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I have been reading a book about statistics for physicists and there was this line given: "The mean of the sum of N independent variables with the same distribution is N times the mean of a single variable". Also, the following equation was given: $$\mu_X\equiv\langle X\rangle=\sum_{i=1}^N\langle x_i\rangle=N\langle x_i\rangle = N\mu$$

So my question is, what does that actually mean? I mean, if $x_i$ can be any number, e.g. $2, 3, 5$ and $6$, than the $N$ would be $4$. The average would be $\frac{16}{4} = 4$. So is that average $4$ actually $\langle x_i\rangle$ or $\langle X\rangle$? Also if $\mu$ is the avarage, than $N\mu$ would be $4 \cdot 4 = 16$. So $\mu_X$ would be $16$. But what does that mean?

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  • $\begingroup$ $N$ is not the number of values $x_i$ can take, but the sample size and here the dimension of $X$ $\endgroup$ – Henry Apr 15 at 21:49
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When the authors refer to the "mean" in this context, they are referring to the expected value of the random variable $X$. (Note also that they are using the symbol $\langle \ \ \rangle$ for this operation, which is common notation in physics.) While the textual statement given in your post is correct, the equation you give does not make sense. It appears to equate the expected value of a single random variable with the expected value of a sum of $N$ of those random variables.


I will try here to state the result correctly and show you how it is derived. For a single random variable $X_i$ with density function $f_X$ we get the expected value:

$$\mu \equiv \langle X_i \rangle = \int \limits_\mathscr{X} x f_X(x) \ dx.$$

Now suppose you have a set of $N$ random variables $X_1,...,X_N$ with this same distribution, and denote the sum of these values as $S_N \equiv \sum_{i=1}^N X_i$. Since the random variables are independent and identically distributed we have $f_\mathbf{X}(x_1,...,x_n) = f_X(x_1) \cdots f_X(x_n)$ so the expected value of the sum of these random variables is:

$$\begin{aligned} \langle S_N \rangle &= \bigg\langle \sum_{i=1}^N X_i \bigg\rangle \\[6pt] &= \int \limits_{\mathscr{X}^N} \Bigg( \sum_{i=1}^N x_i \Bigg) f_\mathbf{X}(x_1,...,x_n) \ dx_1 \cdots dx_n \\[6pt] &= \int \limits_{\mathscr{X}^N} \Bigg( \sum_{i=1}^N x_i \Bigg) f_X(x_1) \cdots f_X(x_n) \ dx_1 \cdots dx_n \\[6pt] &= \sum_{i=1}^N \int \limits_{\mathscr{X}^N} x_i f_X(x_1) \cdots f_X(x_n) \ dx_1 \cdots dx_n \\[6pt] &= \sum_{i=1}^N \int \limits_{\mathscr{X}} x_i f_X(x_i) \ dx_i \\[6pt] &= \sum_{i=1}^N \langle X_i \rangle \\[6pt] &= \sum_{i=1}^N \mu \\[6pt] &= N \mu. \\[6pt] \end{aligned}$$

(Note that in the transition from the first to the second line I have used the law of the unconscious statistician.) This working is really just confirming the linearity property of the expected value operator. Using the linearity property we see that the expected value of the sum of $N$ independent and identically distributed random variables is $N$ times the expected value of an individual random variable.

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Basically, they are telling you this.

If $X_{1}, X_{2},..., X_{N}$ are indenpendent random variables with the same distribution (iid), where $E(X_{i})=\mu$, then:

$E\left[\sum_{i=1}^{N}X_{i} \right] =N\mu$.

If you want to understand why, check out this:

$E\left[\sum_{i=1}^{N}X_{i} \right] =E(X_{1} +X_{2} +...+X_{N})$

$=E(X_{1}) +E(X_{2})+... +E(X_{N})$,

remember that $X_{i}$ has the same distribution, so $E(X_{i}) =\mu$. Then:

$=\mu +\mu +...\mu =N\mu$.

I think it's important to point out that the condition of independence is not necessary

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As also pointed out in the comments, $N$ is the number of terms (RVs) you sum, not the possible values that $X$ can take. For example, the number of possible values $X$ can take could have been infinite as well, as in geometric distribution.

Basically, the statement means if you roll $N$ dice, on average, the sum will be around $7N/2$ where $7/2$ is the average of one dice.

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