Proof of expected loss in linear regression While reading section 1.5.5: Loss functions for regression in PRML, I came across the following derivation that I need help with.
Notation:
$$ \mathbb{E}[L] = \int \int L(t,y(x))p(x,t) \; dx \; dt$$
and $L(t, y(x))$ is chosen as $\{y(x) - t\}^2$. 
Given from the calculus approach, we already know that loss is minimized when $y(x) = \mathbb{E}[t|x]$, so we can write loss as $L(t, y(x)) = \{y(x) - \mathbb{E}[t|x] + \mathbb{E}[t|x] - t\}^2$.
The textbook further mentions that after substituting for $L(t,y(x))$ and integrating over $t$ to compute $\mathbb{E}[L]$, the cross term vanishes. This is where I'm stuck and I've done the following expansion for the cross term.
\begin{align*}
\int \int \{y(x)-\mathbb{E}[t|x]\}\{\mathbb{E}[t|x]-t\} p(t,x) \; dt \;dx &= \int\int y(x)\mathbb{E}[t|x] p(t,x)\;dt\;dx - \int\int t \; y(x) p(t,x)\;dt\;dx \\
& - \int\int \mathbb{E}[t|x]^2  p(t,x)\;dt\;dx + \int\int t \; y(x) p(t,x)\;dt\;dx \\
&= \int y(x)\mathbb{E}[t|x]\; p(x) dx - \int\int t \; y(x) p(t,x)\;dt\;dx \\
& - \int \mathbb{E}[t|x]^2  p(x)\;dx + \int\int t \; y(x) p(t,x)\;dt\;dx
\end{align*}
But I don't see how the cross term vanishes. Could someone help me understand what I'm missing? TIA.
Edit
Here the aim is to prove that $y(x) = \mathbb{E}[t|x]$ is an optimal choice without using calculus. I have also attached below the exact reference from the textbook.

So, the author mentions that the cross term cancels out after substituting for the loss term and integrating over $t$ and this does not require $\mathbb{E}[t|x] = y(x)$ yet. After examining the left over two terms, we finally determine that to minimize the loss, we should have $\mathbb{E}[t|x] = y(x)$.
 A: You are almost there;
if you use the assumption that $\mathbf{y}(\mathbf{x}) = \mathbb{E}[\mathbf{t}|\mathbf{x}]$ and substitute $\mathbf{y}(\mathbf{x})$ in your last step, you will get that the cross-term is zero.
For each of the four terms, we re-write the joint distribution $p(t, \mathbf{x})$ as $p(t|\mathbf{x})p(\mathbf{x})$ and push the integral over $t$ inside; we get following:
$$
\begin{aligned}
&\int\int y(\mathbf{x})\mathbb{E}[t|\mathbf{x}] p(t,\mathbf{x}) \; \mathrm{d}t \; \mathrm{d}\mathbf{x}
- \int\int t  y(\mathbf{x}) p(t,\mathbf{x}) \; \mathrm{d}t \; \mathrm{d}\mathbf{x} \\
- & \int\int \mathbb{E}[t|\mathbf{x}]^2  p(t,\mathbf{x}) \; \mathrm{d}t \; \mathrm{d}\mathbf{x}
+ \int\int t  \mathbb{E}[t|\mathbf{x}] p(t,\mathbf{x}) \; \mathrm{d}t \; \mathrm{d}\mathbf{x} \\
= & \int y(\mathbf{x})\mathbb{E}[t|\mathbf{x}] p(\mathbf{x})\left\{\int p(t|\mathbf{x}) \; \mathrm{d}t\right\} \mathrm{d}\mathbf{x}
- \int y(\mathbf{x}) p(\mathbf{x})\left\{\int t p(t|\mathbf{x}) \; \mathrm{d}t\right\} \mathrm{d}\mathbf{x} \\
- & \int \mathbb{E}[t|\mathbf{x}]^2  p(\mathbf{x}) \left\{\int p(t|\mathbf{x}) \; \mathrm{d}t\right\} \mathrm{d}\mathbf{x}
+ \int  \mathbb{E}[t|\mathbf{x}] p(\mathbf{x}) \left\{\int t p(t|\mathbf{x}) \; \mathrm{d}t\right\} \mathrm{d}\mathbf{x} \\
=& \int y(\mathbf{x})\mathbb{E}[t|\mathbf{x}] p(\mathbf{x}) \; \mathrm{d}\mathbf{x}
- \int y(\mathbf{x}) \mathbb{E}[t|\mathbf{x}] p(\mathbf{x}) \; \mathrm{d}\mathbf{x} \\
-& \int \mathbb{E}[t|\mathbf{x}]^2  p(\mathbf{x}) \; \mathrm{d}\mathbf{x}
+ \int  \mathbb{E}[t|\mathbf{x}]^2 p(\mathbf{x}) \; \mathrm{d}\mathbf{x}.
\end{aligned}
$$
The last equation shows that the four terms cancel each other.
