2)}\\ f_Y(y)=\mathcal{N}(y;{\mu }_Y,{\sigma }_Y^2)=\frac{1}{\sqrt{2\pi }{\sigma }_Y}{e}^{-(y-{\mu }_Y{)}^2/(2{\sigma }_Y^2)}\end{array}}$

Substituting into the convolution:

Defining${\displaystyle {\sigma }_Z=\sqrt{{\sigma }_X^2+{\sigma }_Y^2}}$, andcompleting the square:

The expression in the integral is a normal density distribution onx, and so the integral evaluates to 1. The desired result follows:

${\displaystyle f_Z(z)=\frac{1}{\sqrt{2\pi }{\sigma }_Z}\exp \left[-\frac{(z-({\mu }_X+{\mu }_Y){)}^2}{2{\sigma }_Z^2}\right]}$

It can be shown that theFourier transform of a Gaussian,${\displaystyle f_X(x)=\mathcal{N}(x;{\mu }_X,{\sigma }_X^2)}$, is^[3]

${\displaystyle \mathcal{F}\{f_X\}=F_X(\omega )=\exp \left[-j\omega {\mu }_X\right]\exp \left[-\frac{{\sigma }_X^2{\omega }^2}{2}\right]}$

By theconvolution theorem:

First consider the normalized case whenX,Y ~N(0, 1), so that theirPDFs are

${\displaystyle f(x)=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}}$

and

${\displaystyle g(y)=\frac{1}{\sqrt{2\pi }}{e}^{-y^2/2}.}$

LetZ =X + Y. Then theCDF forZ will be

${\displaystyle z\mapsto {\int }_{x+y\le z}f(x)g(y)dxdy.}$

This integral is over the half-plane which lies under the linex+y =z.

The key observation is that the function

${\displaystyle f(x)g(y)=\frac{1}{2\pi }{e}^{-(x^2+y^2)/2}}$

is radially symmetric. So we rotate the coordinate plane about the origin, choosing new coordinates${\displaystyle x^{\prime },y^{\prime }}$ such that the linex+y =z is described by the equation${\displaystyle x^{\prime }=c}$ where${\displaystyle c=c(z)}$ is determined geometrically. Because of the radial symmetry, we have${\displaystyle f(x)g(y)=f(x^{\prime })g(y^{\prime })}$, and the CDF forZ is

${\displaystyle {\int }_{x^{\prime }\le c,y^{\prime }\in \mathbb{R}}f(x^{\prime })g(y^{\prime })d{x}^{\prime }d{y}^{\prime }.}$

This is easy to integrate; we find that the CDF forZ is

${\displaystyle {\int }_{-\mathrm{\infty }}^{c(z)}f(x^{\prime })d{x}^{\prime }=\mathrm{\Phi }(c(z)).}$

To determine the value${\displaystyle c(z)}$, note that we rotated the plane so that the linex+y =z now runs vertically withx-intercept equal toc. Soc is just the distance from the origin to the linex+y =z along the perpendicular bisector, which meets the line at its nearest point to the origin, in this case${\displaystyle (z/2,z/2)}$. So the distance is${\displaystyle c=\sqrt{(z/2{)}^2+(z/2{)}^2}=z/\sqrt{2}}$, and the CDF forZ is${\displaystyle \mathrm{\Phi }(z/\sqrt{2})}$, i.e.,${\displaystyle Z=X+Y\sim N(0,2).}$

Now, ifa,b are any real constants (not both zero) then the probability that${\displaystyle aX+bY\le z}$ is found by the same integral as above, but with the bounding line${\displaystyle ax+by=z}$. The same rotation method works, and in this more general case we find that the closest point on the line to the origin is located a (signed) distance

