how log likelihood's derivative is equal to zero in maximum log likelihood.
0
$begingroup$
if the log-likelihood function is strictly increasing and it has not horizontal asymptote then how it's derivative is equal to zero in maximum log likelihood. Now since it is strictly increasing every time the parameters will tend to infinity. So why not take infinity as parameters directly.
probability machine-learning maximum-likelihood logistic-regression log-likelihood
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
$endgroup$
add a comment |
0
$begingroup$
if the log-likelihood function is strictly increasing and it has not horizontal asymptote then how it's derivative is equal to zero in maximum log likelihood. Now since it is strictly increasing every time the parameters will tend to infinity. So why not take infinity as parameters directly.
probability machine-learning maximum-likelihood logistic-regression log-likelihood
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
$endgroup$
1
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
1
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28
add a comment |
0
0
0
$begingroup$
if the log-likelihood function is strictly increasing and it has not horizontal asymptote then how it's derivative is equal to zero in maximum log likelihood. Now since it is strictly increasing every time the parameters will tend to infinity. So why not take infinity as parameters directly.
probability machine-learning maximum-likelihood logistic-regression log-likelihood
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
$endgroup$
if the log-likelihood function is strictly increasing and it has not horizontal asymptote then how it's derivative is equal to zero in maximum log likelihood. Now since it is strictly increasing every time the parameters will tend to infinity. So why not take infinity as parameters directly.
probability machine-learning maximum-likelihood logistic-regression log-likelihood
probability machine-learning maximum-likelihood logistic-regression log-likelihood
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
asked Jan 22 at 18:35
Anurag SharmaAnurag Sharma
1
1
1
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
1
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28
add a comment |
1
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
1
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28
1
1
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
1
1
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28
add a comment |
0
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1
$begingroup$
log is strictly increasing, but the likelihood function need not be an increasing function, meaning there are numbers $x<y$ where the likelihood function, say $ell$, applied to these numbers is $ell (x) > ell (y)$. Then the log-likelihood, say $L$, for these numbers is $L(x) > L(y)$ even though $x<y$.
$endgroup$
– gd1035
Jan 22 at 18:51
1
$begingroup$
If the log-likelihood is strictly increasing then its maximum and the likelihood's maximum occurs at the right hand end; if that is at infinity then there is no maximum
$endgroup$
– Henry
Jan 22 at 19:52
$begingroup$
So this means that the log is strictly increasing but log-likelihood is not. Am I correct?
$endgroup$
– Anurag Sharma
Jan 23 at 6:28