# Conditional Expectation as a function of X

I hope someone can clear the doubt on conditional distribution, in a text, it mentions:

To interpret the notation $$f_{X \mid Y}(x \mid y)$$, we need to be clear that there is nothing random about the random variable $$Y$$, since it is already fixed. What remains is a sub-population distribution in terms of $$X$$ (i.e. the reduced sample space of $$\Omega_X$$ where $$Y$$ happened). Consequently, the conditional PMF/PDF $$f_{X \mid Y}(x \mid y)$$ is a distribution in terms of $$X$$.

I have no trouble understanding this, since conditional just means looking into the reduced sample space of $$X$$ where $$Y$$ has happened for a specific set. But soon after, when conditional expectation is mentioned, it says:

There are two points to note here. First, the expectation of $$\mathbb{E}[X \mid Y=y]$$ is taken with respect to $$f_{X \mid Y}(x \mid y)$$. We assume that the random variable $$Y$$ is already fixed at the state $$Y=y$$. Thus, the only source of randomness is $$X$$. Secondly, since the expectation $$\mathbb{E}[X \mid Y=y]$$ has eliminated the randomness of $$X$$, the resulting function is in $$y$$.

A quick read online reveals that the expectation of $$X$$ given $$Y$$ can indeed be a function of $$Y$$ if we treat this conditional expectation as random variable. I am just slightly confused why the conditional expectation is not a function of $$X$$, since the underlying conditional distribution is a function of $$X$$.

Without getting very much into measure theory,consider the random vector $$(X,Y)$$ with density $$f_{X,Y}(\cdot,\cdot)$$ (wrt a dominating measure $$\text d\mu(x,y)$$) decomposed as $$f_{X,Y}(x,y)=f_Y(y)\times f_{X|Y}(x|y)$$ where

• $$f_Y(\cdot)$$ is a probability density (wrt the appropriate dominating measure $$\text d\mu_2(y)$$) attached with the random variable $$Y$$
• $$f_{X|Y}(\cdot|y)$$ is a probability density (wrt the appropriate dominating measure $$\text d\mu_1(x)$$) for (almost) every $$y\in\mathcal Y$$, attached with a random variable $$Z_y$$. In this notation, $$y$$ is a parameter of the density.

For a fixed value of $$y\in\mathcal Y$$, $$f_{X|Y}(\cdot|y)$$ can thus be understood as a regular density over the set $$\mathcal X$$ and $$f_{X|Y}(\cdot|y):\ x\longmapsto f_{X|Y}(x|y)$$ is a non-negative integrable (measurable) function on $$\mathcal X$$ such that $$\int_\mathcal Xf_{X|Y}(x|y)\text d\mu_1(x)=1$$ This means that, for a fixed value of $$y\in\mathcal Y$$, the expectation of $$Z_y\sim f_{X|Y}(\cdot|y)$$ can considered and, provided it exists for this specific value of $$y\in\mathcal Y$$, be defined as $$\mathbb E_y[X] = \int_\mathcal X xf_{X|Y}(x|y)\text d\mu_1(x)\tag{1}$$ It is usually written as $$\mathbb E[X|Y=y]$$. In the event (1) exists for all values of $$y\in\mathcal Y$$, the function $$\varphi:\ y \longmapsto \mathbb E[X|Y=y]$$ is rigorously defined. It can therefore be called to transform the random variable $$Y$$ into the new random variable $$\varphi(Y)$$. It is usually written as $$\mathbb E[X|Y]$$ and is equal to (1) when the realisation of $$Y$$ is equal to $$y$$.

As seen above, this random variable $$\varphi(Y)=\mathbb E[X|Y]$$ is not a function of the random variable $$X$$, even though they may be correlated with one another. Hence, the citation

First, the expectation of $$\mathbb E[X∣Y=y]$$ is taken with respect to $$f_{X∣Y}(x∣y)$$. We assume that the random variable $$Y$$ is already fixed at the state $$Y=y$$. Thus, the only source of randomness is $$X$$.

should be restated as [with highlighted changes]:

First, the expectation $$\mathbb E[X∣Y=y]$$ is taken with respect to the distribution with density $$f_{X∣Y}(x∣y)$$. We assume that the random variable $$Y$$ is already observed at the realisation $$Y=y$$. Thus, the only remaining source of randomness in the expectation is $$X$$ with distribution $$f_{X∣Y}(\cdot∣y)$$, that is, the conditional distribution of $$X$$ given $$Y=y$$.

Similarly,

Secondly, since the expectation $$\mathbb E[X∣Y=y]$$ has eliminated the randomness of $$X$$, the resulting function is in $$y$$.

should state

Secondly, since the expectation $$\mathbb E[X∣Y=y]$$ has eliminated the (remaining) conditional randomness of $$X$$ given $$Y=y$$, the resulting function is a function of $$y$$.

• Thanks for this rigorous explanation! Jan 9 at 2:15
• Thanks. This is rigourous and clarifies my doubt. From what you wrote, the PDF of $X$ given $Y$ is still a function of $X$ right? Just want to make sure that this is still true.
– nan
Jan 9 at 7:12
• The pdf of the rv $X$ given thr rv $Y$ is the function $f_{X|Y}(⋅|⋅)$ and, therefore, is a function/mapping from $\mathcal X$ to $\mathbb R^+$ indexed by the realisation of $Y$ or $Y$ itself, but definitely not a function of the random variable $X$. (Just as the marginal density $f_X$ is not a function of $X$. It determines the distribution of $X$ and thus in a very informal sense depends on $X$, but it is not a mathematical function of the rv $X$.) Jan 9 at 10:04

If you take the expectation of $$X$$, it's not a function of $$X$$. You integrate $$xf(x)$$ over $$x$$, and so $$x$$ is gone, producing only a number.

If you take the expectation of $$X|Y=y$$, it's not a function of $$X$$. You integrate $$xf(x|y)$$ over $$x$$ (think of it as $$xf(x)$$ with a different $$f(x)$$ for different $$y$$), and so $$x$$ is gone, producing only a number, which may be different for different $$y$$s, hence a function of $$y$$.