Why no variance term in Bayesian logistic regression?
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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 Apr 5 at 0:45
Krantz
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
$endgroup$
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 Apr 5 at 0:45
Krantz
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
$endgroup$
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 Apr 5 at 0:45
Krantz
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
$endgroup$
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 Apr 5 at 0:45
Krantz
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
edited Apr 5 at 0:45
Krantz
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
edited Apr 5 at 0:45
Krantz
394111
edited Apr 5 at 0:45
Krantz
394111
edited Apr 5 at 0:45
Krantz
394111
394111
asked Apr 3 at 20:32
PatrickPatrick
1666
asked Apr 3 at 20:32
PatrickPatrick
1666
asked Apr 3 at 20:32
PatrickPatrick
1666
1666
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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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
60.5k9133230
$endgroup$
$begingroup$
@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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
|
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
60.5k9133230
$endgroup$
$begingroup$
@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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
|
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
60.5k9133230
$endgroup$
$begingroup$
@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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
|
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
60.5k9133230
$endgroup$
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
60.5k9133230
answered Apr 3 at 20:46
Tim♦Tim
60.5k9133230
answered Apr 3 at 20:46
Tim♦Tim
60.5k9133230
answered Apr 3 at 20:46
Tim♦Tim
60.5k9133230
60.5k9133230
$begingroup$
@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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
|
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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
$begingroup$
@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 $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
$begingroup$
@Patrick linear regression is $Y sim mathcal{N}(Xbeta, sigma)$ while logistic regression is $Y sim mathcal{B}(h^{-1}(Xbeta))$. 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♦
Apr 4 at 19:12
|
show 2 more comments
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