conditional probability which their condition have $XOR and Z=X+Y$

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Let $X$ and $Y$ be two independent binary random variable with the same alphabet $0,1$,ie,$Pr(0)=Pr(1)=frac12$

Define $I(X;Y|Z)=H(X|Z)-H(X|Y,Z)$

$1.$Let $Z=X+Y$,Find $I(X;Y|Z)$

$2.$Let $Z=X Ã¢ÂŠÂ•Y$, Find $I(X;Y|Z)$. Ã¢ÂŠÂ• means XOR operation

For these two questions,i don't know how to find their probability,can anyone teach me

By the way, i know the $I$(information) and $H$(entropy) are not the probability,$Pr(X;Y|Z)$,but i need the probability before calculating them

edited Jul 29 at 8:41

asked Jul 29 at 2:00

Shine Sun

1258

So you want the joint distribution of $X$ and $Y$ given $Z$? $Z$ can be $0,1$ or $2.$ Conditional on $Z=0,$ $(X,Y) = (0,0)$ with probability one. Conditional on $Z=2,$ $(X,Y)=(1,1)$ with probability one. And conditional on $Z=1,$ $(X,Y) = (1,0)$ with probability $1/2$ and $(X,Y) = (0,1)$ with probability $1/2.$
â€“Â spaceisdarkgreen
Jul 29 at 3:26

oh!you misunderstand something,that 2 is the question two.And in fact, i want the joint distribution of X given Y and Z to calculate the I(X;Y|Z)
â€“Â Shine Sun
Jul 29 at 8:40

Given $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2
â€“Â Henry
Jul 29 at 9:12

@Henry I just think that is H(X|Y,Z)=H(X|Z)? Because P(X|Y,Z) should me the sum of Pr(X=x) given that Y=y and Z=z
â€“Â Shine Sun
Jul 29 at 13:07

add a commentÂ |Â

up vote
0
down vote

favorite

Let $X$ and $Y$ be two independent binary random variable with the same alphabet $0,1$,ie,$Pr(0)=Pr(1)=frac12$

Define $I(X;Y|Z)=H(X|Z)-H(X|Y,Z)$

$1.$Let $Z=X+Y$,Find $I(X;Y|Z)$

$2.$Let $Z=X Ã¢ÂŠÂ•Y$, Find $I(X;Y|Z)$. Ã¢ÂŠÂ• means XOR operation

For these two questions,i don't know how to find their probability,can anyone teach me

By the way, i know the $I$(information) and $H$(entropy) are not the probability,$Pr(X;Y|Z)$,but i need the probability before calculating them

edited Jul 29 at 8:41

asked Jul 29 at 2:00

Shine Sun

1258

So you want the joint distribution of $X$ and $Y$ given $Z$? $Z$ can be $0,1$ or $2.$ Conditional on $Z=0,$ $(X,Y) = (0,0)$ with probability one. Conditional on $Z=2,$ $(X,Y)=(1,1)$ with probability one. And conditional on $Z=1,$ $(X,Y) = (1,0)$ with probability $1/2$ and $(X,Y) = (0,1)$ with probability $1/2.$
â€“Â spaceisdarkgreen
Jul 29 at 3:26

oh!you misunderstand something,that 2 is the question two.And in fact, i want the joint distribution of X given Y and Z to calculate the I(X;Y|Z)
â€“Â Shine Sun
Jul 29 at 8:40

Given $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2
â€“Â Henry
Jul 29 at 9:12

@Henry I just think that is H(X|Y,Z)=H(X|Z)? Because P(X|Y,Z) should me the sum of Pr(X=x) given that Y=y and Z=z
â€“Â Shine Sun
Jul 29 at 13:07

add a commentÂ |Â

up vote
0
down vote

favorite

Let $X$ and $Y$ be two independent binary random variable with the same alphabet $0,1$,ie,$Pr(0)=Pr(1)=frac12$

Define $I(X;Y|Z)=H(X|Z)-H(X|Y,Z)$

$1.$Let $Z=X+Y$,Find $I(X;Y|Z)$

$2.$Let $Z=X Ã¢ÂŠÂ•Y$, Find $I(X;Y|Z)$. Ã¢ÂŠÂ• means XOR operation

For these two questions,i don't know how to find their probability,can anyone teach me

By the way, i know the $I$(information) and $H$(entropy) are not the probability,$Pr(X;Y|Z)$,but i need the probability before calculating them

edited Jul 29 at 8:41

asked Jul 29 at 2:00

Shine Sun

1258

Let $X$ and $Y$ be two independent binary random variable with the same alphabet $0,1$,ie,$Pr(0)=Pr(1)=frac12$

