Why no variance term in Bayesian logistic regression?
$begingroup$
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 2 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 2 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 2 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 2 hours ago
PatrickPatrick
1396
asked 2 hours ago
PatrickPatrick
1396
asked 2 hours ago
PatrickPatrick
1396
asked 2 hours ago
PatrickPatrick
1396
asked 2 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 2 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
2 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins 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 2 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
2 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins 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 2 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
2 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins 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 2 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 2 hours ago
Tim♦Tim
59.7k9131224
answered 2 hours ago
Tim♦Tim
59.7k9131224
answered 2 hours ago
Tim♦Tim
59.7k9131224
answered 2 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
2 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins 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
2 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
2 hours ago
$begingroup$
@patrick for linear regression $y = mx + c + epsilon$, whereas logistic regression p(y=1|x) = logistic(mx +c).
$endgroup$
– seanv507
2 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
1 hour 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
1 hour 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♦
1 hour ago
$begingroup$
@Patrick what would this formulation exactly mean? Could you give an example where would you imagine it to be used?
$endgroup$
– Tim♦
1 hour ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins ago
$begingroup$
@Patrick When you describe the conditional expectation in terms of a distribution there's no error term.
$endgroup$
– Firebug
57 mins ago
add a comment |
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