Do error bars on probabilities have any meaning?
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People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 6 hours ago
mahnamahnamahnamahna
261
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add a comment |
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People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 6 hours ago
mahnamahnamahnamahna
261
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
asked 6 hours ago
mahnamahnamahnamahna
261
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
People often say some event has a 50-60% chance of happening. Sometimes I will even see people give explicit error bars on probability assignments. Do these statements have any meaning or are they just a linguistic quirk of discomfort choosing a specific number for something that is inherently unknowable?
probability
probability
asked 6 hours ago
mahnamahnamahnamahna
261
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
mahnamahnamahnamahna
261
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
mahnamahnamahnamahna
261
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
mahnamahnamahnamahna
261
asked 6 hours ago
mahnamahnamahnamahna
261
261
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mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mahnamahna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
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$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 6 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 5 hours ago
Xi'anXi'an
56.9k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 6 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
add a comment |
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 6 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
add a comment |
$begingroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 6 hours ago
Tim♦Tim
57.5k9126218
$endgroup$
It wouldn't make sense if you were talking about known probabilities, e.g. with fair coin the probability of throwing heads is 0.5 by the laws of physics.
The different story is when you estimate the probabilities from the data, e.g. you observed 13 winning tickets among the 12563 tickets you bought, so from this data you estimate the probability to be 13/12563. Since this is something you estimated from the sample, it is uncertain, since with different sample you could observe different value. The uncertainty estimate is not about the probability, but around the estimate of it.
Another example would be when the probability is not fixed, but depends on other factors. Say that we are talking about probability of dying in car accident. We can consider "global" probability, single value that is marginalized over all the factors that directly and indirectly lead to car accidents. On another hand, you can consider how the probabilities vary among the population given the risk factors.
You can find many more examples where probabilities themselves are considered as random variables, so they vary rather then being fixed.
answered 6 hours ago
Tim♦Tim
57.5k9126218
answered 6 hours ago
Tim♦Tim
57.5k9126218
answered 6 hours ago
Tim♦Tim
57.5k9126218
answered 6 hours ago
Tim♦Tim
57.5k9126218
57.5k9126218
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
add a comment |
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
If the calculation of a probability estimate was done through something like a logistic regression wouldn't be also natural to assume that these "error bars" refer to prediction intervals? (I am asking mostly as a clarification to the first point you raise, +1 obviously)
$endgroup$
– usεr11852
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
$begingroup$
@usεr11852 confidence intervals, prediction intervals, highest density regions etc., depending on actual case. I made the answer very broad, since we have "varying" probabilities in many scenarios and they vary in different ways. Also you can interpret them differently in different scenarios.
$endgroup$
– Tim♦
3 hours ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 5 hours ago
Xi'anXi'an
56.9k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 5 hours ago
Xi'anXi'an
56.9k895357
$endgroup$
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
add a comment |
$begingroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 5 hours ago
Xi'anXi'an
56.9k895357
$endgroup$
A most relevant illustration from xkcd:
with associated caption:
...an effect size of 1.68 (95% CI: 1.56 (95% CI: 1.52 (95% CI: 1.504
(95% CI: 1.494 (95% CI: 1.488 (95% CI: 1.485 (95% CI: 1.482 (95% CI:
1.481 (95% CI: 1.4799 (95% CI: 1.4791 (95% CI: 1.4784...
answered 5 hours ago
Xi'anXi'an
56.9k895357
answered 5 hours ago
Xi'anXi'an
56.9k895357
answered 5 hours ago
Xi'anXi'an
56.9k895357
answered 5 hours ago
Xi'anXi'an
56.9k895357
56.9k895357
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
add a comment |
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
$begingroup$
Does this imply that error bars on probabilities are redundant?
$endgroup$
– BalinKingOfMoria
18 mins ago
add a comment |
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