How to calculate P(X|W,Z) in a Bayesian network? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Bayesian Network for dummiesProbabilities from Bayesian networkhow to calculate unknown probabilities in the bayesian networkBayesian Network, Sprinkler ExampleProbability Bayesian network problemConfusion in a simple Bayesian NetworkBayesian network ProblemBayesian networkIs this Bayesian Network Probability calculation correct?
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How to calculate P(X|W,Z) in a Bayesian network?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Bayesian Network for dummiesProbabilities from Bayesian networkhow to calculate unknown probabilities in the bayesian networkBayesian Network, Sprinkler ExampleProbability Bayesian network problemConfusion in a simple Bayesian NetworkBayesian network ProblemBayesian networkIs this Bayesian Network Probability calculation correct?
$begingroup$
Bayesian net
I can see that $P(X,Y,W,Z)$ = $P(X|Y,Z)P(Y|W,Z)P(W)P(Z)$.
I did the following till now to calculate $P(X|W,Z)$:
$P(X|W,Z)$ = $P(X|Y,W,Z)$ + $P(X|overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y,W,Z)$ + $P(X,overlineY,W,Z)P(overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y|W,Z)P(W)P(Z)$ + $P(X,overlineY,W,Z)P(overlineY|W,Z)P(W)P(Z)$
Am I proceeding in the right direction? Help!
bayesian-network
asked Feb 24 at 22:22
SisyphusSisyphus
124
$endgroup$
add a comment |
$begingroup$
Bayesian net
I can see that $P(X,Y,W,Z)$ = $P(X|Y,Z)P(Y|W,Z)P(W)P(Z)$.
I did the following till now to calculate $P(X|W,Z)$:
$P(X|W,Z)$ = $P(X|Y,W,Z)$ + $P(X|overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y,W,Z)$ + $P(X,overlineY,W,Z)P(overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y|W,Z)P(W)P(Z)$ + $P(X,overlineY,W,Z)P(overlineY|W,Z)P(W)P(Z)$
Am I proceeding in the right direction? Help!
bayesian-network
asked Feb 24 at 22:22
SisyphusSisyphus
124
$endgroup$
add a comment |
$begingroup$
Bayesian net
I can see that $P(X,Y,W,Z)$ = $P(X|Y,Z)P(Y|W,Z)P(W)P(Z)$.
I did the following till now to calculate $P(X|W,Z)$:
$P(X|W,Z)$ = $P(X|Y,W,Z)$ + $P(X|overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y,W,Z)$ + $P(X,overlineY,W,Z)P(overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y|W,Z)P(W)P(Z)$ + $P(X,overlineY,W,Z)P(overlineY|W,Z)P(W)P(Z)$
Am I proceeding in the right direction? Help!
bayesian-network
asked Feb 24 at 22:22
SisyphusSisyphus
124
$endgroup$
Bayesian net
I can see that $P(X,Y,W,Z)$ = $P(X|Y,Z)P(Y|W,Z)P(W)P(Z)$.
I did the following till now to calculate $P(X|W,Z)$:
$P(X|W,Z)$ = $P(X|Y,W,Z)$ + $P(X|overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y,W,Z)$ + $P(X,overlineY,W,Z)P(overlineY,W,Z)$
= $P(X,Y,W,Z)P(Y|W,Z)P(W)P(Z)$ + $P(X,overlineY,W,Z)P(overlineY|W,Z)P(W)P(Z)$
Am I proceeding in the right direction? Help!
bayesian-network
bayesian-network
asked Feb 24 at 22:22
SisyphusSisyphus
124
asked Feb 24 at 22:22
SisyphusSisyphus
124
asked Feb 24 at 22:22
SisyphusSisyphus
124
asked Feb 24 at 22:22
SisyphusSisyphus
124
asked Feb 24 at 22:22
SisyphusSisyphus
124
124
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the D.A.G.; $requireenclosedefPoperatornamesf PbeginarraycenclosecircleWlower2exsearrow&&lower2exswarrowenclosecircleZ\&Y&quaddownarrow\&&raise2exsearrow Xendarray$
$X$ is directly influenced by $Y$ and $Z$, and $Y$ is directly influenced by $Z$ and $W$.
Therefore the way $X$ is indirectly influenced by $Z$ and $W$ is through $Y$.
