Mixing
If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $XY$ are independent?
$begingroup$
If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $Z=XY$ are independent?
It is quite intuivively obvious that $Zsimmathcal N(0,1)$ and that would imply that $Y$ and $Z$ are independent, but how to prove it?
$E(Z)=E(XY)=E(X)E(Y)=0cdot0=0$
$Var(Z)=E((E(XY)-XY)^2)=E(X^2Y^2)=E(X^2)=Var(X)+E(X)^2=1+0=1$
But that merely gives us the expected value and variance, this probably isn't enough to conclude that $Zsimmathcal N(0,1)$
Would $X$ and $Z$ also be independent?
Update:
Greedoid's answer led me to this:
$$P(Zle z)= P(Zle z|Y=1)P(Y=1)+P(Zle z|Y=-1)P(Y=-1) \ = P(Xle z)cdot{1over2}+P(-Xle z)cdot{1over2}\ ={1over2}phi(z)+{1over2}phi(z)=phi(z)$$
where $phi$ is the cumulative distribution function of the $mathcal N(0,1)$ law
And this is true because $P(-Xle z)=P(Xle z)$ because $-Xsimmathcal N(0,1)$
probability-theory normal-distribution independence
edited Jan 27 at 10:28
Did
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
$endgroup$
add a comment |
$begingroup$
If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $Z=XY$ are independent?
It is quite intuivively obvious that $Zsimmathcal N(0,1)$ and that would imply that $Y$ and $Z$ are independent, but how to prove it?
$E(Z)=E(XY)=E(X)E(Y)=0cdot0=0$
$Var(Z)=E((E(XY)-XY)^2)=E(X^2Y^2)=E(X^2)=Var(X)+E(X)^2=1+0=1$
But that merely gives us the expected value and variance, this probably isn't enough to conclude that $Zsimmathcal N(0,1)$
Would $X$ and $Z$ also be independent?
Update:
Greedoid's answer led me to this:
$$P(Zle z)= P(Zle z|Y=1)P(Y=1)+P(Zle z|Y=-1)P(Y=-1) \ = P(Xle z)cdot{1over2}+P(-Xle z)cdot{1over2}\ ={1over2}phi(z)+{1over2}phi(z)=phi(z)$$
where $phi$ is the cumulative distribution function of the $mathcal N(0,1)$ law
And this is true because $P(-Xle z)=P(Xle z)$ because $-Xsimmathcal N(0,1)$
probability-theory normal-distribution independence
edited Jan 27 at 10:28
Did
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
$endgroup$
$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59
add a comment |
$begingroup$
If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $Z=XY$ are independent?
It is quite intuivively obvious that $Zsimmathcal N(0,1)$ and that would imply that $Y$ and $Z$ are independent, but how to prove it?
$E(Z)=E(XY)=E(X)E(Y)=0cdot0=0$
$Var(Z)=E((E(XY)-XY)^2)=E(X^2Y^2)=E(X^2)=Var(X)+E(X)^2=1+0=1$
But that merely gives us the expected value and variance, this probably isn't enough to conclude that $Zsimmathcal N(0,1)$
Would $X$ and $Z$ also be independent?
Update:
Greedoid's answer led me to this:
$$P(Zle z)= P(Zle z|Y=1)P(Y=1)+P(Zle z|Y=-1)P(Y=-1) \ = P(Xle z)cdot{1over2}+P(-Xle z)cdot{1over2}\ ={1over2}phi(z)+{1over2}phi(z)=phi(z)$$
where $phi$ is the cumulative distribution function of the $mathcal N(0,1)$ law
And this is true because $P(-Xle z)=P(Xle z)$ because $-Xsimmathcal N(0,1)$
probability-theory normal-distribution independence
edited Jan 27 at 10:28
Did
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
$endgroup$
If $X$ standard normal and $Y$ symmetric Bernoulli are independent then $Y$ and $Z=XY$ are independent?
It is quite intuivively obvious that $Zsimmathcal N(0,1)$ and that would imply that $Y$ and $Z$ are independent, but how to prove it?
$E(Z)=E(XY)=E(X)E(Y)=0cdot0=0$
$Var(Z)=E((E(XY)-XY)^2)=E(X^2Y^2)=E(X^2)=Var(X)+E(X)^2=1+0=1$
But that merely gives us the expected value and variance, this probably isn't enough to conclude that $Zsimmathcal N(0,1)$
Would $X$ and $Z$ also be independent?
Update:
Greedoid's answer led me to this:
$$P(Zle z)= P(Zle z|Y=1)P(Y=1)+P(Zle z|Y=-1)P(Y=-1) \ = P(Xle z)cdot{1over2}+P(-Xle z)cdot{1over2}\ ={1over2}phi(z)+{1over2}phi(z)=phi(z)$$
where $phi$ is the cumulative distribution function of the $mathcal N(0,1)$ law
And this is true because $P(-Xle z)=P(Xle z)$ because $-Xsimmathcal N(0,1)$
probability-theory normal-distribution independence
probability-theory normal-distribution independence
edited Jan 27 at 10:28
Did
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
edited Jan 27 at 10:28
Did
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
edited Jan 27 at 10:28
Did
248k23226466
edited Jan 27 at 10:28
Did
248k23226466
edited Jan 27 at 10:28
Did
248k23226466
248k23226466
asked Jan 27 at 8:36
H. WalterH. Walter
1047
asked Jan 27 at 8:36
H. WalterH. Walter
1047
asked Jan 27 at 8:36
H. WalterH. Walter
1047
1047
$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59
add a comment |
$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59
$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $f$ and $g$ be bounded measurable functions.
