Sums of Independent Normal Variables

Introduction

The sum of two independent normal variables is again normal, with a mean equal to the sum of the means and a variance equal to the sum of the variances. The parameter rules are easy to remember, but the convolution integral explains why they hold: multiplying the two densities and integrating produces another normal density.

The derivation depends on independence, because independence gives a product density. It also depends on the variables having normal densities with positive variances. We will evaluate the integral directly, then check the resulting parameters using expectation and variance.

Set-Up: The Convolution Density

Let \(X\) and \(Y\) be independent continuous random variables with densities \(f_X\) and \(f_Y\). For their sum \(S=X+Y\), the continuous convolution formula gives a density

f_S(s)=\int_{-\infty}^{\infty} f_X(x)f_Y(s-x)\,dx.

For a fixed sum \(s\), the integrand accounts for all possible values \(x\) of \(X\), paired with the value \(s-x\) of \(Y\). The integral adds their joint density along the set of pairs whose sum is \(s\). Independence is what allows that joint density to be written as \(f_X(x)f_Y(s-x)\).

Definition: A normal random variable \(X\sim N(\mu,\sigma^2)\), with \(\sigma^2>0\), has density

f_X(x)=\frac{1}{\sqrt{2\pi\sigma^2}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \qquad -\infty<x<\infty.

Here \(\mu\) is the mean and \(\sigma^2\) is the variance.

Deriving the Distribution of the Sum

Suppose \(X\sim N(\mu_1,\sigma_1^2)\) and \(Y\sim N(\mu_2,\sigma_2^2)\) are independent, where \(\sigma_1^2>0\) and \(\sigma_2^2>0\). Write \(v_1=\sigma_1^2\), \(v_2=\sigma_2^2\), \(V=v_1+v_2\), and \(d=s-\mu_1-\mu_2\). Substituting the normal densities into the convolution formula gives

f_S(s)=\frac{1}{2\pi\sqrt{v_1v_2}} \int_{-\infty}^{\infty} \exp\!\left\{ -\frac{1}{2}\left[ \frac{(x-\mu_1)^2}{v_1} +\frac{(s-x-\mu_2)^2}{v_2} \right]\right\}\,dx.

To evaluate this integral, let \(t=x-\mu_1\). Then \(s-x-\mu_2=d-t\). The quadratic in the exponent can be reorganized by completing the square:

\begin{aligned} \frac{t^2}{v_1}+\frac{(d-t)^2}{v_2} &=\frac{v_2t^2+v_1(d-t)^2}{v_1v_2}\\ &=\frac{V}{v_1v_2}\left(t-\frac{v_1d}{V}\right)^2+\frac{d^2}{V}. \end{aligned}

Indeed, expanding the right-hand side gives a coefficient of \(t^2\) equal to \(V/(v_1v_2)=1/v_1+1/v_2\), a coefficient of \(t\) equal to \(-2d/v_2\), and a constant equal to \(d^2/v_2\), matching the left-hand side. The term \(d^2/V\) does not depend on \(x\), so it can be taken outside the integral. The remaining integral is a centered Gaussian integral. With \(a=V/(v_1v_2)>0\), translation of the integration variable and the standard Gaussian integral give

\int_{-\infty}^{\infty} \exp\!\left[-\frac{a}{2} \left(t-\frac{v_1d}{V}\right)^2\right]\,dt =\sqrt{\frac{2\pi}{a}} =\sqrt{\frac{2\pi v_1v_2}{V}}.

Theorem: If \(X\sim N(\mu_1,\sigma_1^2)\) and \(Y\sim N(\mu_2,\sigma_2^2)\) are independent, with \(\sigma_1^2,\sigma_2^2>0\), then \(S=X+Y\sim N(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2)\).

Substituting the evaluated integral into the convolution expression proves the theorem:

\begin{aligned} f_S(s) &=\frac{1}{2\pi\sqrt{v_1v_2}} \exp\!\left(-\frac{d^2}{2V}\right) \sqrt{\frac{2\pi v_1v_2}{V}}\\ &=\frac{1}{\sqrt{2\pi V}} \exp\!\left(-\frac{(s-\mu_1-\mu_2)^2}{2V}\right). \end{aligned}

This is the density of a normal distribution with mean \(\mu_1+\mu_2\) and variance \(V=v_1+v_2\). The integral is valid for every real \(s\); the completed-square integral is finite because \(V/(v_1v_2)>0\). Thus the calculation identifies the density of the sum over its entire support. \(\square\)

Checking the Mean and Variance

The convolution calculation identifies the full distribution. The parameters can also be checked directly from the definitions of expectation and variance. Because normal variables have finite second moments, the following expectations and covariances exist.

Result: Under the theorem’s assumptions,

E[X+Y]=\mu_1+\mu_2, \qquad \operatorname{Var}(X+Y)=\sigma_1^2+\sigma_2^2.

