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Just so that this could be useful for people who refer to this post later on: Please refer to Sutton's reinforcement learning book (2nd edition) example 11.2. It provides an example for why full gradient wouldn't work.


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Let's say we have $a$ - constant and $\epsilon \sim \mathcal{N}(0,\sigma)$, then: $$\mathbb{E}\left[(a+\epsilon)^2\right] = \mathbb{E}\left[a^2\right] + 2 \mathbb{E}\left[a\right]\mathbb{E}\left[\epsilon\right] + \mathbb{E}\left[\epsilon^2\right] $$ Expectations of constants are just the constants: $\mathbb{E}[a] = a$ and $\mathbb{E}[a^2] = a^2$ The mean of ...


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Let me try to show this. The only (non-constant) random variable here is $\epsilon$, while $f(X)$ and $\hat{Y} = \hat{f}(X)$ are constant random variables (so their expectations is equal to their only value). So, we start with the following expression. \begin{align} \mathbb{E} \left[ (Y - \hat{Y})^2 \right] \tag{1}\label{1} \end{align} Now, we just apply the ...


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