Consider $2\times3$.
The above multiplication is just $2\times3=2+2+2$
or more generally $a\times n= \underbrace{a+a+.....}_{n -times}$
Now consider $$\bigg \langle\frac{1}{r^2}\bigg\rangle=\bigg \langle\frac{1}{r}\times\frac{1}{r}\bigg\rangle=\bigg \langle\underbrace{\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+....}_{\frac{1}{r} \ \ times}\bigg\rangle=\underbrace{\bigg \langle\frac{1}{r}\bigg\rangle+\bigg \langle\frac{1}{r}\bigg\rangle+\bigg \langle\frac{1}{r}\bigg\rangle+...}_{\frac{1}{r}\ \ times}=\frac{1}{r}\bigg \langle\frac{1}{r}\bigg\rangle$$
and therefore,
$$\bigg \langle\frac{1}{r^2}\bigg\rangle=\frac{1}{r}\bigg \langle\frac{1}{r}\bigg\rangle$$
What's up with this?
Edit: The $\langle\rangle$ denotes average.
(I will use $E[\cdot]$ instead of $\langle \cdot \rangle$ to denote expectation.)
By the same logic, if $N$ is an integer-valued random variable, you could argue $$E[N] = E[\underbrace{1 + \cdots + 1}_{\text{$N$ times}}] \overset{*}{=} \underbrace{E[1] + \cdots + E[1]}_{\text{$N$ times}} = N.$$
The starred equality is the problem: this does not actually follow from the linearity of expectation property, since the number of addends is random.