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Conditional entropy on race outcome
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30UTC (7:30pm US/Eastern)Non-zero Conditional Differential Entropy between a random variable and a function of itInformation Entropy Applied to Complexity TheoryWhat does the $-log[P(X)]$ term mean in the calculation of entropy?Why can we use entropy to measure the quality of a language model?Conditional entropy of repetition code over BSCProof that the inequality with mutual information and conditional mutual information is not true always.Entropy and Mutual InformationConditional Expectation and EntropyUnderstanding information entropyEntropy of roulette
$begingroup$
The problem is:
9 guys are racing.
The favorite has a probability of 3/4 to win the race.
Each other competitor has an equal chance to win.
If it becomes known that the favorite did not win the race, what is the uncertainty of the result?
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win. My trouble is on how to model the P(X|Y) and P(X,Y) needed to find the entropy.
information-theory
asked Apr 1 at 23:57
JohnDoughJohnDough
134
$endgroup$
add a comment |
$begingroup$
The problem is:
9 guys are racing.
The favorite has a probability of 3/4 to win the race.
Each other competitor has an equal chance to win.
If it becomes known that the favorite did not win the race, what is the uncertainty of the result?
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win. My trouble is on how to model the P(X|Y) and P(X,Y) needed to find the entropy.
information-theory
asked Apr 1 at 23:57
JohnDoughJohnDough
134
$endgroup$
add a comment |
$begingroup$
The problem is:
9 guys are racing.
The favorite has a probability of 3/4 to win the race.
Each other competitor has an equal chance to win.
If it becomes known that the favorite did not win the race, what is the uncertainty of the result?
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win. My trouble is on how to model the P(X|Y) and P(X,Y) needed to find the entropy.
information-theory
asked Apr 1 at 23:57
JohnDoughJohnDough
134
$endgroup$
The problem is:
9 guys are racing.
The favorite has a probability of 3/4 to win the race.
Each other competitor has an equal chance to win.
If it becomes known that the favorite did not win the race, what is the uncertainty of the result?
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win. My trouble is on how to model the P(X|Y) and P(X,Y) needed to find the entropy.
information-theory
information-theory
asked Apr 1 at 23:57
JohnDoughJohnDough
134
asked Apr 1 at 23:57
JohnDoughJohnDough
134
edited Apr 2 at 0:26
JohnDough
asked Apr 1 at 23:57
JohnDoughJohnDough
134
asked Apr 1 at 23:57
JohnDoughJohnDough
134
asked Apr 1 at 23:57
JohnDoughJohnDough
134
134
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given that the winner is one of the 8 equi-probable participants, the entropy of the result is $log 8 = 3$ bits.
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
$endgroup$
add a comment |
$begingroup$
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win.
When learning conditional entropy, you need to distinguish between $H(X | Y)$ and $H(X | Y =y)$. In the first one, the condition is not with respect to an event, but with respect with the distribution of the other variable; that's why $H(X |Y)$ is a plain number. Instead, $H(X | Y =y)$ conditions with respect of an event (in this case, the value of $Y$), hence the result depends on $y$.
(In other words, the notation $H(X|Y)$ is not analogous to other conditionals such as $E(X|Y)$ )
In your case you are interested in the latter, you are conditioning on an event: the winner is not the player (say) 1, that is $H(X mid X ne 1)$. Now, the conditional probability on that event is a uniform over eight values, hence the entropy is $3$ bits.
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
Given that the winner is one of the 8 equi-probable participants, the entropy of the result is $log 8 = 3$ bits.
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
$endgroup$
add a comment |
$begingroup$
Given that the winner is one of the 8 equi-probable participants, the entropy of the result is $log 8 = 3$ bits.
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
$endgroup$
add a comment |
$begingroup$
Given that the winner is one of the 8 equi-probable participants, the entropy of the result is $log 8 = 3$ bits.
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
$endgroup$
Given that the winner is one of the 8 equi-probable participants, the entropy of the result is $log 8 = 3$ bits.
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
answered Apr 2 at 1:18
ChargeShiversChargeShivers
1,261613
1,261613
add a comment |
add a comment |
$begingroup$
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win.
When learning conditional entropy, you need to distinguish between $H(X | Y)$ and $H(X | Y =y)$. In the first one, the condition is not with respect to an event, but with respect with the distribution of the other variable; that's why $H(X |Y)$ is a plain number. Instead, $H(X | Y =y)$ conditions with respect of an event (in this case, the value of $Y$), hence the result depends on $y$.
(In other words, the notation $H(X|Y)$ is not analogous to other conditionals such as $E(X|Y)$ )
In your case you are interested in the latter, you are conditioning on an event: the winner is not the player (say) 1, that is $H(X mid X ne 1)$. Now, the conditional probability on that event is a uniform over eight values, hence the entropy is $3$ bits.
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
$endgroup$
add a comment |
$begingroup$
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win.
When learning conditional entropy, you need to distinguish between $H(X | Y)$ and $H(X | Y =y)$. In the first one, the condition is not with respect to an event, but with respect with the distribution of the other variable; that's why $H(X |Y)$ is a plain number. Instead, $H(X | Y =y)$ conditions with respect of an event (in this case, the value of $Y$), hence the result depends on $y$.
(In other words, the notation $H(X|Y)$ is not analogous to other conditionals such as $E(X|Y)$ )
In your case you are interested in the latter, you are conditioning on an event: the winner is not the player (say) 1, that is $H(X mid X ne 1)$. Now, the conditional probability on that event is a uniform over eight values, hence the entropy is $3$ bits.
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
$endgroup$
add a comment |
$begingroup$
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win.
When learning conditional entropy, you need to distinguish between $H(X | Y)$ and $H(X | Y =y)$. In the first one, the condition is not with respect to an event, but with respect with the distribution of the other variable; that's why $H(X |Y)$ is a plain number. Instead, $H(X | Y =y)$ conditions with respect of an event (in this case, the value of $Y$), hence the result depends on $y$.
(In other words, the notation $H(X|Y)$ is not analogous to other conditionals such as $E(X|Y)$ )
In your case you are interested in the latter, you are conditioning on an event: the winner is not the player (say) 1, that is $H(X mid X ne 1)$. Now, the conditional probability on that event is a uniform over eight values, hence the entropy is $3$ bits.
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
$endgroup$
My intuition would be a conditional entropy approach H(X|Y) where X denotes the competitor and Y the information that the champion did not win.
When learning conditional entropy, you need to distinguish between $H(X | Y)$ and $H(X | Y =y)$. In the first one, the condition is not with respect to an event, but with respect with the distribution of the other variable; that's why $H(X |Y)$ is a plain number. Instead, $H(X | Y =y)$ conditions with respect of an event (in this case, the value of $Y$), hence the result depends on $y$.
(In other words, the notation $H(X|Y)$ is not analogous to other conditionals such as $E(X|Y)$ )
In your case you are interested in the latter, you are conditioning on an event: the winner is not the player (say) 1, that is $H(X mid X ne 1)$. Now, the conditional probability on that event is a uniform over eight values, hence the entropy is $3$ bits.
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
answered Apr 2 at 18:26
leonbloyleonbloy
42.5k647108
42.5k647108
add a comment |
add a comment |
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