I have come across the following in my Computational Finance slides but I'm unsure on the final equality.
$ \bar{r}_P = \sum_{i=1}^{n}w_i\bar{r_i} = \sum_{i=1}^{n}\frac{1}{n}\bar{r} = \bar{r}$
I have come across the following in my Computational Finance slides but I'm unsure on the final equality.
$ \bar{r}_P = \sum_{i=1}^{n}w_i\bar{r_i} = \sum_{i=1}^{n}\frac{1}{n}\bar{r} = \bar{r}$
Keep in mind that $n$ is independent of the index sum, so it is constant wrt $i$: $\sum_{i=1}^n \frac{1}{n}=\frac{1}{n}\sum_{i=1}^n1=\frac{1}{n}\cdot n=1 $
See:
$1=1$ $\frac{1}{2}+\frac{1}{2}=1 $ $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1$ $\cdots\cdots\cdots$
$\sum_{i=1}^{n}\frac{1}{n} =\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+ \cdots + \frac{1}{n}$ (n terms)
Cut a pie into $n$ equal pieces. What proportion of the pie is in each piece?