Annual variance from monthly data (returns)

Question

I can't determine how we establish equality between the two bracketed expressions in the image below. The square of the sum is being equated with the sum of the squares, which in general isn't true. What am I missing? Thank you.

Answer 1

We can write the squared sum of residual returns as

$$\left(\sum_{i=1}^{12} (r_i - \bar{r})\right)^2 = \sum_{i=1}^{12} (r_i - \bar{r})\sum_{j=1}^{12} (r_j - \bar{r}) = \sum_{i=1}^{12} \sum_{j=1}^{12}(r_i - \bar{r}) (r_j - \bar{r}) \\ = \sum_{i=1}^{12} (r_i - \bar{r})^2 + \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r}) $$

Taking the expected value, we get

$$\sigma_y^2 = \mathbb{E} \left(\sum_{i=1}^{12} (r_i - \bar{r})^2 + \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r}) \right) \\= \mathbb{E}\left(\sum_{i=1}^{12} (r_i - \bar{r})^2\right) + \mathbb{E}\left(\underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r})\right) $$

For the second term on the right-hand side, we have

$$\mathbb{E}\left(\underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}(r_i - \bar{r}) (r_j - \bar{r})\right) = \underset{i\neq j}{\sum_{i=1}^{12} \sum_{j=1}^{12}}\mathbb{E}\left((r_i - \bar{r}) (r_j - \bar{r})\right)$$

You are given "... the monthly returns ... are mutually uncorrelated," which means

$$\mathbb{E}\left((r_i - \bar{r}) (r_j - \bar{r})\right) = 0 \,\, \text{ for } i \neq j$$

Thus,

$$\sigma_y^2 = \mathbb{E}\left(\sum_{i=1}^{12} (r_i - \bar{r})^2\right) = \sum_{i=1}^{12} \mathbb{E}\left((r_i - \bar{r})^2\right) = 12 \sigma^2$$