How can I prove that the sum of $X_1, X_2, \ldots,X_n$ random variables, all of which have normal distributions $N(\mu_i, \sigma_i)$, is a random variable that is itself normally distributed with mean $$\mu =\sum_{i=1}^n \mu_i$$

and variance

$$\sigma^2 = \sum_{i=1}^n \sigma_i^2$$

Edit: I forgot to add that this was with the assumption that all $X_1, X_2,\ldots,X_n$ are independent.

  • $\begingroup$ As did said, in general your claim is wrong. But it holds if you additionally assume that the $X_i$ are independent. Note that it suffices to establish your claim for $n=2$ (you can use induction for the general case then). $\endgroup$ – martini Nov 16 '12 at 10:27
  • $\begingroup$ Quite many proofs can be found in Wikipedia - Sum of normally distributed random variables $\endgroup$ – Golob Nov 16 '12 at 10:42
  • $\begingroup$ For the case $n=2$ (as suggested by @martini followed by induction), a proof without using moment-generating functions can be found here on this site. $\endgroup$ – Dilip Sarwate Nov 16 '12 at 17:28
  • $\begingroup$ One way is two explicitly compute the convolution of the density functions. See my answer below. $\endgroup$ – Michael Hardy Nov 16 '12 at 18:11

I figured this out on the transit ride to work this morning. I used the moment-generating function, which may be the same thing as the "Probability-generating function" that is recommended in the answer above? I'm not sure. Anyway...

If we let $Y = X_{1} + X_{2} + \space \cdots \space + X_{n}$ , then the moment-generating function of $Y$ is given by:

$$ M_{Y}(t) = E[e^{t(X_{1} \space + \space X_{2} \space + \space \cdots \space + \space X_{n})}] = \prod_{i=1}^{n}E[e^{tX_{i}}] = \prod_{i=1}^{n}M_{X_{i}}(t) $$

If we then find

$$M_{X_{i}}(t) = \exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}]$$

then we have

$$ M_{Y}(t) = \prod_{i=1}^{n}\exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}] $$

then by properties of exponents we have

$$ M_{Y}(t) = \prod_{i=1}^{n}\exp[\mu_{i}t + \frac{\sigma_{i}^2t^2}{2}] = \prod_{i=1}^{n}\exp[\mu_{i}t]\exp[\frac{\sigma_{i}^2t^2}{2}] = \exp[t(\sum_{i=1}^{n}\mu_{i}) + \frac{t^2}{2}(\sum_{i=1}^{n}\sigma_{i}^2)] $$


$$ M_{Y}(t) = \exp[t(\sum_{i=1}^{n}\mu_{i}) + \frac{t^2}{2}(\sum_{i=1}^{n}\sigma_{i}^2)] $$ this implies that Y is normally distributed with mean $\space\sum_{i=1}^{n}\mu_{i}$ and variance $ \space \sum_{i=1}^{n}\sigma_{i}^2\space$


Use Probability-generating functions.


The result is false: consider $n=2$, $\mu_1=0\ne\sigma_1^2$ and $X_2=SX_1$ where $S=\pm1$ is Bernoulli centered and independent of $X_1$. Then $X_1$ and $X_2$ are normal $(0,\sigma_1^2)$ but $X_1+X_2$ is not normal $(0,2\sigma_1^2)$ since $X_1+X_2$ is not normal.

  • $\begingroup$ that 's because $X_1$ and $X_2$ are not independent. I guees Cory Gross forgot to add that assumption $\endgroup$ – Golob Nov 16 '12 at 10:40

There are many proofs by mathematical induction that take the following form:

  • The case $n=1$ is vacuously true.
  • Case $n+1$ follows easily from case $n$ if $n\ge 2$, but that step is impossible when $n=1$, and the proof of the induction step uses both case $n$ and case $2$.
  • The substantial part of the proof is case $2$.

This is one of those. Prove it for a sum of just two random variables and the rest is easy.

Suppose $X_i\sim N(\mu_i,\sigma_i^2)$ for $i=1,2$, and these are independent.

If you know that the density of $X_1+X_2$ is the convolution of the two separate densities, then just evaluate the integral: \begin{align} & f_{X_1+X_2}(x) \\[10pt] & =\int_{-\infty}^\infty f_{X_1}(w)f_{X_2}(x-w)\, dw \\[10pt] & = \text{constant}\cdot \int_{-\infty}^\infty \exp\left(-\frac12\left(\frac{w-\mu_1}{\sigma_1}\right)^2\right)\exp\left(-\frac12\left(\frac{x-w-\mu_2}{\sigma_2}\right)^2\right) \, dw \end{align}

The product of the two "exp"s is the "exp" of \begin{align} & -\frac12\cdot\frac{(\sigma_1^2+\sigma_2^2)w^2 - 2w(\sigma_2^2\mu_1+\sigma_1^2(x-\mu_2))+ \sigma_2^2\mu_1^2+\sigma_1^2(x-\mu_2)^2}{\sigma_1\sigma_2} \\[12pt] & = -\frac12 \cdot \frac{w^2 - (\text{a certain weighted average of $\mu_1$ and $x-\mu_2$}) + \cdots}{\frac{\sigma_1\sigma_2}{\sigma_1^2+\sigma_2^2}} \end{align}

At this point the algebra gets moderately messy, but you complete the square and get $$ \text{constant}\cdot\int_{-\infty}^\infty \exp(\text{expression}_1)\cdot\exp(\text{expression}_2) \, dw $$

Now $\text{expression}_2$ should not depend on $w$, so you can pull a factor out: $$ \text{constant}\cdot\exp(\text{expression}_2)\cdot\int_{-\infty}^\infty \exp(\text{expression}_1)\,dw. $$ Then $\text{expression}_1$ will be $\text{constant}\cdot(w-\bullet)^2$, where "$\bullet$" is something that depends on $x$, so it may look as if the value of this integral depends on $x$. But it doesn't! Just let $u=w-\bullet$ so that $du=dw$, and "$\bullet$" is gone. Thus as a function of $x$, this integral is a constant. The whole thing reduces to a constant times $\exp(\text{expression}_2)$. And $\text{expression}_2$ ends up being a quadratic polynomial in $x$, with a negative leading coefficient, so we've got a normal (or "Gaussian") density function.


For a geometric proof, see Bennett Eisenberg & Rosemary Sullivan, Why Is the Sum of Independent Normal Random Variables Normal?, Mathematics Magazine, Dec. 2008, 362-366, available at http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/why-is-the-sum-of-independent-normal-random-variables-normal


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.