${\displaystyle \frac{z}{\sqrt{a^2+b^2}}}$

away, so that

${\displaystyle aX+bY\sim N(0,a^2+b^2).}$

The same argument in higher dimensions shows that if

${\displaystyle X_i\sim N(0,{\sigma }_i^2),i=1,\dots ,n,}$

then

${\displaystyle X_1+\cdots +X_n\sim N(0,{\sigma }_1^2+\cdots +{\sigma }_n^2).}$

Now we are essentially done, because

${\displaystyle X\sim N(\mu ,{\sigma }^2)\iff \frac{1}{\sigma }(X-\mu )\sim N(0,1).}$

So in general, if

${\displaystyle X_i\sim N({\mu }_i,{\sigma }_i^2),i=1,\dots ,n,}$

then

${\displaystyle \sum _{i=1}^n{a}_i{X}_i\sim N\left(\sum _{i=1}^n{a}_i{\mu }_i,\sum _{i=1}^n(a_i{\sigma }_i{)}^2\right).}$

In the event that the variablesX andY are jointly normally distributed random variables, thenX + Y is still normally distributed (seeMultivariate normal distribution) and the mean is the sum of the means. However, the variances are not additive due to the correlation. Indeed,

${\displaystyle {\sigma }_{X+Y}=\sqrt{{\sigma }_X^2+{\sigma }_Y^2+2\rho {\sigma }_X{\sigma }_Y},}$

where ρ is thecorrelation. In particular, whenever ρ < 0, then the variance is less than the sum of the variances ofX andY.

Extensions of this result can be made for more than two random variables, using thecovariance matrix.

Note that the condition thatX andY are known to be jointly normally distributed is necessary for the conclusion that their sum is normally distributed to apply. It is possible to have variablesX andY which are individually normally distributed, but have a more complicated joint distribution. In that instance,X + Y may of course have a complicated, non-normal distribution. In some cases, this situation can be treated usingcopulas.

In this case (withX andY having zero means), one needs to consider

${\displaystyle \frac{1}{2\pi {\sigma }_x{\sigma }_y\sqrt{1-{\rho }^2}}{\iint }_{xy}\exp \left[-\frac{1}{2(1-{\rho }^2)}\left(\frac{x^2}{{\sigma }_x^2}+\frac{y^2}{{\sigma }_y^2}-\frac{2\rho xy}{{\sigma }_x{\sigma }_y}\right)\right]\delta (z-(x+y))\mathrm{d}x\mathrm{d}y.}$

As above, one makes the substitution${\displaystyle y\to z-x}$

This integral is more complicated to simplify analytically, but can be done easily using a symbolic mathematics program. The probability distributionf_Z(z) is given in this case by

${\displaystyle f_Z(z)=\frac{1}{\sqrt{2\pi }{\sigma }_{+}}\exp \left(-\frac{z^2}{2{\sigma }_{+}^2}\right)}$

where

${\displaystyle {\sigma }_{+}=\sqrt{{\sigma }_x^2+{\sigma }_y^2+2\rho {\sigma }_x{\sigma }_y}.}$

If one considers insteadZ =X − Y, then one obtains

${\displaystyle f_Z(z)=\frac{1}{\sqrt{2\pi ({\sigma }_x^2+{\sigma }_y^2-2\rho {\sigma }_x{\sigma }_y)}}\exp \left(-\frac{z^2}{2({\sigma }_x^2+{\sigma }_y^2-2\rho {\sigma }_x{\sigma }_y)}\right)}$

which also can be rewritten with

${\displaystyle {\sigma }_{X-Y}=\sqrt{{\sigma }_x^2+{\sigma }_y^2-2\rho {\sigma }_x{\sigma }_y}.}$

The standard deviations of each distribution are obvious by comparison with the standard normal distribution.

• Propagation of uncertainty
• Algebra of random variables
• Stable distribution
• Standard error (statistics)
• Ratio distribution
• Product distribution
• Slash distribution
• List of convolutions of probability distributions

• Normal distribution
• Theory of probability distributions

• Articles with short description
• Short description is different from Wikidata