Define $I(X;Y|Z)=H(X|Z)-H(X|Y,Z)$

$1.$Let $Z=X+Y$,Find $I(X;Y|Z)$

$2.$Let $Z=X Ã¢ÂŠÂ•Y$, Find $I(X;Y|Z)$. Ã¢ÂŠÂ• means XOR operation

For these two questions,i don't know how to find their probability,can anyone teach me

By the way, i know the $I$(information) and $H$(entropy) are not the probability,$Pr(X;Y|Z)$,but i need the probability before calculating them

edited Jul 29 at 8:41

asked Jul 29 at 2:00

Shine Sun

1258

edited Jul 29 at 8:41

asked Jul 29 at 2:00

Shine Sun

1258

asked Jul 29 at 2:00

Shine Sun

1258

asked Jul 29 at 2:00

Shine Sun

1258

So you want the joint distribution of $X$ and $Y$ given $Z$? $Z$ can be $0,1$ or $2.$ Conditional on $Z=0,$ $(X,Y) = (0,0)$ with probability one. Conditional on $Z=2,$ $(X,Y)=(1,1)$ with probability one. And conditional on $Z=1,$ $(X,Y) = (1,0)$ with probability $1/2$ and $(X,Y) = (0,1)$ with probability $1/2.$
â€“Â spaceisdarkgreen
Jul 29 at 3:26

oh!you misunderstand something,that 2 is the question two.And in fact, i want the joint distribution of X given Y and Z to calculate the I(X;Y|Z)
â€“Â Shine Sun
Jul 29 at 8:40

Given $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2
â€“Â Henry
Jul 29 at 9:12

@Henry I just think that is H(X|Y,Z)=H(X|Z)? Because P(X|Y,Z) should me the sum of Pr(X=x) given that Y=y and Z=z
â€“Â Shine Sun
Jul 29 at 13:07

add a commentÂ |Â

So you want the joint distribution of $X$ and $Y$ given $Z$? $Z$ can be $0,1$ or $2.$ Conditional on $Z=0,$ $(X,Y) = (0,0)$ with probability one. Conditional on $Z=2,$ $(X,Y)=(1,1)$ with probability one. And conditional on $Z=1,$ $(X,Y) = (1,0)$ with probability $1/2$ and $(X,Y) = (0,1)$ with probability $1/2.$
â€“Â spaceisdarkgreen
Jul 29 at 3:26

oh!you misunderstand something,that 2 is the question two.And in fact, i want the joint distribution of X given Y and Z to calculate the I(X;Y|Z)
â€“Â Shine Sun
Jul 29 at 8:40

Given $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2
â€“Â Henry
Jul 29 at 9:12

@Henry I just think that is H(X|Y,Z)=H(X|Z)? Because P(X|Y,Z) should me the sum of Pr(X=x) given that Y=y and Z=z
â€“Â Shine Sun
Jul 29 at 13:07

So you want the joint distribution of $X$ and $Y$ given $Z$? $Z$ can be $0,1$ or $2.$ Conditional on $Z=0,$ $(X,Y) = (0,0)$ with probability one. Conditional on $Z=2,$ $(X,Y)=(1,1)$ with probability one. And conditional on $Z=1,$ $(X,Y) = (1,0)$ with probability $1/2$ and $(X,Y) = (0,1)$ with probability $1/2.$
â€“Â spaceisdarkgreen
Jul 29 at 3:26

oh!you misunderstand something,that 2 is the question two.And in fact, i want the joint distribution of X given Y and Z to calculate the I(X;Y|Z)
â€“Â Shine Sun
Jul 29 at 8:40

Given $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2
â€“Â Henry
Jul 29 at 9:12

@Henry I just think that is H(X|Y,Z)=H(X|Z)? Because P(X|Y,Z) should me the sum of Pr(X=x) given that Y=y and Z=z
â€“Â Shine Sun
Jul 29 at 13:07

add a commentÂ |Â

1 Answer
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0
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You asked for the probabilities. These are:

X Y X+Y XÃ¢ÂŠÂ•Y Prob
0 0 0 0 1/4 
0 1 1 1 1/4
1 0 1 1 1/4
1 1 2 0 1/4

For $mathbb P(X=x mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

$mathbb P(X=0 mid X+Y=0)=1$, $,mathbb P(X=1 mid X+Y=0)=0$,

$mathbb P(X=0 mid X+Y=1)=frac12$, $mathbb P(X=1 mid X+Y=1)=frac12$,

$mathbb P(X=0 mid X+Y=2)=0$, $,mathbb P(X=1 mid X+Y=1)=1$

In Part 2 you have

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=0)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=0)=frac12$,

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=1)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=1)=frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=ZÃ¢ÂŠÂ•Y$

answered Jul 29 at 9:20

Henry

92.8k469147

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

add a commentÂ |Â

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You asked for the probabilities. These are:

