1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.