# A step of a proof regarding the Nadaraya-Watson estimator

Let the data be $(y_i , X_i)$ where $y_i$ is real valued and $X_i$ is a q-vector.

The regression function for $y_i$ on $X_i$ is $g(x) = E(y_i | X_i = x)$, we can write this as: $$y_i = g(X_i) + e_i$$ where $E(e_i | X_i ) = 0$.

Now since $E(e_i | X_i ) = 0$ then it follows that $E(k(\frac{X_i - x}{h})e_i) = 0$

where $k$ is a kernel function. What is the reason of this implication?

If $X_i$ is known, then this implies that the function $k\left(\frac{X_i - x}{h}\right)$ is known. In other words we have that for the conditional expectation:

$$\operatorname{E}\left( k\left(\frac{X_i - x}{h} \right) e_i \mid X_i \right) = k\left(\frac{X_i - x}{h}\right) \operatorname{E}(e_i \mid X_i)=0$$

Therefore, we have by the law of iterated expectations that:

$$\operatorname{E}\left( k\left(\frac{X_i - x}{h} \right) e_i \right) = \operatorname{E}\left( \operatorname{E}\left(k\Big(\frac{X_i - x}{h}\Big) e_i \mid X_i \right) \right) = \mathrm{E}(0) = 0$$

as required.

$E(k(\frac{X_{i} - x}{h})e_{i}) = E(E(k(\frac{X_{i} - x}{h})e_{i}|X_{i}))$ by tower property

$= E(k(\frac{X_{i} - x}{h})E(e_{i}|X_{i}))$ because $k(.)$ is deterministic given $Xi$

$=E(k(\frac{X_{i} - x}{h})0)=0$ since $E(e_{i}|X_{i})=0$