X Y X+Y XÃ¢ÂŠÂ•Y Prob
0 0 0 0 1/4 
0 1 1 1 1/4
1 0 1 1 1/4
1 1 2 0 1/4

For $mathbb P(X=x mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

$mathbb P(X=0 mid X+Y=0)=1$, $,mathbb P(X=1 mid X+Y=0)=0$,

$mathbb P(X=0 mid X+Y=1)=frac12$, $mathbb P(X=1 mid X+Y=1)=frac12$,

$mathbb P(X=0 mid X+Y=2)=0$, $,mathbb P(X=1 mid X+Y=1)=1$

In Part 2 you have

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=0)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=0)=frac12$,

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=1)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=1)=frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=ZÃ¢ÂŠÂ•Y$

answered Jul 29 at 9:20

Henry

92.8k469147

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

add a commentÂ |Â

up vote
0
down vote

You asked for the probabilities. These are:

X Y X+Y XÃ¢ÂŠÂ•Y Prob
0 0 0 0 1/4 
0 1 1 1 1/4
1 0 1 1 1/4
1 1 2 0 1/4

For $mathbb P(X=x mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

$mathbb P(X=0 mid X+Y=0)=1$, $,mathbb P(X=1 mid X+Y=0)=0$,

$mathbb P(X=0 mid X+Y=1)=frac12$, $mathbb P(X=1 mid X+Y=1)=frac12$,

$mathbb P(X=0 mid X+Y=2)=0$, $,mathbb P(X=1 mid X+Y=1)=1$

In Part 2 you have

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=0)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=0)=frac12$,

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=1)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=1)=frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=ZÃ¢ÂŠÂ•Y$

answered Jul 29 at 9:20

Henry

92.8k469147

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

add a commentÂ |Â

up vote
0
down vote

You asked for the probabilities. These are:

X Y X+Y XÃ¢ÂŠÂ•Y Prob
0 0 0 0 1/4 
0 1 1 1 1/4
1 0 1 1 1/4
1 1 2 0 1/4

For $mathbb P(X=x mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

$mathbb P(X=0 mid X+Y=0)=1$, $,mathbb P(X=1 mid X+Y=0)=0$,

$mathbb P(X=0 mid X+Y=1)=frac12$, $mathbb P(X=1 mid X+Y=1)=frac12$,

$mathbb P(X=0 mid X+Y=2)=0$, $,mathbb P(X=1 mid X+Y=1)=1$

In Part 2 you have

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=0)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=0)=frac12$,

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=1)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=1)=frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=ZÃ¢ÂŠÂ•Y$

answered Jul 29 at 9:20

Henry

92.8k469147

You asked for the probabilities. These are:

X Y X+Y XÃ¢ÂŠÂ•Y Prob
0 0 0 0 1/4 
0 1 1 1 1/4
1 0 1 1 1/4
1 1 2 0 1/4

For $mathbb P(X=x mid Z=z)$, in Part 1 you have, as spaceisdarkgreen says,

$mathbb P(X=0 mid X+Y=0)=1$, $,mathbb P(X=1 mid X+Y=0)=0$,

$mathbb P(X=0 mid X+Y=1)=frac12$, $mathbb P(X=1 mid X+Y=1)=frac12$,

$mathbb P(X=0 mid X+Y=2)=0$, $,mathbb P(X=1 mid X+Y=1)=1$

In Part 2 you have

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=0)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=0)=frac12$,

$mathbb P(X=0 mid XÃ¢ÂŠÂ•Y=1)=frac12$, $mathbb P(X=1 mid XÃ¢ÂŠÂ•Y=1)=frac12$

Given both $Y$ and $Z$, the value of $X$ is certain in both parts 1 and 2, since in part 1 you have $X=Z-Y$ while in part 2 you have $X=ZÃ¢ÂŠÂ•Y$

answered Jul 29 at 9:20

Henry

92.8k469147

answered Jul 29 at 9:20

Henry

92.8k469147

answered Jul 29 at 9:20

Henry

92.8k469147

answered Jul 29 at 9:20

Henry

92.8k469147

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

add a commentÂ |Â

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

Why is the $P(X=0|X+Y=0)=1$,not $frac14$ ?
â€“Â Shine Sun
Jul 29 at 13:11

@ShineSun Because the only case where $X+Y=0$ is the top row of the table. You need to calculate the conditional probability $mathbb P(X=0 mid X+Y=0)= dfracmathbb P(X=0 , X+Y=0)mathbb P(X+Y=0) =dfrac1/41/4=1$
â€“Â Henry
Jul 29 at 14:16

get it! can i ask how to calculate the P(X|Y,Z) according to your table?because i think P(X|Y,Z) may be equal to P(X,Z)
â€“Â Shine Sun
Jul 29 at 14:26

add a commentÂ |Â

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