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Consider three random variables $\epsilon_1, \epsilon_2, X$. Let $F_{\epsilon_i}(\cdot| x)$ denote the cumulative distribution function (cdf) of $\epsilon_i$ conditional on $X=x$ for any $i\in \{1,2\}$. Suppose $X$ has support $\mathcal{X}$.

In what follows I will impose restrictions on $F_{\epsilon_i}(\cdot| x)$ $\forall i \in \{1,2\}$ and I want to investigate the impact of these restrictions on the cdf of $\epsilon_1-\epsilon_2$ conditional on $X=x$, denoted by $F_{\epsilon_1-\epsilon_2}(\cdot| x)$.

Case 1 (zero median): Suppose $F_{\epsilon_1}(0|x)=F_{\epsilon_2}(0|x)=\frac{1}{2}$ $\forall x \in \mathcal{X}$. What can we say about $F_{\epsilon_1-\epsilon_2}(0|x)$ for any $x \in \mathcal{X}$? Can we claim that $F_{\epsilon_1-\epsilon_2}(0|x)=\frac{1}{2}$ $\forall x \in \mathcal{X}$?

Case 2 (symmetry): Suppose $F_{\epsilon_1}(u|x)=1-F_{\epsilon_1}(-u|x)$ and $F_{\epsilon_2}(u|x)=1-F_{\epsilon_2}(-u|x)$ $\forall u\in \mathbb{R}\cup\{-\infty, \infty\}$ $\forall x \in \mathcal{X}$. Can we claim that $F_{\epsilon_1-\epsilon_2}(u|x)=1-F_{\epsilon_1-\epsilon_2}(-u|x)$ $\forall u\in \mathbb{R}\cup\{-\infty, \infty\}$ $\forall x \in \mathcal{X}$?

Case 3 (independence): Suppose $F_{\epsilon_1}(\cdot|x)=F_{\epsilon_1}(\cdot|x')$ $\forall x,x'\in \mathcal{X}$ and $F_{\epsilon_2}(\cdot|x)=F_{\epsilon_2}(\cdot|x')$ $\forall x,x'\in \mathcal{X}$. Can we claim that $F_{\epsilon_1-\epsilon_2}(\cdot|x)=F_{\epsilon_1-\epsilon_2}(\cdot|x')$ $\forall x,x'\in \mathcal{X}$?

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The answer to the first two "can we claim" questions is no.

Case 1: Consider the case where $\epsilon_1$ is $-1$ and $1$ with probability $1/2$, but that $\epsilon_2 = \epsilon_1$ with probability $1$. Then $F_{\epsilon_1 - \epsilon_2}(0 \mid x) = 1$ for any $x$, regardless of what the variable $X$ is or what its relationship is to $\epsilon_1, \epsilon_2$. On the other hand, if all the variables are independent of one another, then you can easily cook up a case where the claim would hold. Consequently, not much can be said about this case in general.

Case 2: Consider a case where the sample space for the variables $\epsilon_1, \epsilon_2$ is $\{a, b, c\}$, each with probability $1/3$. Let $\epsilon_1(a) = -1, \epsilon_1(b) = 0, \epsilon_1(c) = 1$, while $\epsilon_1(a) = 0, \epsilon_2(b) = 1, \epsilon_2(c) = -1$. Let $X$ be independent of $\epsilon_1$ and $\epsilon_2$. Then the variable $\epsilon_1 - \epsilon_2$ is $-1$ with probability $2/3$ and is $1$ with probability $1/3$, so it is not symmetric.

Case 3: The answer to this claim is yes. If $X$ is independent of $\epsilon_1$ and $\epsilon_2$, then it is independent of any function of those (such as $\epsilon_1 - \epsilon_2$) by a classical theorem.

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