# How did the variance and double summation of the covariance come to the L2 minimization equation?

I am trying to understand the last two lines of this math notation (from this paper).

How did the Var and double summation of the Cov come to the equation?

The first two lines I understood something like $$(a-b)^2 = a^2 -2ab +b^2$$.

As in the second line, the first two terms are $$\mathbb{E}_{\hat{y}}$$, it means the variable of the expectation is $$\hat{y}_i$$ and you can take out $$y_i$$s from $$\mathbb{E}_{\hat{y}}$$s. Now we can use $$\mathbb{E}\{\hat{y}_i\} = y_i$$ and rewrite the terms of the second line likes the following:

$$\mathbb{E}_{\hat{y}}(\sum_i y_i)^2 = (\sum_i y_i)^2 \\ \mathbb{E}_{\hat{y}}\left[(\sum_i y_i)(\sum_i \hat{y}_i)\right] = (\sum_i y_i) \mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i) = (\sum_i y_i)^2$$

The second line is written based on linearity of the expectation and $$\mathbb{E}\{\hat{y}_i\} = y_i$$. Hence, we can rewrite the second line like the following as $$\sum_i y_i = \mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i)$$:

$$\frac{1}{N^2}\left[\mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i)^2 - \left(\mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i)\right)^2\right]$$

And the final step is using this formula $$Var(X) = E(X^2) - (E(X))^2$$ and take $$X = \sum_i \hat{y}_i$$:

$$\frac{1}{N^2}\left[\mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i)^2 - \left(\mathbb{E}_{\hat{y}}(\sum_i \hat{y}_i)\right)^2\right] = \frac{1}{N^2} Var(\sum_i \hat{y}_i)$$

From variance to covariance, you can use this formula: $$Var(\sum_{i=1}^nX_i) = \sum_{i=1}^n\sum_{j=1}^n cov(X_i, X_j)$$