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?
$begingroup$
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 variance
asked 7 hours ago
PatrickPatrick
1396
$endgroup$
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 variance
asked 7 hours ago
PatrickPatrick
1396
$endgroup$
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 variance
asked 7 hours ago
PatrickPatrick
1396
$endgroup$
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 variance
logistic bayesian variance
asked 7 hours ago
PatrickPatrick
1396
asked 7 hours ago
PatrickPatrick
1396
asked 7 hours ago
PatrickPatrick
1396
asked 7 hours ago
PatrickPatrick
1396
asked 7 hours ago
PatrickPatrick
1396
1396
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 7 hours ago
Tim♦Tim
59.7k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
add a comment |
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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 7 hours ago
Tim♦Tim
59.7k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
add a comment |
$begingroup$
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 7 hours ago
Tim♦Tim
59.7k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
add a comment |
$begingroup$
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 7 hours ago
Tim♦Tim
59.7k9131224
$endgroup$
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 7 hours ago
Tim♦Tim
59.7k9131224
answered 7 hours ago
Tim♦Tim
59.7k9131224
answered 7 hours ago
Tim♦Tim
59.7k9131224
answered 7 hours ago
Tim♦Tim
59.7k9131224
59.7k9131224
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
add a comment |
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
1
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
7 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
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♦
6 hours ago
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
6 hours ago
add a comment |
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