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Proof of $P(X|Y) = sum_z P(X,z | Y)$ (check my solution)
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Let X,Y,Z be a random variables. Prove that $P(X= x|Y = y) = sum_z P(X= x,Z = z | Y = y)$
First of all on the LHD we know, that $P(X = x|Y = y) = fracP(X = x,Y = y)P(Y = y)$.
The same we can do on the RHD: $sum_z P(X = x,Z = z | Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
We get: $fracP(X = x,Y = y)P(Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
$P(X = x,Y =y) = sum_z P(X = x ,Z = z , Y = y)$
then on the RHD we use the chain rule of the probabilities:
$P(X = x,Y = Y) = sum_z P(X = x|Z = z,Y = y) P(Y =y|Z = z) P(Z = z)$
In the case of $sum_z$ $P(Z = z) = 1$ and $P(X = x|Z= z,Y = y) = P(X =x|Y = y)$
I am not sure about $P(Y = y|Z = z)$ but I suppose it's equal to $P(Y = y)$ as well.
Hence we get $P(X = x,Y = y) = P(X = x|Y = Y) P(Y = y)$ which is correct by definition.
probability proof-verification
asked Jul 20 at 9:17
Daniel Yefimov
508417
 |Â
show 2 more comments
up vote
0
down vote
favorite
Let X,Y,Z be a random variables. Prove that $P(X= x|Y = y) = sum_z P(X= x,Z = z | Y = y)$
First of all on the LHD we know, that $P(X = x|Y = y) = fracP(X = x,Y = y)P(Y = y)$.
The same we can do on the RHD: $sum_z P(X = x,Z = z | Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
We get: $fracP(X = x,Y = y)P(Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
$P(X = x,Y =y) = sum_z P(X = x ,Z = z , Y = y)$
then on the RHD we use the chain rule of the probabilities:
$P(X = x,Y = Y) = sum_z P(X = x|Z = z,Y = y) P(Y =y|Z = z) P(Z = z)$
In the case of $sum_z$ $P(Z = z) = 1$ and $P(X = x|Z= z,Y = y) = P(X =x|Y = y)$
I am not sure about $P(Y = y|Z = z)$ but I suppose it's equal to $P(Y = y)$ as well.
Hence we get $P(X = x,Y = y) = P(X = x|Y = Y) P(Y = y)$ which is correct by definition.
probability proof-verification
asked Jul 20 at 9:17
Daniel Yefimov
508417
$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let X,Y,Z be a random variables. Prove that $P(X= x|Y = y) = sum_z P(X= x,Z = z | Y = y)$
First of all on the LHD we know, that $P(X = x|Y = y) = fracP(X = x,Y = y)P(Y = y)$.
The same we can do on the RHD: $sum_z P(X = x,Z = z | Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
We get: $fracP(X = x,Y = y)P(Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
$P(X = x,Y =y) = sum_z P(X = x ,Z = z , Y = y)$
then on the RHD we use the chain rule of the probabilities:
$P(X = x,Y = Y) = sum_z P(X = x|Z = z,Y = y) P(Y =y|Z = z) P(Z = z)$
In the case of $sum_z$ $P(Z = z) = 1$ and $P(X = x|Z= z,Y = y) = P(X =x|Y = y)$
I am not sure about $P(Y = y|Z = z)$ but I suppose it's equal to $P(Y = y)$ as well.
Hence we get $P(X = x,Y = y) = P(X = x|Y = Y) P(Y = y)$ which is correct by definition.
probability proof-verification
asked Jul 20 at 9:17
Daniel Yefimov
508417
Let X,Y,Z be a random variables. Prove that $P(X= x|Y = y) = sum_z P(X= x,Z = z | Y = y)$
First of all on the LHD we know, that $P(X = x|Y = y) = fracP(X = x,Y = y)P(Y = y)$.
The same we can do on the RHD: $sum_z P(X = x,Z = z | Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
We get: $fracP(X = x,Y = y)P(Y = y) = fracsum_z P(X = x,Z = z , Y = y)P(Y = y)$
$P(X = x,Y =y) = sum_z P(X = x ,Z = z , Y = y)$
then on the RHD we use the chain rule of the probabilities:
$P(X = x,Y = Y) = sum_z P(X = x|Z = z,Y = y) P(Y =y|Z = z) P(Z = z)$
In the case of $sum_z$ $P(Z = z) = 1$ and $P(X = x|Z= z,Y = y) = P(X =x|Y = y)$
I am not sure about $P(Y = y|Z = z)$ but I suppose it's equal to $P(Y = y)$ as well.
