Why are the local extrema of a log-transformed function equal to local extrema of the original function?
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I am studying maximum likelihood and to simplify taking the derivative of the likelihood function, it is often transformed by the natural log before taking the derivative.
I have read in other posts that this is because the logarithm is a monotonic function, so its extrema will be the same as the original function. However, I do not understand why this is the case. Can someone explain intuitively why the transformation does not affect the local extrema?
derivatives logarithms maximum-likelihood monotone-functions
edited Dec 26 '18 at 17:53
Bernard
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
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add a comment |
$begingroup$
I am studying maximum likelihood and to simplify taking the derivative of the likelihood function, it is often transformed by the natural log before taking the derivative.
I have read in other posts that this is because the logarithm is a monotonic function, so its extrema will be the same as the original function. However, I do not understand why this is the case. Can someone explain intuitively why the transformation does not affect the local extrema?
derivatives logarithms maximum-likelihood monotone-functions
edited Dec 26 '18 at 17:53
Bernard
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
$endgroup$
add a comment |
$begingroup$
I am studying maximum likelihood and to simplify taking the derivative of the likelihood function, it is often transformed by the natural log before taking the derivative.
I have read in other posts that this is because the logarithm is a monotonic function, so its extrema will be the same as the original function. However, I do not understand why this is the case. Can someone explain intuitively why the transformation does not affect the local extrema?
derivatives logarithms maximum-likelihood monotone-functions
edited Dec 26 '18 at 17:53
Bernard
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
$endgroup$
I am studying maximum likelihood and to simplify taking the derivative of the likelihood function, it is often transformed by the natural log before taking the derivative.
I have read in other posts that this is because the logarithm is a monotonic function, so its extrema will be the same as the original function. However, I do not understand why this is the case. Can someone explain intuitively why the transformation does not affect the local extrema?
derivatives logarithms maximum-likelihood monotone-functions
derivatives logarithms maximum-likelihood monotone-functions
edited Dec 26 '18 at 17:53
Bernard
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
edited Dec 26 '18 at 17:53
Bernard
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
edited Dec 26 '18 at 17:53
Bernard
122k741116
edited Dec 26 '18 at 17:53
Bernard
122k741116
edited Dec 26 '18 at 17:53
Bernard
122k741116
122k741116
asked Dec 26 '18 at 17:29
hlineehlinee
755
asked Dec 26 '18 at 17:29
hlineehlinee
755
asked Dec 26 '18 at 17:29
hlineehlinee
755
755
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $f(x)$ be a positive function and suppose that $x_0$ is the local maximum of $f(x)$ in the interval $[a,b]$. This means that for any $yin[a,b]$, $f(y)le f(x_0)$.
Logarithms is a monotonically increasing function: if $zle w$, then $log zle log w$. So for any $yin[a,b]$, since $f(y)le f(x_0)$, we have $log f(y) le log f(x_0)$. Hence, $x_0$ is also the local maximum for $log f(x)$.
The other direction can be proven by noting that the inverse of log is also monotonically increasing.
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Let $f(x)$ be a positive function and suppose that $x_0$ is the local maximum of $f(x)$ in the interval $[a,b]$. This means that for any $yin[a,b]$, $f(y)le f(x_0)$.
Logarithms is a monotonically increasing function: if $zle w$, then $log zle log w$. So for any $yin[a,b]$, since $f(y)le f(x_0)$, we have $log f(y) le log f(x_0)$. Hence, $x_0$ is also the local maximum for $log f(x)$.
The other direction can be proven by noting that the inverse of log is also monotonically increasing.
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be a positive function and suppose that $x_0$ is the local maximum of $f(x)$ in the interval $[a,b]$. This means that for any $yin[a,b]$, $f(y)le f(x_0)$.
Logarithms is a monotonically increasing function: if $zle w$, then $log zle log w$. So for any $yin[a,b]$, since $f(y)le f(x_0)$, we have $log f(y) le log f(x_0)$. Hence, $x_0$ is also the local maximum for $log f(x)$.
The other direction can be proven by noting that the inverse of log is also monotonically increasing.
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be a positive function and suppose that $x_0$ is the local maximum of $f(x)$ in the interval $[a,b]$. This means that for any $yin[a,b]$, $f(y)le f(x_0)$.
Logarithms is a monotonically increasing function: if $zle w$, then $log zle log w$. So for any $yin[a,b]$, since $f(y)le f(x_0)$, we have $log f(y) le log f(x_0)$. Hence, $x_0$ is also the local maximum for $log f(x)$.
The other direction can be proven by noting that the inverse of log is also monotonically increasing.
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
$endgroup$
Let $f(x)$ be a positive function and suppose that $x_0$ is the local maximum of $f(x)$ in the interval $[a,b]$. This means that for any $yin[a,b]$, $f(y)le f(x_0)$.
Logarithms is a monotonically increasing function: if $zle w$, then $log zle log w$. So for any $yin[a,b]$, since $f(y)le f(x_0)$, we have $log f(y) le log f(x_0)$. Hence, $x_0$ is also the local maximum for $log f(x)$.
The other direction can be proven by noting that the inverse of log is also monotonically increasing.
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
edited Dec 26 '18 at 21:32
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
answered Dec 26 '18 at 17:38
DirkGentlyDirkGently
1,49858
1,49858
add a comment |
add a comment |
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