Why no variance term in Bayesian logistic regression?Bayesian logit model - intuitive explanation?Are logistic regression coefficient estimates biased when the predictor has large variance?logistic regression with slackClosed form for the variance of a sum of two estimates in logistic regression?Evaluate posterior predictive distribution in Bayesian linear regressionClassical vs Bayesian logistic regression assumptionsCase Control Sampling in Logistic RegressionFit logistic regression with linear constraints on coefficients in RIs Bayesian Ridge Regression another name of Bayesian Linear Regression?Bayesian Inference for More Than Linear RegressionEconometrics: What are the assumptions of logistic regression for causal inference?
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Why no variance term in Bayesian logistic regression?
Bayesian logit model - intuitive explanation?Are logistic regression coefficient estimates biased when the predictor has large variance?logistic regression with slackClosed form for the variance of a sum of two estimates in logistic regression?Evaluate posterior predictive distribution in Bayesian linear regressionClassical vs Bayesian logistic regression assumptionsCase Control Sampling in Logistic RegressionFit logistic regression with linear constraints on coefficients in RIs Bayesian Ridge Regression another name of Bayesian Linear Regression?Bayesian Inference for More Than Linear RegressionEconometrics: What are the assumptions of logistic regression for causal inference?
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$begingroup$
I've read here that
... (Bayesian linear regression) is most similar to Bayesian inference
in logistic regression, but in some ways logistic regression is even
simpler, because there is no variance term to estimate, only the
regression parameters.
Why is it the case, why no variance term in Bayesian logistic regression?
logistic bayesian generalized-linear-model variance bernoulli-distribution
edited 2 days ago
Krantz
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
$endgroup$
add a comment |
$begingroup$
I've read here that
... (Bayesian linear regression) is most similar to Bayesian inference
in logistic regression, but in some ways logistic regression is even
simpler, because there is no variance term to estimate, only the
regression parameters.
Why is it the case, why no variance term in Bayesian logistic regression?
logistic bayesian generalized-linear-model variance bernoulli-distribution
edited 2 days ago
Krantz
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
$endgroup$
add a comment |
$begingroup$
I've read here that
... (Bayesian linear regression) is most similar to Bayesian inference
in logistic regression, but in some ways logistic regression is even
simpler, because there is no variance term to estimate, only the
regression parameters.
Why is it the case, why no variance term in Bayesian logistic regression?
logistic bayesian generalized-linear-model variance bernoulli-distribution
edited 2 days ago
Krantz
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
$endgroup$
I've read here that
... (Bayesian linear regression) is most similar to Bayesian inference
in logistic regression, but in some ways logistic regression is even
simpler, because there is no variance term to estimate, only the
regression parameters.
Why is it the case, why no variance term in Bayesian logistic regression?
logistic bayesian generalized-linear-model variance bernoulli-distribution
logistic bayesian generalized-linear-model variance bernoulli-distribution
edited 2 days ago
Krantz
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
edited 2 days ago
Krantz
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
edited 2 days ago
Krantz
35211
edited 2 days ago
Krantz
35211
edited 2 days ago
Krantz
35211
35211
asked Apr 3 at 20:32
PatrickPatrick
1546
asked Apr 3 at 20:32
PatrickPatrick
1546
asked Apr 3 at 20:32
PatrickPatrick
1546
1546
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
|
show 2 more comments
$begingroup$
Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
|
show 2 more comments
$begingroup$
Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
$endgroup$
Logistic regression, Bayesian or not, is a model defined in terms of Bernoulli distribution. The distribution is parametrized by "probability of success" $p$ with mean $p$ and variance $p(1-p)$, i.e. the variance directly follows from the mean. So there is no "separate" variance term, this is what the quote seems to say.
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
answered Apr 3 at 20:46
Tim♦Tim
59.8k9132225
59.8k9132225
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
|
show 2 more comments
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
Apr 3 at 21:02
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ or not? If not, is it because $p()$ is a probability and already includes some uncertainty?
$endgroup$
– Patrick
Apr 3 at 21:37
1
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
Apr 3 at 21:39
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
Apr 3 at 22:12
1
1
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
$begingroup$
@Patrick linear regression is $Y sim mathcalN(Xbeta, sigma)$ while logistic regression is $Y sim mathcalB(h^-1(Xbeta))$. Error term is alternative way of writing linear regression, but in both cases you estimate mean of some distribution. So error term inside logistic regression would mean that you assume $h(p)$ to be normally distributed. This does not have to be nonsense, but I doubt that this is what you meant.
$endgroup$
– Tim♦
2 days ago
|
show 2 more comments
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