I might just be slow (or too drunk), but I'm seeing a conflict in the equations for adding two normals and scaling a normal. According to page 2 of this, if $X_1 \sim N(\mu_1,\sigma_1^2)$ and $X_2 \sim N(\mu_2,\sigma_2^2)$, then $X_1 + X_2 \sim N(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$, and for some $c \in \mathbb{R}$, $cX_1 = N(c\mu_1, c^2\sigma_1^2)$.

Then for $X \sim N(\mu,\sigma^2)$, we have $X + X = N(\mu + \mu,\sigma^2 + \sigma^2) = N(2\mu,2\sigma^2)$, but also $X + X = 2X = N(2\mu,2^2\sigma^2) = N(2\mu,4\sigma^2)$ ? Ie, the variances disagree.

edit: Oh, am I mistaken in saying that $2X = X + X$? Is the former "rolling" $X$ just once and doubling it while the latter "rolls" twice and adds them?

  • $\begingroup$ If you have transcribed "page 2" correctly, then it contains numerous false statements, and you should stop reading it. Look carefully again. Does the word "independent" occur anywhere in the vicinity of $X_1$ and $X_2$? Does "for some $c$" actually read "for any $c$"? (emphasis added) $\endgroup$ Feb 20, 2014 at 0:14
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    $\begingroup$ $X_{1}+X_{2} \sim \mathcal{N}(\mu_{1}+\mu_{2},\sigma_{1}^{2}+\sigma_{2}^{2})$ is for independent $X_{1}$ and $X_{2}$. $\endgroup$
    – pitchounet
    Feb 20, 2014 at 0:14
  • $\begingroup$ @jibounet Ok this makes more sense. This was unstated in the linked pdf but it was more of a "recall that..." thing than trying to introduce the concept. $\endgroup$
    – alecbz
    Feb 20, 2014 at 0:17

4 Answers 4


On the first page of the cited document, $X_1$ and $X_2$ were previously defined to be two (distinct) independent, identically distributed random variables. For your purposes, the "identically distributed" part is not important, but the "independent" part is.

On the second page, where $X_1$ and $X_2$ are considered to be normal variables, there's still the assumption that they're independent. Possibly this could have been stated more clearly, but in context this assumption makes sense.

When you consider $2X = X + X$, you are not dealing with two independent variables. The two "copies" of $X$ are correlated (in fact, as correlated as any two variables can be). The formula for the sum of two independent normal variables therefore does not apply.

  • $\begingroup$ I think this is the correct answer why no people up vote this one. $\endgroup$
    – Deep North
    Aug 14, 2015 at 1:39

Expectation is always linear. So for any two variables $X,Y$, we have $E[X+Y]=E[X]+E[Y]$. And $E[\underbrace{X+X+\ldots+X}_{k \text{ times}}]=E[kX]=kE[X]$

Variance is linear when the variables are independent. In this case, $V[X+Y]=V[X]+V[Y]$. However when the variables are the same, i.e. when we scale, we have $V[kX]=k^2 V[X]$.

These are true no matter what the distribution. Determining the distribution of the sum of random variables is, in general, difficult. However when $X$ and $Y$ are independent normal variables, then the sum of $X$ and $Y$ is also normally distributed (and the means and variances add as above).

In my opinion your question is a good one and it is very easy to become confused about what is adding and what is scaling. A very important example which involves adding independent distributions and scaling is when you compute the variance of the mean $\bar{X}$ of independent identically distributed (iid) samples from the same distribution.

To keep things simple we have $n$ samples and let's say each sample is normally distributed with variance $\sigma^2$. So each sample is drawn from $X_i \sim N(\mu,\sigma^2)$. Then adding independent random variables the variance of the sum is $V(S) = V\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n V\left(X_i\right) = n\sigma^2$. But the mean $\bar{X}=S/n$ and so by scaling $V(\bar{X}) = V(S/n) = n\sigma^2 / n^2 = \sigma^2/n$. It is this combination of adding and scaling which leads to the famous relationship that standard deviation of the sum increases according to the square root of $n$, and of the mean as $1/\sqrt{n}$.

  • $\begingroup$ For the last part, maybe "standard deviation decreases as $\frac{1}{\sqrt{n}}$" would make more sense. $\endgroup$
    – Grayscale
    Apr 21, 2020 at 12:57
  • $\begingroup$ At least if you talking about standard deviation of the sample mean. $\endgroup$
    – Grayscale
    Apr 21, 2020 at 13:15
  • $\begingroup$ @Grayscale thx, clarified last sentence $\endgroup$
    – TooTone
    Apr 21, 2020 at 20:12

Suppose $X_1, X_2,..., X_k$ are any k random variables (not necessarily independent).

Then variance of $X_1+X_2+...+X_n$ is calculated by:

$Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)+2*\sum_{1\le i\le k}Cov(X_i,X_j)$

When $X_1, X_2,...X_k$ are independent, then $Cov(X_j, X_j)=0$

$\therefore$ $Var(X_1+X_2+...+X_n)=\sum_{i=1}^kVar(X_i)$

When you have $X, X,...,X$, i.e $kX$ variables, $X$ and $X$ are no more independent then $Cov(X_i,X_j)$ will not be zero.

For example, you want to calculate $Var(X+X+X)$ it is not $3*Var(X)$ it will be $9*Var(X)$

Since $Var(X+X+X)=\sum_{i=1}^3Var(X)+2*\sum_{1 \le i \le j} Cov(X,X) =3*Var(X)+6*Cov(X,X)$ and $Cov(X,X)=Var(X,X)$

Therefore, $Var(X+X+X)=9*Var(X)$

The key is "When you consider 2X=X+X, you are not dealing with two independent variables" as David K points out.


You make a mistake assuming that $X+X = 2X$ like in arithmetic. This does not hold for random variables. As a simple counterexample take $X$ to be a Bernoulli trial with succes probability $0.5$. If $X+X=2X$ then you cannot flip two coins and get heads on one and tails on the other. Summing random variables is given by a convolution.

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    $\begingroup$ Sorry but $X+X=2X$ "like in arithmetic" is perfectly valid for random variables. No idea what you are talking about. $\endgroup$
    – Did
    Mar 7, 2017 at 23:44

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