Show that if $Y$ is $sigma(X)$-measurable, $Xsimoperatorname B(p)$ then $Y=aX+b$

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Let $X$ a Bernoulli random variable with parameter $pin [0,1]$ : $$ P(X=1)=p,quad P(X=0)=1-p.$$ Let $Y$ a real r.v. $sigma(X)$-measurable. Then there exists $a,bin mathbbR$ such that almost surely: $$Y=aX+b.$$
I would like to prove this; any hint/help would be appreciated.




probability-theory measure-theory




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  • Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
    – GEdgar
    Aug 6 at 16:55














up vote
1
down vote

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Let $X$ a Bernoulli random variable with parameter $pin [0,1]$ : $$ P(X=1)=p,quad P(X=0)=1-p.$$ Let $Y$ a real r.v. $sigma(X)$-measurable. Then there exists $a,bin mathbbR$ such that almost surely: $$Y=aX+b.$$
I would like to prove this; any hint/help would be appreciated.




probability-theory measure-theory




share|cite|improve this question





















  • Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
    – GEdgar
    Aug 6 at 16:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $X$ a Bernoulli random variable with parameter $pin [0,1]$ : $$ P(X=1)=p,quad P(X=0)=1-p.$$ Let $Y$ a real r.v. $sigma(X)$-measurable. Then there exists $a,bin mathbbR$ such that almost surely: $$Y=aX+b.$$
I would like to prove this; any hint/help would be appreciated.




probability-theory measure-theory




share|cite|improve this question













Let $X$ a Bernoulli random variable with parameter $pin [0,1]$ : $$ P(X=1)=p,quad P(X=0)=1-p.$$ Let $Y$ a real r.v. $sigma(X)$-measurable. Then there exists $a,bin mathbbR$ such that almost surely: $$Y=aX+b.$$
I would like to prove this; any hint/help would be appreciated.





probability-theory measure-theory





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share|cite|improve this question




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edited Aug 6 at 18:15









Michael Hardy

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asked Aug 6 at 16:43









anonymus

854312




854312











  • Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
    – GEdgar
    Aug 6 at 16:55
















  • Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
    – GEdgar
    Aug 6 at 16:55















Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
– GEdgar
Aug 6 at 16:55




Perhaps: any function defined on the set $0,1$ is of the form $ax+b$.
– GEdgar
Aug 6 at 16:55










1 Answer
1






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2
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accepted










Since $Y$ is $sigma(X)$-measurable, $Y$ is constant on the atoms of $sigma(X)$, which are $X=1$ and $X=0$. Let $y_1 = Y(omega)$ if $omega in X=1$ and $b = Y(omega)$ if $omega in X=0$. Let $a = y_1 - b$. Then,
$$Y(omega) = aX(omega) + b$$
for almost all $omega$.






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  • I wasn't expecting that much, thanks alot !
    – anonymus
    Aug 6 at 17:32










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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Since $Y$ is $sigma(X)$-measurable, $Y$ is constant on the atoms of $sigma(X)$, which are $X=1$ and $X=0$. Let $y_1 = Y(omega)$ if $omega in X=1$ and $b = Y(omega)$ if $omega in X=0$. Let $a = y_1 - b$. Then,
$$Y(omega) = aX(omega) + b$$
for almost all $omega$.






share|cite|improve this answer





















  • I wasn't expecting that much, thanks alot !
    – anonymus
    Aug 6 at 17:32














up vote
2
down vote



accepted










Since $Y$ is $sigma(X)$-measurable, $Y$ is constant on the atoms of $sigma(X)$, which are $X=1$ and $X=0$. Let $y_1 = Y(omega)$ if $omega in X=1$ and $b = Y(omega)$ if $omega in X=0$. Let $a = y_1 - b$. Then,
$$Y(omega) = aX(omega) + b$$
for almost all $omega$.






share|cite|improve this answer





















  • I wasn't expecting that much, thanks alot !
    – anonymus
    Aug 6 at 17:32












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Since $Y$ is $sigma(X)$-measurable, $Y$ is constant on the atoms of $sigma(X)$, which are $X=1$ and $X=0$. Let $y_1 = Y(omega)$ if $omega in X=1$ and $b = Y(omega)$ if $omega in X=0$. Let $a = y_1 - b$. Then,
$$Y(omega) = aX(omega) + b$$
for almost all $omega$.






share|cite|improve this answer













Since $Y$ is $sigma(X)$-measurable, $Y$ is constant on the atoms of $sigma(X)$, which are $X=1$ and $X=0$. Let $y_1 = Y(omega)$ if $omega in X=1$ and $b = Y(omega)$ if $omega in X=0$. Let $a = y_1 - b$. Then,
$$Y(omega) = aX(omega) + b$$
for almost all $omega$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 6 at 17:03









aduh

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  • I wasn't expecting that much, thanks alot !
    – anonymus
    Aug 6 at 17:32
















  • I wasn't expecting that much, thanks alot !
    – anonymus
    Aug 6 at 17:32















I wasn't expecting that much, thanks alot !
– anonymus
Aug 6 at 17:32




I wasn't expecting that much, thanks alot !
– anonymus
Aug 6 at 17:32












 

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