Hence we get $P(X = x,Y = y) = P(X = x|Y = Y) P(Y = y)$ which is correct by definition.
probability proof-verification
asked Jul 20 at 9:17
Daniel Yefimov
508417
edited Jul 20 at 9:59
asked Jul 20 at 9:17
Daniel Yefimov
508417
asked Jul 20 at 9:17
Daniel Yefimov
508417
asked Jul 20 at 9:17
Daniel Yefimov
508417
508417
$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47
 |Â
show 2 more comments
$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47
$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47
 |Â
show 2 more comments
1 Answer
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1
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We have: $$P(X=xmid Y=y)P(Y=y)=P(X=x, Y=y)$$and: $$P(X=x, Z=zmid Y=y)P(Y=y)=P(X=x, Z=z, Y=y)$$So if $P(Y=y)>0$ then proving that: $$P(X=xmid Y=y)=sum_zP(X=x,Z=zmid Y=y)$$ comes to the same as proving that $$P(X=x, Y=y)=sum_zP(X=x, Z=z, Y=z)tag1$$
Here $(1)$ is an immediate consequence of:$$X=x,Y=y=bigcup_z{X=x,Z=z,Y=y)$$(preassuming that $Z$ is a discrete random variable and above $z$ ranges over a countable set that serves as support of $Z$).
answered Jul 20 at 10:10
Vera
1,706413
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have: $$P(X=xmid Y=y)P(Y=y)=P(X=x, Y=y)$$and: $$P(X=x, Z=zmid Y=y)P(Y=y)=P(X=x, Z=z, Y=y)$$So if $P(Y=y)>0$ then proving that: $$P(X=xmid Y=y)=sum_zP(X=x,Z=zmid Y=y)$$ comes to the same as proving that $$P(X=x, Y=y)=sum_zP(X=x, Z=z, Y=z)tag1$$
Here $(1)$ is an immediate consequence of:$$X=x,Y=y=bigcup_z{X=x,Z=z,Y=y)$$(preassuming that $Z$ is a discrete random variable and above $z$ ranges over a countable set that serves as support of $Z$).
answered Jul 20 at 10:10
Vera
1,706413
add a comment |Â
up vote
1
down vote
accepted
We have: $$P(X=xmid Y=y)P(Y=y)=P(X=x, Y=y)$$and: $$P(X=x, Z=zmid Y=y)P(Y=y)=P(X=x, Z=z, Y=y)$$So if $P(Y=y)>0$ then proving that: $$P(X=xmid Y=y)=sum_zP(X=x,Z=zmid Y=y)$$ comes to the same as proving that $$P(X=x, Y=y)=sum_zP(X=x, Z=z, Y=z)tag1$$
Here $(1)$ is an immediate consequence of:$$X=x,Y=y=bigcup_z{X=x,Z=z,Y=y)$$(preassuming that $Z$ is a discrete random variable and above $z$ ranges over a countable set that serves as support of $Z$).
answered Jul 20 at 10:10
Vera
1,706413
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have: $$P(X=xmid Y=y)P(Y=y)=P(X=x, Y=y)$$and: $$P(X=x, Z=zmid Y=y)P(Y=y)=P(X=x, Z=z, Y=y)$$So if $P(Y=y)>0$ then proving that: $$P(X=xmid Y=y)=sum_zP(X=x,Z=zmid Y=y)$$ comes to the same as proving that $$P(X=x, Y=y)=sum_zP(X=x, Z=z, Y=z)tag1$$
Here $(1)$ is an immediate consequence of:$$X=x,Y=y=bigcup_z{X=x,Z=z,Y=y)$$(preassuming that $Z$ is a discrete random variable and above $z$ ranges over a countable set that serves as support of $Z$).
answered Jul 20 at 10:10
Vera
1,706413
We have: $$P(X=xmid Y=y)P(Y=y)=P(X=x, Y=y)$$and: $$P(X=x, Z=zmid Y=y)P(Y=y)=P(X=x, Z=z, Y=y)$$So if $P(Y=y)>0$ then proving that: $$P(X=xmid Y=y)=sum_zP(X=x,Z=zmid Y=y)$$ comes to the same as proving that $$P(X=x, Y=y)=sum_zP(X=x, Z=z, Y=z)tag1$$
Here $(1)$ is an immediate consequence of:$$X=x,Y=y=bigcup_z{X=x,Z=z,Y=y)$$(preassuming that $Z$ is a discrete random variable and above $z$ ranges over a countable set that serves as support of $Z$).
answered Jul 20 at 10:10
Vera
1,706413
edited Jul 20 at 10:33
answered Jul 20 at 10:10
Vera
1,706413
answered Jul 20 at 10:10
Vera
1,706413
answered Jul 20 at 10:10
Vera
1,706413
1,706413
add a comment |Â
add a comment |Â
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$P(X|z,Y)$: what is this?
– zoli
Jul 20 at 9:40
What exactly denote $P(Xmid Y)$, $P(X,zmid,Y)$ and $P(Y)$?
– Vera
Jul 20 at 9:40
@zoli Added the descriprion
– Daniel Yefimov
Jul 20 at 9:43
@Vera Added the description
– Daniel Yefimov
Jul 20 at 9:43
I assume that $X$ denotes a random variable, but what is $P(X)$? I am familiar with expressions like $P(X=x)$, $P(Xin B)$ or $P_X$ (standing for the distribution induced by $X$), but not with just $P(X)$ as if the $X$ is a argument of probability measure $P$.
– Vera
Jul 20 at 9:47