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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}$

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    The terms of the sum in question are all the same, and so the value of the sum is just the number of terms times the value of the first term.2012-02-22

3 Answers 3

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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 $ $\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$ $\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1$ $\cdots\cdots\cdots$

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$\sum_{i=1}^{n}\frac{1}{n} =\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+ \cdots + \frac{1}{n}$ (n terms)

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Cut a pie into $n$ equal pieces. What proportion of the pie is in each piece?