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.


1 Answer 1


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.


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