Parzen Window convergence of the mean

Question

I'm studying the convergence of the mean in Parzen Window estimates, and am having trouble figuring out the intuition behind one particular step in the derivation. It goes from

$$ \frac{1}{n}\sum_{i=1}^n E\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg]$$ to $$\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

What I'm confused about is the relationship between $\mathbf{x}_i$ and $\mathbf{v}$, and why it is possible to directly substitute $\mathbf{x}_i$ with $\mathbf{v}$.

Answer 1

Assuming $\mathbf x_i$ are iid:

$$\frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_1},\dots, \mathbf{x_n})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n E_{f(\mathbf{x_i})}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg]$$

Again, assuming $\mathbf x_i$ are iid and letting $\bf v$ follow the distribution of $\mathbf x_i$, call this distribution $p(\cdot)$:

$$\frac{1}{n}\sum_{i=1}^nE_{f(\mathbf x_i)}\bigg[\frac{1}{V_n}\;\varphi\bigg(\frac{\mathbf{x} - \mathbf{x}_i}{h_n}\bigg)\bigg] \\ = \frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$

We see that the summand does not depend on i. $$\frac{1}{n}\sum_{i=1}^n\int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v} \\ = \int\frac{1}{V_n}\varphi\bigg(\frac{\mathbf{x} - \mathbf{v}}{h_n}\bigg)\;p(\mathbf{v})\;d\mathbf{v}$$