Let $Y,X,W$ be real-valued random variables, respectively with supports denoted by $\mathcal{Y},\mathcal{X},\mathcal{W}$.

(A1) Assume that $\mathcal{X},\mathcal{W}$ are finite. Without loss of generality, assume that $\mathcal{X}\equiv \{x_1,x_2\}$ and $\mathcal{W}\equiv \{w_1,w_2\}$.

(A2) For each realisation $x\in \mathcal{X}$ of $X$, let $\epsilon_x$ be another random variable. Assume that, $\forall x \in \mathcal{X}$, $\epsilon_x$ is stochastically independent of $X,W$.

(A3) Assume that the following relation holds $$ Y=h(X,W)+\epsilon_{X} $$

Consider now the cdf $F(\cdot)$ of $Y$ evaluated at $y\in \mathcal{Y}$. Following here, we can write

$$F(y)=p(x_1,w_1)\times F(y| X=x_1, W=w_1) $$ $$ +p(x_2,w_1)\times F(y| X=x_2, W=w_1) $$ $$ +p(x_1,w_2)\times F(y| X=x_1, W=w_2)$$ $$ +p(x_2,w_2)\times F(y| X=x_2, W=w_2) $$

where $p(x,w)$ is the probability mass function of $(X,W)$ evaluated at $(x,w)$ and $F(\cdot| X=x, W=w)$ is the cdf of $Y$ conditional on $X=x,W=w$.

The lines above highlight that $F(\cdot)$ can be expressed as a finite mixture.

Let's focus on the relation between $F(\cdot| X=x, W=w)$, (A3), and the cdf of $\epsilon_x$.

For any $(x,w)$, $F(\cdot| X=x, W=w)$ is determined by (A3) and the cdf of $\epsilon_x$.

For example, if $\epsilon_x\sim \mathcal{N}(\alpha_x,\sigma^2_x)$, then $Y|X=x, W=w\sim N(h(x,w)+\alpha_x,\sigma^2_x)$.

In my exercise, I want to remain non-parametric about the distribution of $\epsilon_x$ and I'm looking for non-parametric features of $F(\cdot |X=x,W=w)$ that are compatible with (A3). Specifically, this is my question:

Question: is (A3) compatible with writing $F(\cdot)$ at any $y\in \mathcal{Y}$ as $$ F(y)=\sum_{x\in \mathcal{X}, w\in \mathcal{W}} p(x,w) G(y-\mu_{x,w}) $$ where $G: \mathbb{R}\rightarrow [0,1]$ is a cdf symmetric around zero [i.e., $G(y)=1-G(-y)$] and $\{\mu_{x,w}\}_{x,w}$ are real numbers all different between each other?

In other words, I'm wondering whether the differences across $$ F(y| X=x_1, W=w_1), F(y| X=x_1, W=w_2),F(y| X=x_2, W=w_1) ,F(y| X=x_2, W=w_2) $$ could be captured by a location shift $\mu_{x,w}$ differing across $(x,w)$.

Further thoughts: notice that, as explained here, the cdf's $\{H_{x,w}\}_{x,w}$ with $$ H_{x,w}: t\in \mathbb{R}\mapsto G(t-\mu_{x,w}) $$ are characterised by equivalent central moments.


$$F\left(y\mid X=x_{i},W=w_{j}\right)=P\left(h\left(X,W\right)+\epsilon_{X}\leq y\mid X=x_{i},W=w_{j}\right)=$$$$P\left(h\left(x_{i},w_{j}\right)+\epsilon_{x_{i}}\leq y\mid X=x_{i},W=w_{j}\right)=P\left(h\left(x_{i},w_{j}\right)+\epsilon_{x_{i}}\leq y\right)=G_{i}\left(y-h\left(x_{i},w_{j}\right)\right)$$ where $G_{i}$ denotes the CDF of $\epsilon_{x_{i}}$.

The third equality is based on independence.

Only if the distribution of $\epsilon_{x_{i}}$ does not depend on $i$ then you can write $$F\left(y\mid X=x_{i},W=w_{j}\right)=G\left(y-h\left(x_{i},w_{j}\right)\right)==G\left(y-h\left(x_{i},w_{j}\right)\right)$$and:$$ F(y)=\sum_{x\in \mathcal{X}, w\in \mathcal{W}} p(x,w) G(y-h(x,w)) $$ where $G$ denotes the common CDF of the $\epsilon_{x_{i}}$.

  • $\begingroup$ Thanks. Just to clarify "Only if the distribution of $\epsilon_{x_i}$ does not depend on $i$": is this equivalent to say "Only if $\epsilon_{x_1}, \epsilon_{x_2}$ are identically distributed"? $\endgroup$ – STF Dec 28 '18 at 16:50
  • $\begingroup$ Yes. That is correct. $\endgroup$ – drhab Dec 28 '18 at 17:10
  • $\begingroup$ Also, maybe there is a typo in your answer: what do you mean by $G(y-h(x_i, w_j))==G(y-h(x_i, w_j))$? $\endgroup$ – STF Dec 28 '18 at 17:16

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