Want to show that $$ E_X op = \frac{2}{n} \sum_{i=1}^n Cov_X(\hat{Y}, Y_i)$$

For 0-1 loss function with 0-1 response.

Want I've done

$$op = l_{in} - l=\frac{1}{n}\sum_{i=1} ^n Loss(Y_i', \hat{Y})-\frac{1}{n}\sum_{i=1} ^n Loss(Y_i, \hat{Y})$$ where each response $Y_i'$ is drawn as a random sample from the density function. $$Loss(Y_i', \hat{Y}) = \left\{\begin{matrix} 1 \; \; \; \, , \; if\; Y_i \neq \hat{Y} \\ 0 \; \; \; \, , \; if\; Y_i = \hat{Y} \end{matrix}\right.=\mathbf{1}_{\{Y_i \neq \hat{Y}\}}$$ where $\mathbf{1}$ is the indicator function.

Given that $E_X op = E [op | X_1 = x_1, ..., X_n = x_n]$ where $E$ is the expected value

$$E_X op = \frac{1}{n}\sum_{i=1} ^n E [\mathbf{1}_{\{Y_i' \neq \hat{Y}\}} - \mathbf{1}_{\{Y_i \neq \hat{Y}\}}]$$

I'm not sure how to proceed from here.


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