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

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

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


1 Answer 1


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: $$Var(\sum_{i=1}^nX_i) = \sum_{i=1}^n\sum_{j=1}^n cov(X_i, X_j)$$


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