From independence to conditional independence
$begingroup$
Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.
Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.
probability probability-theory random-variables independence
asked Apr 27 '17 at 14:55
STFSTF
681420
$endgroup$
|
show 7 more comments
$begingroup$
Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.
Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.
probability probability-theory random-variables independence
asked Apr 27 '17 at 14:55
STFSTF
681420
$endgroup$
2
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
1
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09
|
show 7 more comments
$begingroup$
Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.
Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.
probability probability-theory random-variables independence
asked Apr 27 '17 at 14:55
STFSTF
681420
$endgroup$
Consider two random variables $X,Y, Z$. Suppose $Xperp Z$ and $Yperp Z$, where $perp $ denotes independence.
Is it true that ($Xperp Y$) implies ($Xperp Y$ conditional on $Z$)? I know that in general independence does not imply conditional independence, but I was wondering whether the fact that $X$ and $Y$ are independent of $Z$ simplifies things.
probability probability-theory random-variables independence
probability probability-theory random-variables independence
asked Apr 27 '17 at 14:55
STFSTF
681420
asked Apr 27 '17 at 14:55
STFSTF
681420
asked Apr 27 '17 at 14:55
STFSTF
681420
asked Apr 27 '17 at 14:55
STFSTF
681420
asked Apr 27 '17 at 14:55
STFSTF
681420
681420
2
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
1
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09
|
show 7 more comments
2
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
1
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09
2
2
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
1
1
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
$endgroup$
add a comment |
$begingroup$
$Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.
And further we can obtain:
$$begin{align}
begin{array}{rc}
Xperp Y | Z
implies&
p(X cap Y|Z) = p(X|Z)p(Y|Z) \
implies&
frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
implies&
p(X|Y,Z) = p(X|Z)
end{array}
end{align}$$
And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:
$$begin{align}
p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
& = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
& = frac{p(Y)p(Z|X,Y)}{p(Z)}
end{align}$$
So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.
I hope the above reduction and reasoning hold.
answered Jan 13 at 3:16
lernerlerner
312216
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add a comment |
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2 Answers
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$begingroup$
It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
$endgroup$
add a comment |
$begingroup$
It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
$endgroup$
add a comment |
$begingroup$
It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
$endgroup$
It is well known that there exist events $A,B,C$ such that any two of them are independent but $P(Acap Bcap C) neq P(A)P(B)P(C)$. Take $X=I_A.Y=I_B,Z=I_C$ to get a counterexample.
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
answered Jan 13 at 12:36
Kavi Rama MurthyKavi Rama Murthy
56.5k42159
56.5k42159
add a comment |
add a comment |
$begingroup$
$Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.
And further we can obtain:
$$begin{align}
begin{array}{rc}
Xperp Y | Z
implies&
p(X cap Y|Z) = p(X|Z)p(Y|Z) \
implies&
frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
implies&
p(X|Y,Z) = p(X|Z)
end{array}
end{align}$$
And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:
$$begin{align}
p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
& = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
& = frac{p(Y)p(Z|X,Y)}{p(Z)}
end{align}$$
So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.
I hope the above reduction and reasoning hold.
answered Jan 13 at 3:16
lernerlerner
312216
$endgroup$
add a comment |
$begingroup$
$Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.
And further we can obtain:
$$begin{align}
begin{array}{rc}
Xperp Y | Z
implies&
p(X cap Y|Z) = p(X|Z)p(Y|Z) \
implies&
frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
implies&
p(X|Y,Z) = p(X|Z)
end{array}
end{align}$$
And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:
$$begin{align}
p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
& = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
& = frac{p(Y)p(Z|X,Y)}{p(Z)}
end{align}$$
So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.
I hope the above reduction and reasoning hold.
answered Jan 13 at 3:16
lernerlerner
312216
$endgroup$
add a comment |
$begingroup$
$Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.
And further we can obtain:
$$begin{align}
begin{array}{rc}
Xperp Y | Z
implies&
p(X cap Y|Z) = p(X|Z)p(Y|Z) \
implies&
frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
implies&
p(X|Y,Z) = p(X|Z)
end{array}
end{align}$$
And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:
$$begin{align}
p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
& = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
& = frac{p(Y)p(Z|X,Y)}{p(Z)}
end{align}$$
So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.
I hope the above reduction and reasoning hold.
answered Jan 13 at 3:16
lernerlerner
312216
$endgroup$
$Xperp Y | Z$ implies that $p(X cap Y|Z) = p(X|Z)p(Y|Z)$.
And further we can obtain:
$$begin{align}
begin{array}{rc}
Xperp Y | Z
implies&
p(X cap Y|Z) = p(X|Z)p(Y|Z) \
implies&
frac{p(X cap Y|Z)}{p(Y|Z)}=p(X|Z) \
implies&
p(X|Y,Z) = p(X|Z)
end{array}
end{align}$$
And if $X perp Z$, $Yperp Z$ and $Xperp Y$ we can get:
$$begin{align}
p(X|Y,Z) &= frac{p(X, Y, Z)}{p(X,Z)} \
& = frac{p(X)p(Y|X)p(Z|Y, X)}{p(X)p(Z|X)} \
& = frac{p(Y)p(Z|X,Y)}{p(Z)}
end{align}$$
So the first case $X perp Z$, $Yperp Z$ and $Xperp Y$ cannot imply the second $Xperp Y | Z$.
I hope the above reduction and reasoning hold.
answered Jan 13 at 3:16
lernerlerner
312216
edited Jan 13 at 8:02
answered Jan 13 at 3:16
lernerlerner
312216
answered Jan 13 at 3:16
lernerlerner
312216
answered Jan 13 at 3:16
lernerlerner
312216
312216
add a comment |
add a comment |
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2
$begingroup$
No. Say you are tossing two fair coins. $X$ is the event "the first coin comes up $H$". $Y$ is the event "the second coin comes up $H$". $Z$ is the event "the two coins come up the same"
$endgroup$
– lulu
Apr 27 '17 at 15:03
$begingroup$
Is in your example $X$ independent of $Z$?
$endgroup$
– STF
Apr 27 '17 at 15:06
1
$begingroup$
Of course. Knowing that the two coins match doesn't tell me what the first coin came up.
$endgroup$
– lulu
Apr 27 '17 at 15:07
$begingroup$
OK, what additional condition I would need to go from independence to conditional independence?
$endgroup$
– STF
Apr 27 '17 at 15:08
$begingroup$
Well, you could just assume conditional independence. Not sure there's some natural intermediate assumption....
$endgroup$
– lulu
Apr 27 '17 at 15:09