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


 * $$Y \sim N(\nu, \tau^2)\,$$

and X and Y are independent, then


 * $$Z = X + Y \sim N(\mu + \nu, \sigma^2 + \tau^2).\,$$

Proofs
This proposition may be proved by any of several methods.

Proof using convolutions
By the total probability theorem, we have


 * $$f_Z(z) = \iint_{x\,y} f_{X,Y,Z}(x,y,z)\, dx\,dy$$

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 fZ(z|x,y) is trivially equal to


 * $$f_Z(z) = \iint_{x\,y} f_X(x) f_Y(y) \delta(z - (x+y))\, dx\,dy$$

where &delta; is Dirac's delta function. We substitute (z &minus; x) for y


 * $$f_Z(z) = \int_{x} f_X(x) f_Y(z-x)\, dx$$

which we recognize as a convolution of $$f_X$$ with $$f_Y$$.

Therefore the probability density function of the sum of two independent random variables X and Y with probability density functions f 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 &mu; and &nu; are zero. Thus the two densities are


 * $$f(x) = {1 \over \sigma\sqrt{2\pi}} \exp\left({-x^2 \over 2\sigma^2}\right)$$ and $$g(x) = {1 \over \tau\sqrt{2\pi}} \exp\left({-x^2 \over 2\tau^2}\right).$$

The convolution is


 * $$[\mathrm{constant}]\cdot\int_{-\infty}^\infty \exp\left({-u^2 \over 2\sigma^2}\right) \exp\left({-(x-u)^2 \over 2\tau^2}\right)\,du$$


 * $$=[\mathrm{constant}]\cdot\int_{-\infty}^\infty \exp\left({-(\tau^2 u^2 + \sigma^2(x-u)^2) \over 2\sigma^2 \tau^2} \right)\,du. $$

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 &minus; A, dw = du, and the bounds of integration remain &minus;&infin; and +&infin;.

Now we have


 * $$[\mathrm{constant}]\cdot\int_{-\infty}^\infty \exp\left({-(\tau^2 u^2 + \sigma^2(x-u)^2) \over 2\sigma^2 \tau^2} \right)\,du$$


 * $$=[\mathrm{constant}]\cdot\int_{-\infty}^\infty \exp\left({-(\tau^2+\sigma^2)(u-{\sigma^2 \over \sigma^2+\tau^2}x)^2 \over 2\sigma^2\tau^2} + {-x^2 \over 2(\sigma^2 + \tau^2)}\right) \,du$$


 * $$=[\mathrm{constant}]\cdot \exp\left({-x^2 \over 2(\sigma^2 + \tau^2)}\right) \cdot \int_{-\infty}^\infty \exp\left({-(\tau^2+\sigma^2)(u-{\sigma^2 \over \sigma^2+\tau^2}x)^2 \over 2\sigma^2\tau^2}\right) \,du$$


 * $$=[\mathrm{constant}]\cdot \exp\left({-x^2 \over 2(\sigma^2 + \tau^2)}\right) \cdot [\mathrm{constant}], $$

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


 * $$\exp\left({-x^2 \over 2(\sigma^2 + \tau^2)}\right)$$

is the density of a normal distribution with variance &sigma;2 + &tau;2. Although we did not explicitly develop the constant in this derivation, this is indeed the case.

Proof using characteristic functions
The characteristic function


 * $$\varphi_{X+Y}(t) = \operatorname{E}\left(e^{it(X+Y)}\right)\,$$

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


 * $$\varphi_X (t) = \operatorname{E}\left(e^{itX}\right)\,$$

and


 * $$\varphi_Y(t) = \operatorname{E}\left(e^{itY}\right)\,$$

of X and Y.

The characteristic function of the normal distribution with expected value &mu; and variance &sigma;2 is


 * $$\varphi_X(t) = \exp\left(it\mu - {\sigma^2 t^2 \over 2}\right).$$

So


 * $$\varphi_{X+Y}(t)=\varphi_X(t) \varphi_Y(t)

=\exp\left(it\mu - {\sigma^2 t^2 \over 2}\right) \cdot \exp\left(it\nu - {\tau^2 t^2 \over 2}\right)$$


 * $$=\exp\left(it(\mu+\nu) - {(\sigma^2 + \tau^2) t^2 \over 2}\right).$$

This is the characteristic function of the normal distribution with expected value &mu; + &nu; and variance &sigma;2 + &tau;2.

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.