Sum of normally distributed random variables

In probability theory, if "X" and "Y" are independent random variables that are normally distributed, then "X" + "Y" is also normally distributed; i.e. if

:$X\; sim\; N(mu,\; sigma^2),$

and

:$Y\; sim\; N(\; u,\; au^2),$

and "X" and "Y" are independent, then

:$Z\; =\; X\; +\; Y\; sim\; N(mu\; +\; u,\; sigma^2\; +\; au^2).,$

Proofs

This proposition may be proved by any of several methods.

Proof using convolutions

By the total probability theorem, we have the probability density function of "z"

:$f\_Z(z)\; =\; iint\_\{x,y\}\; f\_\{X,Y,Z\}(x,y,z),\; dx,dy$

and since "X" and "Y" are independent, we get

:$,\; f\_Z(z)\; =\; iint\_\{x,y\}\; f\_X(x)\; f\_Y(y)\; f\_Z(z|x,y),\; dx,dy.$

But ƒ_"Z"("z"|"x,y") is trivially equal to "δ"("z" − ("x" + "y")), so

:$f\_Z(z)\; =\; iint\_\{x,y\}\; f\_X(x)\; f\_Y(y)\; delta(z\; -\; (x+y)),\; dx,dy$

where "δ" is Dirac's delta function. We substitute ("z" − "x") for "y":

:$f\_Z(z)\; =\; int\_\{x\}\; f\_X(x)\; f\_Y(z-x),\; dx$

which we recognize as a convolution of ƒ_"X" with ƒ_"Y".

Therefore the probability density function of the sum of two independent random variables "X" and "Y" with probability density functions ƒ and "g" is the convolution

:$(f*g)(x)=int\_\{-infty\}^infty\; f(u)\; g(x-u),du.,$

No generality is lost by assuming the two expected values μ and ν are zero. Thus the two densities are

:$f(x)\; =\; \{1\; over\; sigmasqrt\{2pi\; expleft(\{-x^2\; over\; 2sigma^2\}\; ight)\; ext\{\; and\; \}g(x)\; =\; \{1\; over\; ausqrt\{2pi\; expleft(\{-x^2\; over\; 2\; au^2\}\; ight).$

The convolution is

::$[mathrm\{constant\}]\; cdotint\_\{-infty\}^infty\; expleft(\{-u^2\; over\; 2sigma^2\}\; ight)\; expleft(\{-(x-u)^2\; over\; 2\; au^2\}\; ight),du$

:$=\; [mathrm\{constant\}]\; cdotint\_\{-infty\}^infty\; expleft(\{-(\; au^2\; u^2\; +\; sigma^2(x-u)^2)\; over\; 2sigma^2\; au^2\}\; ight),du.$

In simplifying this expression it saves some effort to recall this obvious fact that the context might later make easy to forget: The integral

:$int\_\{-infty\}^infty\; exp(-(u-A)^2),du$

actually does not depend on "A". This is seen be a simple substitution: "w" = "u" − "A", "dw" = "du", and the bounds of integration remain −∞ and +∞.

Now we have

:

where "constant" in this context means not depending on "x". The last integral does not depend on "x" because of the "obvious fact" mentioned above.

A probability density function that is a constant multiple of

:$expleft(\{-x^2\; over\; 2(sigma^2\; +\; au^2)\}\; ight)$

is the density of a normal distribution with variance σ² + τ². Although we did not explicitly develop the constant in this derivation, this is indeed the case.

Proof using characteristic functions

The characteristic function

:$varphi\_\{X+Y\}(t)\; =\; operatorname\{E\}left(e^\{it(X+Y)\}\; ight),$

of the sum of two independent random variables "X" and "Y" is just the product of the two separate characteristic functions:

:$varphi\_X\; (t)\; =\; operatorname\{E\}left(e^\{itX\}\; ight),$

and

:$varphi\_Y(t)\; =\; operatorname\{E\}left(e^\{itY\}\; ight),$

of "X" and "Y".

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

:$varphi\_X(t)\; =\; expleft(itmu\; -\; \{sigma^2\; t^2\; over\; 2\}\; ight).$

So

::$varphi\_\{X+Y\}(t)=varphi\_X(t)\; varphi\_Y(t)=expleft(itmu\; -\; \{sigma^2\; t^2\; over\; 2\}\; ight)cdot\; expleft(it\; u\; -\; \{\; au^2\; t^2\; over\; 2\}\; ight)$

:$=expleft(it(mu+\; u)\; -\; \{(sigma^2\; +\; au^2)\; t^2\; over\; 2\}\; ight).$

This is the characteristic function of the normal distribution with expected value μ + ν and variance σ² + τ².

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.

Correlated random variables

In the event that the variables "X" and "Y" are jointly normally distributed random variables, then "X" + "Y" is still normally distributed (see Multivariate normal distribution). In this case, one needs to consider

:$frac\{1\}\{2\; pi\; sigma\_x\; sigma\_y\; sqrt\{1-\; ho^2\; iint\_\{x,y\}\; exp\; left\; [\; -frac\{1\}\{2(1-\; ho^2)\}\; left(frac\{x^2\}\{sigma\_x^2\}\; +\; frac\{y^2\}\{sigma\_y^2\}\; -\; frac\{2\; ho\; x\; y\}\{sigma\_xsigma\_y\}\; ight)\; ight]\; delta(z\; -\; (x+y)),\; dx,dy.$

As above, one makes the substitution $y\; ightarrow\; z-x$

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

:$f\_Z(z)=expleft(-frac\{z^2\}\{2(sigma\_x^2+sigma\_y^2+2\; hosigma\_x\; sigma\_y)\}\; ight)\; dz$

If one considers instead "Z" = "X" − "Y", then one obtains :$f\_Z(z)=expleft(-frac\{z^2\}\{2(sigma\_x^2+sigma\_y^2-2\; hosigma\_x\; sigma\_y)\}\; ight)\; dz$

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

ee also

:*List of convolutions of probability distributions