I did the following till now to calculate $P(Xmid W,Z)$:
$$P(Xmid W,Z) = P(Xmid Y,W,Z) + P(Xmid overline Y,W,Z) ...$$
Error! That is not how marginalisation works. The principle is that $X=(Xcap Y)cup(Xcapoverline Y)$
$$beginalignP(Xmid W,Z)&=P(X,Ymid W,Z)+P(X,overline Ymid W,Z)\[1ex]&=P(Xmid Y,W,Z)P(Ymid W,Z)+P(Xmid overline Y,W,Z)P(overline Ymid W,Z)endalign$$
Now, because $W$ only influences $X$ through $Y$, it is redundant in the conditioning when $Y$ is present. (Whereas $Z$ still directly influences $X$.)
$$beginalignP(Xmid W,Z)&=P(Xmid Y,Z)P(Ymid W,Z)+P(Xmid overline Y,Z)P(overline Ymid W,Z)endalign$$
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
The definition of conditional probability is $P(Amid B)=fracP(Acap B)P(B)$.
So, $P(Xmid W,Z)=fracP(X,W,Z)P(W,Z)$. The denominator is easy to calculate. For the numerator, the event $Xcap Wcap Z$ is stating that $X,W$, and $Z$ all occur. There are 2 cases: Either $Y$ happens or it does not, and you can compute the probability of $Xcap Wcap Z$ both under the presence of $Y$ and in the absence of $Y$.
Can you take it from here?
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
$endgroup$
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Using the D.A.G.; $requireenclosedefPoperatornamesf PbeginarraycenclosecircleWlower2exsearrow&&lower2exswarrowenclosecircleZ\&Y&quaddownarrow\&&raise2exsearrow Xendarray$
$X$ is directly influenced by $Y$ and $Z$, and $Y$ is directly influenced by $Z$ and $W$.
Therefore the way $X$ is indirectly influenced by $Z$ and $W$ is through $Y$.
I did the following till now to calculate $P(Xmid W,Z)$:
$$P(Xmid W,Z) = P(Xmid Y,W,Z) + P(Xmid overline Y,W,Z) ...$$
Error! That is not how marginalisation works. The principle is that $X=(Xcap Y)cup(Xcapoverline Y)$
$$beginalignP(Xmid W,Z)&=P(X,Ymid W,Z)+P(X,overline Ymid W,Z)\[1ex]&=P(Xmid Y,W,Z)P(Ymid W,Z)+P(Xmid overline Y,W,Z)P(overline Ymid W,Z)endalign$$
Now, because $W$ only influences $X$ through $Y$, it is redundant in the conditioning when $Y$ is present. (Whereas $Z$ still directly influences $X$.)
$$beginalignP(Xmid W,Z)&=P(Xmid Y,Z)P(Ymid W,Z)+P(Xmid overline Y,Z)P(overline Ymid W,Z)endalign$$
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
Using the D.A.G.; $requireenclosedefPoperatornamesf PbeginarraycenclosecircleWlower2exsearrow&&lower2exswarrowenclosecircleZ\&Y&quaddownarrow\&&raise2exsearrow Xendarray$
$X$ is directly influenced by $Y$ and $Z$, and $Y$ is directly influenced by $Z$ and $W$.
Therefore the way $X$ is indirectly influenced by $Z$ and $W$ is through $Y$.
I did the following till now to calculate $P(Xmid W,Z)$:
$$P(Xmid W,Z) = P(Xmid Y,W,Z) + P(Xmid overline Y,W,Z) ...$$
Error! That is not how marginalisation works. The principle is that $X=(Xcap Y)cup(Xcapoverline Y)$
$$beginalignP(Xmid W,Z)&=P(X,Ymid W,Z)+P(X,overline Ymid W,Z)\[1ex]&=P(Xmid Y,W,Z)P(Ymid W,Z)+P(Xmid overline Y,W,Z)P(overline Ymid W,Z)endalign$$
Now, because $W$ only influences $X$ through $Y$, it is redundant in the conditioning when $Y$ is present. (Whereas $Z$ still directly influences $X$.)
$$beginalignP(Xmid W,Z)&=P(Xmid Y,Z)P(Ymid W,Z)+P(Xmid overline Y,Z)P(overline Ymid W,Z)endalign$$
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
$endgroup$
add a comment |
$begingroup$
Using the D.A.G.; $requireenclosedefPoperatornamesf PbeginarraycenclosecircleWlower2exsearrow&&lower2exswarrowenclosecircleZ\&Y&quaddownarrow\&&raise2exsearrow Xendarray$
$X$ is directly influenced by $Y$ and $Z$, and $Y$ is directly influenced by $Z$ and $W$.
Therefore the way $X$ is indirectly influenced by $Z$ and $W$ is through $Y$.