Linearity of expectation gives \(E[X+Y]=E[X]+E[Y]=\mu_1+\mu_2\); this part does not require independence. For the variance, expand around the mean:

\begin{aligned} \operatorname{Var}(X+Y) &=E\!\left[\{(X-\mu_1)+(Y-\mu_2)\}^2\right]\\ &=E[(X-\mu_1)^2]+E[(Y-\mu_2)^2] +2E[(X-\mu_1)(Y-\mu_2)]. \end{aligned}

The first two terms are \(\sigma_1^2\) and \(\sigma_2^2\). Since \(X\) and \(Y\) are independent, their joint density factors, and the cross term is

\begin{aligned} E[(X-\mu_1)(Y-\mu_2)] &=\int_{\mathbb R}\int_{\mathbb R} (x-\mu_1)(y-\mu_2)f_X(x)f_Y(y)\,dy\,dx\\ &=\left\{\int_{\mathbb R}(x-\mu_1)f_X(x)\,dx\right\} \left\{\int_{\mathbb R}(y-\mu_2)f_Y(y)\,dy\right\}\\ &=0\cdot 0=0. \end{aligned}

The factorization is justified because the variables are independent and the centered variables have finite absolute first moments; the product is integrable under the product density. Consequently, the variance is \(\sigma_1^2+\sigma_2^2\), as found from the density. \(\square\)

Worked Examples

Worked Example: Unequal Means and Variances

Let \(X\sim N(2,4)\) and \(Y\sim N(5,9)\) be independent. Find the distribution of \(S=X+Y\).

The means are \(\mu_1=2\) and \(\mu_2=5\), while the variances are \(v_1=4\) and \(v_2=9\). The theorem gives

E[S]=2+5=7, \qquad \operatorname{Var}(S)=4+9=13.

Therefore \(S\sim N(7,13)\). In particular, the second parameter is \(13\), not the standard deviation. The standard deviation is \(\sqrt{13}\).

The integral confirms these values: its exponent contains \((s-2-5)^2/(2(4+9))=(s-7)^2/26\), and its normalizing constant is \(1/\sqrt{2\pi(13)}\). These are exactly the exponent and normalizing constant of the \(N(7,13)\) density.

Worked Example: Adding Measurement Error

Suppose a measurement \(X\) has distribution \(N(120,16)\), and an independent additive error \(Y\) has distribution \(N(0,9)\). Find the distribution and standard deviation of the recorded value \(S=X+Y\).

The recorded value has mean \(120+0=120\). Its variance is \(16+9=25\), so

S\sim N(120,25), \qquad \operatorname{SD}(S)=\sqrt{25}=5.

The error has mean zero, so it does not shift the expected measurement. It does add variance: the original variance is \(16\), and the error contributes another \(9\). Adding standard deviations instead would give \(4+3=7\), which is not the standard deviation of this sum.

Worked Example: Two Identically Distributed Normal Variables

Let \(X\) and \(Y\) be independent, each with distribution \(N(1,4)\). Evaluate the convolution density for \(S=X+Y\).

Here \(v_1=v_2=4\), \(V=8\), and \(d=s-2\). In the completed-square identity, \(v_1d/V=d/2\). Thus

\frac{(x-1)^2}{4}+\frac{(s-x-1)^2}{4} =\frac{(x-1-d/2)^2}{2}+\frac{d^2}{8}.

The convolution integral, including its density constant, becomes

\begin{aligned} f_S(s) &=\frac{1}{8\pi}\exp\!\left(-\frac{d^2}{16}\right) \int_{-\infty}^{\infty} \exp\!\left[-\frac{(x-1-d/2)^2}{4}\right]\,dx\\ &=\frac{1}{8\pi}\exp\!\left(-\frac{d^2}{16}\right)\sqrt{4\pi}\\ &=\frac{1}{\sqrt{16\pi}}\exp\!\left(-\frac{(s-2)^2}{16}\right). \end{aligned}

Since \(1/\sqrt{16\pi}=1/\sqrt{2\pi(8)}\), this is the density of \(N(2,8)\). The mean is \(1+1=2\), and the variance is \(4+4=8\).

Common Pitfalls and Why the Result Matters

The convolution derivation is useful because it establishes the whole distribution of the sum, not only its first two moments. Knowing a mean and a variance alone does not generally establish that a variable is normal. Here the normal shape follows from the explicit integral and its completed square.

Adding standard deviations instead of variances. For independent normal variables, add \(\sigma_1^2\) and \(\sigma_2^2\), then take a square root if a standard deviation is wanted.
Using the product-density integrand without independence. The convolution formula here uses \(f_X(x)f_Y(s-x)\) because the variables are independent. Without independence, that product need not be their joint density.
Assuming mean and variance prove normality. The moment calculation verifies the parameters, but it does not by itself prove the distributional claim. The convolution calculation supplies that proof.
Confusing the normal variance parameter with its standard deviation. In \(N(\mu,\sigma^2)\), the second argument is the variance. For example, \(N(7,13)\) has standard deviation \(\sqrt{13}\).

The practical rule is concise: for two independent normal variables, add their means and add their variances. The convolution proof explains why the result remains normal, while the moment calculation gives an independent check on the two parameters.

Check Your Understanding

Try these questions using the convolution and parameter rules developed above.

If independent \(X\sim N(3,2)\) and \(Y\sim N(-1,7)\), what is the distribution of \(X+Y\)?
In the completed-square derivation, which quantity is taken outside the integral because it does not depend on the integration variable?
Why does independence make the cross term \(E[(X-\mu_1)(Y-\mu_2)]\) equal to zero?
Independent normal measurements have variances \(1.5\) and \(2.5\). What are the variance and standard deviation of their sum?
Why would knowing only the mean and variance of a sum not be enough to conclude that it is normal?