Show that $\Sigma^{n}_{r=1}(x_{r}-x_{mean})f_{r}=0$

Question

For the frequency distribution:

$$x_{i} : x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$$ $$f_{i} : f_{1}, f_{2}, f_{3}, f_{4}, f_{5}$$

Show that:

$$\Sigma^{n}_{r=1}(x_{r}-x_{mean})f_{r}=0$$

My understanding is that: $$\Sigma^{n}_{r=1}(x_{r}-x_{mean})=0$$ But when one introduces the weighted values how is it still equal to zero?

I can't find a proof anywhere.

Answer 1

$\sum_{i=1}^{5} (x_i-m)f_i= \sum_{i=1}^{5}x_if_i- m\sum_{i=1}^{5}f_i=*$. Now, $m=\frac{\sum_{i=1}^{5}x_if_i}{\sum_{i=1}^{5}f_i}\to *= \sum_{i=1}^{5}x_if_i- \sum_{i=1}^{5}x_if_i=0$

Answer 2

We can say that $$x_{mean}=\frac{\sum_{r=1}^{5}x_rf_r}{\sum_{r=1}^{5}f_r}=\frac{\sum_{r=1}^{5}x_rf_r}{N}$$ where $N=\sum_{r=1}^{5}f_r$

For the weighted values, we have that $$\sum_{r=1}^{5}(x_r-x_{mean})f_r$$ $$=\sum_{r=1}^{5}(x_rf_r-x_{mean}f_r)$$ $$=\sum_{r=1}^{5}x_rf_r-\sum_{r=1}^{5}x_{mean}f_r$$ $$=\sum_{r=1}^{5}x_rf_r-x_{mean}\sum_{r=1}^{5}f_r$$ $$=Nx_{mean}-x_{mean}\cdot N$$ $$=0$$

Hope this helps.

Answer 3

You have by definition $$ x_{\rm mean} = \sum_{r=1}^n f_r x_r $$ (this is the "weighted average" of the $x_r$'s by their frequencies $f_r$, which satisfy $\sum_{r=1}^n f_r = 1$). It follows that $$ \sum_{r=1}^n f_r (x_r - x_{\rm mean}) = \sum_{r=1}^n f_r x_r - \sum_{r=1}^n f_r x_{\rm mean} = x_{\rm mean} - x_{\rm mean}\sum_{r=1}^n f_r = x_{\rm mean}-x_{\rm mean} = 0 $$ since we could factor $x_{\rm mean}$ out of the second sum, as it does not depend on $r$.

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Note that your second statement is wrong: $\sum_{r=1}^n (x_r - x_{\rm mean}) \neq 0$ in general. This would be true, however, for a regular mean (non-weighted), i.e. if all $f_r=\frac{1}{n}$.