Working Through Expected Value I am working through an expectation and have something that I want to be true, and appears to be true in simulation, but I am having a hard time writing a proof that the idea can be derived properly.
Suppose I write $x = t + u$ where $u \sim \mathcal{N}(0,s^2)$ and assume $t$ is fixed but $x$ would of course be random. Then let $\mathbb{E}(x) = t$, $\mathbb{E}(u) = 0$, and $\mathbb{E}(u'u) = s^2$. Last, assume $t \bot u$.
I can write with these assumptions that $\mathbb{E}(x'u) = \mathbb{E}([t+u]'u) = \mathbb{E}(u'u)$ because of the orthogonlity between $t$ and $u$.
What I want to be true is to also be able to show that $x'\mathbb{E}(u|x) = \mathbb{E}(u'u)$. This is where I am struggling to generate a true proof, but is seemingly true with simulation.
Is there a suggestion someone might see that would allow for me to end up with a proof that $x'\mathbb{E}(u|x) = \mathbb{E}(u'u)$ is true?
 A: I'll assume $x$, $t$, and $u$ are all one-dimensional, for simplicity.
You are correct that, under your assumptions, $\mathbb{E}(x u) = \mathbb{E}(t u) + \mathbb{E}(u^2) = \mathbb{E}(u^2)$. The expectation here is over the random variable $u$.
But when you're computing $x \, \mathbb{E}(u \mid x)$, where's the uncertainty? If $t$ is fixed, and you're conditioning on $x$, then you know $u$: it's just $x - t$. So $x \, \mathbb{E}(u \mid x) = x (x - t)$.
What you might be looking for isn't $x \, \mathbb{E}(u \mid x)$, but its expectation. In this case, apply the law of iterated expectations to the statement you derived before:
$$
\mathbb{E}(u^2) = \mathbb{E}(xu) = \mathbb{E}\left( \mathbb{E}(xu \mid x) \right) = \mathbb{E}\left( x \, \mathbb{E}(u \mid x) \right).
$$
Is this what you were looking for?
A: I don't think your formula is true. Can you edit the post to include the simulations you think prove it true? 
First, it is not clear what your use of  $'$ means, I guessed transpose, but there are no vectors there, so I will interpret this as a question about scalar variables, so $x'u = xu$ and $u'u=u^2$. 
Note that since $t$ is assumed constant, in fact $u$ and $t$ are independent, so that $(u,x)=(u,t+u)$ has a singular joint distribution, that is, no density. In fact the joint distribution is 
$$
\mathcal{N}_2\begin{pmatrix}\left(\begin{smallmatrix} 0\\t\end{smallmatrix}\right),\left(\begin{smallmatrix} s^2 & s^2 \\ s^2 & s^2 \end{smallmatrix} \right)\end{pmatrix}
$$
and since the variance-covariance matrix has rank 1, this binormal distribution do not have a density (in two-dimensional space). 
So we can understand you have problems calculating the conditional expectation, since it is not obvious.  (At this point @Frank posted his answer, but I continue).
In the abstract definition of conditional expectations, we condition with respect to $\sigma$-algebras, not random variables. The two random variables $u$ and $x=u+t$ generate the same $\sigma$-algebra, so then the same conditional expectation. So conditioning with respect to $x$ and with respect to $u$ is the same:
$$
  x \mathbb{E}(u|x) = x\mathbb{E}(u | u) = xu \quad\text{which is different from} \\
\mathbb{E}(u^2)
$$
(but taking the expectation of this we recover the other answer)
