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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
PatrickPatrick
1396
asked 1 hour ago
PatrickPatrick
1396
asked 1 hour ago
PatrickPatrick
1396
asked 1 hour ago
PatrickPatrick
1396
asked 1 hour 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 51 mins ago
Tim♦Tim
59.6k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs 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 51 mins ago
Tim♦Tim
59.6k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs 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 51 mins ago
Tim♦Tim
59.6k9131224
$endgroup$
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs 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 51 mins ago
Tim♦Tim
59.6k9131224
$endgroup$
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 51 mins ago
Tim♦Tim
59.6k9131224
answered 51 mins ago
Tim♦Tim
59.6k9131224
answered 51 mins ago
Tim♦Tim
59.6k9131224
answered 51 mins ago
Tim♦Tim
59.6k9131224
59.6k9131224
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs 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
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
34 mins ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs ago
$begingroup$
@seanv507 and would it make sense to have $p(y=1|x)=logistic(mx+c+epsilon)$ ?
$endgroup$
– Patrick
54 secs ago
add a comment |
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