Expected value of the inner product equals l2-norm?

Question

Referring to Butucea (14), I have given $(Y_i)_{i\in \mathbb N}$ iid. with Lebesgue-density $p$, a kernel $K$ and the scalar product $\langle f,g\rangle=\int_{-\infty}^\infty f(x)\overline{g}(x)\mathrm d x$. In (14) it states: $$\mathbb E(d_n)=\mathbb E(\langle K(\cdot-Y_1),K(\cdot-Y_2)\rangle)=\|K\ast p\|_2^2$$ I have a problem with taking the expected value of the scalar product. \begin{align*} \mathbb E(\langle K(\cdot-Y_1),K(\cdot-Y_2)\rangle)&=\mathbb E(\int_{-\infty}^\infty K(x-Y_1)\overline{K(x-Y_2)}\mathrm d x)\\ &=\int_{-\infty}^\infty p(y)\int_{-\infty}^\infty K^2(x-y) \mathrm d x \mathrm d y\\ &\overset{?}{=}\int_{-\infty}^\infty\int_{-\infty}^\infty (p(x) K(x-y))^2 \mathrm d x \mathrm d y=\|K\ast p\|_2^2 \end{align*} I'm not sure how the $p^2(x)$ appears at the "?".

Answer 1

Bruno B is right. Notice that $Y_1$ and $Y_2$ are iid random variables.

\begin{align*} &\mathbb{E} \langle K(\cdot -Y_1), K(\cdot -Y_2) \rangle \\ &= \int_{\mathbb{R}^3} p(y_1) p(y_2) K(x-y_1) \overline{K(x-y_2)} dx dy_1 dy_2 \end{align*}

You should arrive at the desired result by factoring this into two integrals (like $\int (\int \cdots dy_1)(\int \cdots dy_2) dx$ ), one with $y_1$ and one with $y_2$.