Note that $$begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\
&=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) tag1\
&=[E(f(X))frac 12+E(f(X))frac 12]E(g(Y)) tag2\
&= E(f(X))E(g(Y))
end{align}
$$
$(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric
and $$begin{align}
E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\
&= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])tag 1\
&= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])tag2\
&= E(f(X)) E(g(Y))
end{align}$$
$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=int f(yx)dP_{X|Y=y}(x)=int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric
Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.
Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
$endgroup$
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $f$ and $g$ be bounded measurable functions.
Note that $$begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\
&=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) tag1\
&=[E(f(X))frac 12+E(f(X))frac 12]E(g(Y)) tag2\
&= E(f(X))E(g(Y))
end{align}
$$
$(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric
and $$begin{align}
E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\
&= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])tag 1\
&= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])tag2\
&= E(f(X)) E(g(Y))
end{align}$$
$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=int f(yx)dP_{X|Y=y}(x)=int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric
Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.
Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
$endgroup$
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
add a comment |
$begingroup$
Let $f$ and $g$ be bounded measurable functions.
Note that $$begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\
&=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) tag1\
&=[E(f(X))frac 12+E(f(X))frac 12]E(g(Y)) tag2\
&= E(f(X))E(g(Y))
end{align}
$$
$(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric
and $$begin{align}
E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\
&= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])tag 1\
&= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])tag2\
&= E(f(X)) E(g(Y))
end{align}$$
$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=int f(yx)dP_{X|Y=y}(x)=int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric
Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.
Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
$endgroup$
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
add a comment |
$begingroup$
Let $f$ and $g$ be bounded measurable functions.
Note that $$begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\
&=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) tag1\
&=[E(f(X))frac 12+E(f(X))frac 12]E(g(Y)) tag2\
&= E(f(X))E(g(Y))
end{align}
$$
$(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric
and $$begin{align}
E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\
&= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])tag 1\
&= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])tag2\
&= E(f(X)) E(g(Y))
end{align}$$
$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=int f(yx)dP_{X|Y=y}(x)=int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric
Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.
Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
$endgroup$
Let $f$ and $g$ be bounded measurable functions.
Note that $$begin{align}E(f(XY))E(g(Y))&=E(f(X)1_{Y=1}+f(-X)1_{Y=-1})E(g(Y))\
&=[E(f(X))E(1_{Y=1})+E(f(-X))E(1_{Y=-1})]E(g(Y)) tag1\
&=[E(f(X))frac 12+E(f(X))frac 12]E(g(Y)) tag2\
&= E(f(X))E(g(Y))
end{align}
$$
$(1)$: $X$ and $Y$ are independent
$(2)$: $X$ is symmetric
and $$begin{align}
E(f(XY)g(Y))&= E(E(f(XY)g(Y)|Y)) = E(g(Y) E(f(XY)|Y))\
&= E(g(Y)[1_{Y=1}E(f(X)) + 1_{Y=-1}E(f(-X))])tag 1\
&= E(f(X)) [E(g(Y)(1_{Y=1}+1_{Y=-1})])tag2\
&= E(f(X)) E(g(Y))
end{align}$$
$(1)$: $E(f(XY)|Y=y)=E(f(yX)|Y=y)=int f(yx)dP_{X|Y=y}(x)=int f(yx)dP_{X}(x)=E(yX)$
$(2)$: $X$ is symmetric
Thus $E(f(XY)g(Y))= E(f(XY))E(g(Y))$, hence $XY$ and $Y$ are independent.
Note that the only property of $X$ I used is that its distribution is symmetric. Normality does not matter.
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
edited Jan 27 at 10:19
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
answered Jan 27 at 10:13
Gabriel RomonGabriel Romon
18.1k53387
18.1k53387
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
add a comment |
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
Thanks, what definition of independence are you using?
$endgroup$
– H. Walter
Jan 27 at 10:36
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
$begingroup$
@H.Walter See Lemma 2.7 in this or check this question
$endgroup$
– Gabriel Romon
Jan 27 at 10:40
add a comment |
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$begingroup$
are you asking about the independence of $X$ and $Z$ or $Y$ and $Z$?
$endgroup$
– pointguard0
Jan 27 at 8:57
$begingroup$
"Z∼N(0,1) ... would imply that Y and Z are independent" How would this imply that?
$endgroup$
– Did
Jan 27 at 10:29
$begingroup$
BTW "symmetric Bernoulli" is called Rademacher. en.wikipedia.org/wiki/Rademacher_distribution
$endgroup$
– J.G.
Jan 27 at 10:59