On the properties of a finite mixture
$begingroup$
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 $xin mathcal{X}$ 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 $yin mathcal{Y}$. 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 $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.
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 $yin mathcal{Y}$ as
$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
$$
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
$$
H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
$$
are characterised by equivalent central moments.
probability probability-theory probability-distributions random-variables conditional-probability
$endgroup$
add a comment |
$begingroup$
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 $xin mathcal{X}$ 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 $yin mathcal{Y}$. 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 $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.
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 $yin mathcal{Y}$ as
$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
$$
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
$$
H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
$$
are characterised by equivalent central moments.
probability probability-theory probability-distributions random-variables conditional-probability
$endgroup$
add a comment |
$begingroup$
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 $xin mathcal{X}$ 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 $yin mathcal{Y}$. 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 $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.
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 $yin mathcal{Y}$ as
$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
$$
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
$$
H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
$$
are characterised by equivalent central moments.
probability probability-theory probability-distributions random-variables conditional-probability
$endgroup$
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 $xin mathcal{X}$ 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 $yin mathcal{Y}$. 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 $epsilon_xsim mathcal{N}(alpha_x,sigma^2_x)$, then $Y|X=x, W=wsim N(h(x,w)+alpha_x,sigma^2_x)$.
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 $yin mathcal{Y}$ as
$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-mu_{x,w})
$$
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
$$
H_{x,w}: tin mathbb{R}mapsto G(t-mu_{x,w})
$$
are characterised by equivalent central moments.
probability probability-theory probability-distributions random-variables conditional-probability
probability probability-theory probability-distributions random-variables conditional-probability
edited Dec 28 '18 at 16:28
STF
asked Dec 28 '18 at 16:12
STFSTF
541422
541422
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
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
$$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
$$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
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
$$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
$$
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
add a comment |
$begingroup$
$$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
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
$$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
$$
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
add a comment |
$begingroup$
$$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
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
$$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
$$
where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.
$endgroup$
$$Fleft(ymid X=x_{i},W=w_{j}right)=Pleft(hleft(X,Wright)+epsilon_{X}leq ymid X=x_{i},W=w_{j}right)=$$$$Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq ymid X=x_{i},W=w_{j}right)=Pleft(hleft(x_{i},w_{j}right)+epsilon_{x_{i}}leq yright)=G_{i}left(y-hleft(x_{i},w_{j}right)right)$$
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
$$Fleft(ymid X=x_{i},W=w_{j}right)=Gleft(y-hleft(x_{i},w_{j}right)right)==Gleft(y-hleft(x_{i},w_{j}right)right)$$and:$$
F(y)=sum_{xin mathcal{X}, win mathcal{W}} p(x,w) G(y-h(x,w))
$$
where $G$ denotes the common CDF of the $epsilon_{x_{i}}$.
answered Dec 28 '18 at 16:38
drhabdrhab
103k545136
103k545136
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
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StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
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Post as a guest
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Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown