# Does conditional expectation agree with joint expectation?

Suppose I have a vector valued random variable $$(X_1, X_2)$$ in $$\mathbb{R}^2$$, with density $$\pi(x_1, x_2)$$. Let $$\mathbf{\mu} = (\mu_{X_1}, \mu_{X_2})$$ denote the mean vector of this bivariate distribution.

Question: Does $$\mathbb{E}[X_1 | X_2 = \mu_{X_2}] = \mu_{X_1}?$$

Intuitively I would think this should be true, but I am having trouble showing this. I would think it to be analogous to how if $$(x_1^\dagger, x_2^\dagger) = \text{argmax}_{(x_1,x_2)} \pi (x_1,x_2)$$, then

$$\text{argmax}_{x_1} \pi(x_1, x_2^\dagger) = x_1^\dagger,$$ assuming uniqueness of the maximizer.

What I have tried: We can write the expectation as $$\mathbb{E}[X_1 | X_2 = \mu_{X_2}] = \int_{\mathbb{R}} x_1 \pi(x_1 | x_2=\mu_{X_2}) dx_1,$$ and I am unsure how to proceed from this.

If we define the functions $$f(x_2) = \mathbb{E}[X_1 | X_2=x_2], \quad g(x_1) = \mathbb{E}[X_2 | X_1 = x_1],$$ then we can show that $$\mu_{X_2} = \mathbb{E}[g(X_1)], \quad \mu_{X_1} = \mathbb{E}[f(X_2)].$$ For example, $$\mu_{X_1} = \int_{\mathbb{R}} \int_{\mathbb{R}} x_1 \pi(x_1,x_2) dx_1 dx_2 = \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} x_1 \pi(x_1 | x_2) dx_1 \right] \pi(x_2) dx_2 = \int_{\mathbb{R}} f(x_2) \pi(x_2) dx_2.$$ However, does inserting $$f(x_2) = \mathbb{E}[X_1 | X_2=x_2]$$ into $$\mathbb{E}[f(X_2)]$$ then just immediately give us $$\mu_{X_1} = \mathbb{E}[ \mathbb{E}[X_1 | X_2] ],$$ the law of iterated expectations? I'm not sure if this is useful.

This isn't true, because you can change the conditional density of $$X_1$$ for values very close to $$X_2=\mu_2$$ by a lot, but leave $$E[X_1]$$ almost unchanged.
Let's say you start off with $$X_1$$ and $$X_2$$ independent standard Normal, so $$\mu_2=0$$ and the equality holds. Now add 100 to $$X_1$$ if $$|X_2|<10^{-6}$$. The overall mean $$E[X_1]$$ is almost unchanged, but $$E[X_1|X_2=0]$$ is 100 instead of 0.