Is a conditional distribution independent of the variable being conditioned on? [closed]

Question

This might seem like a bit of a silly question, but for some reason I'm having trouble convincing myself of it, so I thought I might solicit opinions on this site. Suppose random variables $(X,Y)$ have some joint distribution (bivariate normal if you'd like). Is the conditional distribution $(Y|X=x)$ independent of random variable $X$? I feel like it's vacuously true because $\text{P}(Y|X,X)=\text{P}(Y|X)$ but for some reason the intuition just isn't coming to me.

Answer 1

It depends what you mean by the "conditional distribution"

There are a number of ways this can be interpreted, and the answer depends on the interpretation. To facilitate this analysis, suppose we define the conditional distribution function $F$ as the bivariate function:

$$F(y|x) = \mathbb{P}(Y \leqslant y | X = x) \quad \quad \quad \text{for all } x \in \mathscr{X}.$$

If we consider the conditional distribution as the full bivariate function $F( \cdot | \cdot )$ then this object is fixed by the joint distribution of $X$ and $Y$, so it is a constant (and therefore trivially independent of $X$). Similarly, if we substitute any constant arguments to reduce the function to a univariate function or scalar value, it is still a constant (and therefore trivially independent of $X$). However, if we substitute one of the original random variables into its argument then it becomes a random variable, and it is now statistically dependent with the substituted random variable (and possibly also the other random variable). This is summarised below.

$$\begin{matrix} F( \cdot | \cdot ) & & & \text{Constant bivariate function} & & & \text{Independent of } X \text{ and } Y, \\[6pt] F( \cdot | x ) & & & \text{Constant univariate function} & & & \text{Independent of } X \text{ and } Y, \\[6pt] F( y | \cdot ) & & & \text{Constant univariate function} & & & \text{Independent of } X \text{ and } Y, \\[6pt] F( y | x ) & & & \text{Constant scalar value} & & & \text{Independent of } X \text{ and } Y, \\[6pt] F( \cdot | X ) & & & \text{Random univariate function} & & & \text{Not independent of } X, \\[6pt] F( Y | \cdot ) & & & \text{Random univariate function} & & & \text{Not independent of } Y, \\[6pt] F( y | X ) & & &\text{Random variable} & & & \text{Not independent of } X, \\[6pt] F( Y | x ) & & &\text{Random variable} & & & \text{Not independent of } Y. \\[6pt] F( Y | X ) & & &\text{Random variable} & & & \text{Not independent of } X \text{ or } Y. \\[6pt] \end{matrix}$$

(Note that this is slightly simplified, insofar as I have not specified dependence on the other random variable; whether there is dependency with the other random variable depends on the original joint distribution and the conditional distribution.)

As you can see, the conditional distribution is independent of the random variable being conditioned on, so long as we take the conditional distribution to be a constant object. If we condition on the random variable $X$ as a random variable then the conditional distribution becomes a random function (or random variable) that is not independent of $X$.

Answer 2

The immediate answer is that, if one considers the random variable $Z_k$ with cdf $$F_k(z)=\mathbb P(Y\le z|X=k)$$ it is defined independently of $X$ and hence $Z$ is independent from $X$.

However, if the question is about the "random variable" $(Y|X=x)$, this object does not exist as such, i.e., separately from $Y$, marginal from the pair $(X,Y)$. That is, $Y$ enjoys both a marginal distribution and a conditional distribution, but remains the same random variable. The cdf $\mathbb P(Y\le z|X=k)$ is thus a characterisation of the conditional distribution of $Y$ when the realisation $X=k$ occurs. Since the conditioning event is $X=k$, the conditional distribution depends on it (or not when $X$ and $Y$ are independent), but it makes no sense to introduce independence once $X=k$ has happened.