# Conditional distribution of conditional expectation

Suppose that $$\theta \sim N(\mu, \nu^2)$$ and $$x \, | \, \theta \sim N(\theta, \sigma^2)$$.

We can then compute the posterior distribution $$\theta \, | \, x$$ using the conjugate prior property of the normal distribution, $$\theta \, | \, x \sim N \left(\frac{\nu^2}{\nu^2 + \sigma^2} x + \frac{\sigma^2}{\nu^2 + \sigma^2} \mu, \frac{\nu^2 \sigma^2}{\nu^2 + \sigma^2} \right)$$ and so the conditional expectation $$\mathbb{E}(\theta | x)$$ is just the mean of this distribution. I understand this.

What I don't understand is the following. It is then stated that the conditional distribution of the conditional expectation is given by, $$\mathbb{E}(\theta \, | \, x) \, | \, \theta \sim N \left(\frac{\nu^2}{\nu^2 + \sigma^2} \theta + \frac{\sigma^2}{\nu^2 + \sigma^2} \mu, \frac{\nu^4 \sigma^2}{(\nu^2 + \sigma^2)^2} \right).$$

I think my problems are two-fold. First, I don't really understand what the conditional distribution of the conditional expectation is conceptually. Shouldn't this just be a function of $$\theta$$? What is $$\mu$$ doing in the above equation? Secondly, I have no idea how this can be computed. Could someone explain how this is derived?

Context: this follows from the notes of an economist, shared privately.

$$\operatorname E(\theta \mid x)=\frac{\nu^2}{\nu^2 + \sigma^2} x + \frac{\sigma^2}{\nu^2 + \sigma^2} \mu$$
This is a linear function of the random variable $$x$$. Since $$x$$ given $$\theta$$ is normal, the above also has a normal distribution given $$\theta$$.
The mean and variance of $$\operatorname E(\theta \mid x)$$ given $$\theta$$ are
$$\operatorname E\left[\operatorname E(\theta \mid x)\mid \theta\right]=\frac{\nu^2}{\nu^2 + \sigma^2} \operatorname E\left[x\mid \theta\right]+\frac{\sigma^2}{\nu^2 + \sigma^2} \mu=\frac{\nu^2}{\nu^2 + \sigma^2} \theta+\frac{\sigma^2}{\nu^2 + \sigma^2} \mu$$
$$\operatorname{Var}\left[\operatorname E(\theta \mid x)\mid \theta\right]=\left(\frac{\nu^2}{\nu^2 + \sigma^2}\right)^2 \operatorname{Var}(x\mid \theta)=\frac{\nu^4 \sigma^2}{(\nu^2 + \sigma^2)^2}$$