I did the following till now to calculate $P(Xmid W,Z)$:
$$P(Xmid W,Z) = P(Xmid Y,W,Z) + P(Xmid overline Y,W,Z) ...$$
Error! That is not how marginalisation works. The principle is that $X=(Xcap Y)cup(Xcapoverline Y)$
$$beginalignP(Xmid W,Z)&=P(X,Ymid W,Z)+P(X,overline Ymid W,Z)\[1ex]&=P(Xmid Y,W,Z)P(Ymid W,Z)+P(Xmid overline Y,W,Z)P(overline Ymid W,Z)endalign$$
Now, because $W$ only influences $X$ through $Y$, it is redundant in the conditioning when $Y$ is present. (Whereas $Z$ still directly influences $X$.)
$$beginalignP(Xmid W,Z)&=P(Xmid Y,Z)P(Ymid W,Z)+P(Xmid overline Y,Z)P(overline Ymid W,Z)endalign$$
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
$endgroup$
Using the D.A.G.; $requireenclosedefPoperatornamesf PbeginarraycenclosecircleWlower2exsearrow&&lower2exswarrowenclosecircleZ\&Y&quaddownarrow\&&raise2exsearrow Xendarray$
$X$ is directly influenced by $Y$ and $Z$, and $Y$ is directly influenced by $Z$ and $W$.
Therefore the way $X$ is indirectly influenced by $Z$ and $W$ is through $Y$.
I did the following till now to calculate $P(Xmid W,Z)$:
$$P(Xmid W,Z) = P(Xmid Y,W,Z) + P(Xmid overline Y,W,Z) ...$$
Error! That is not how marginalisation works. The principle is that $X=(Xcap Y)cup(Xcapoverline Y)$
$$beginalignP(Xmid W,Z)&=P(X,Ymid W,Z)+P(X,overline Ymid W,Z)\[1ex]&=P(Xmid Y,W,Z)P(Ymid W,Z)+P(Xmid overline Y,W,Z)P(overline Ymid W,Z)endalign$$
Now, because $W$ only influences $X$ through $Y$, it is redundant in the conditioning when $Y$ is present. (Whereas $Z$ still directly influences $X$.)
$$beginalignP(Xmid W,Z)&=P(Xmid Y,Z)P(Ymid W,Z)+P(Xmid overline Y,Z)P(overline Ymid W,Z)endalign$$
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
edited Apr 2 at 4:48
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
answered Apr 2 at 2:02
Graham KempGraham Kemp
88.1k43579
88.1k43579
add a comment |
add a comment |
$begingroup$
The definition of conditional probability is $P(Amid B)=fracP(Acap B)P(B)$.
So, $P(Xmid W,Z)=fracP(X,W,Z)P(W,Z)$. The denominator is easy to calculate. For the numerator, the event $Xcap Wcap Z$ is stating that $X,W$, and $Z$ all occur. There are 2 cases: Either $Y$ happens or it does not, and you can compute the probability of $Xcap Wcap Z$ both under the presence of $Y$ and in the absence of $Y$.
Can you take it from here?
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
$endgroup$
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
add a comment |
$begingroup$
The definition of conditional probability is $P(Amid B)=fracP(Acap B)P(B)$.
So, $P(Xmid W,Z)=fracP(X,W,Z)P(W,Z)$. The denominator is easy to calculate. For the numerator, the event $Xcap Wcap Z$ is stating that $X,W$, and $Z$ all occur. There are 2 cases: Either $Y$ happens or it does not, and you can compute the probability of $Xcap Wcap Z$ both under the presence of $Y$ and in the absence of $Y$.
Can you take it from here?
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
$endgroup$
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
add a comment |
$begingroup$
The definition of conditional probability is $P(Amid B)=fracP(Acap B)P(B)$.
So, $P(Xmid W,Z)=fracP(X,W,Z)P(W,Z)$. The denominator is easy to calculate. For the numerator, the event $Xcap Wcap Z$ is stating that $X,W$, and $Z$ all occur. There are 2 cases: Either $Y$ happens or it does not, and you can compute the probability of $Xcap Wcap Z$ both under the presence of $Y$ and in the absence of $Y$.
Can you take it from here?
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
$endgroup$
The definition of conditional probability is $P(Amid B)=fracP(Acap B)P(B)$.
So, $P(Xmid W,Z)=fracP(X,W,Z)P(W,Z)$. The denominator is easy to calculate. For the numerator, the event $Xcap Wcap Z$ is stating that $X,W$, and $Z$ all occur. There are 2 cases: Either $Y$ happens or it does not, and you can compute the probability of $Xcap Wcap Z$ both under the presence of $Y$ and in the absence of $Y$.
Can you take it from here?
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
edited Feb 25 at 23:44
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
answered Feb 24 at 22:55
alphacapturealphacapture
2,220425
2,220425
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
add a comment |
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
$begingroup$
The reason of my confusion is the presence of the node Y and the fact that Z influences both X and Y.
$endgroup$
– Sisyphus
Feb 25 at 0:02
add a comment |
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