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1

$begingroup$

In here, i want to show this entropy exist or not exist, namely i
should calculate the integral of $int_0^cfrac1xlog^2fracexfrac12 logfracex,dx$. If the result is $ <infty$, we can say the entropy exists, otherwise it does not exist.
beginequation*
int_0^1f(x)log f(x),dx geq int_0^c frac1xlog^2fracexfrac12 logfracex , dx
endequation* where $x in (0,c)$

mathematical-statistics entropy nonparametric-density

share|cite|improve this question

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $int_b^cfrac1xlogfracexfrac12dx = -log(log(1/x)+1)bigvert_b^c$ this won't converge when $b=0$ for a proof you might try to change variables $t=e/x$
  $endgroup$
  – Martijn Weterings
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

1

$begingroup$

In here, i want to show this entropy exist or not exist, namely i
should calculate the integral of $int_0^cfrac1xlog^2fracexfrac12 logfracex,dx$. If the result is $ <infty$, we can say the entropy exists, otherwise it does not exist.
beginequation*
int_0^1f(x)log f(x),dx geq int_0^c frac1xlog^2fracexfrac12 logfracex , dx
endequation* where $x in (0,c)$

mathematical-statistics entropy nonparametric-density

share|cite|improve this question

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $int_b^cfrac1xlogfracexfrac12dx = -log(log(1/x)+1)bigvert_b^c$ this won't converge when $b=0$ for a proof you might try to change variables $t=e/x$
  $endgroup$
  – Martijn Weterings
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

In here, i want to show this entropy exist or not exist, namely i
should calculate the integral of $int_0^cfrac1xlog^2fracexfrac12 logfracex,dx$. If the result is $ <infty$, we can say the entropy exists, otherwise it does not exist.
beginequation*
int_0^1f(x)log f(x),dx geq int_0^c frac1xlog^2fracexfrac12 logfracex , dx
endequation* where $x in (0,c)$

mathematical-statistics entropy nonparametric-density

share|cite|improve this question

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

$endgroup$

share|cite|improve this question

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

share|cite|improve this question

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

asked 9 hours ago

mhmtmhmt

1333 bronze badges

2 Answers
2

active

oldest

votes

2

$begingroup$

$$int_0^cfrac1xlogfracexfrac12dx $$

substitute $t=e/x$ (and use $dt/dx=-e/x^2$)

$$int_e/c^inftyfrace^2tlog tfrac12dt = log(log(t)) bigvert_e/c^infty$$

which diverges because $log(log(t))$ becomes infinite as $t to infty$

share|cite|improve this answer

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

By u-substitution,

beginaligned
& int_0^c frac1xlog^2 fracex frac12 logfracex , dx
= int_0^c frac12 left(logfracex right)^-1 frac1x dx \[8pt]
& u=logfracex=1-log(x), qquad du=-fracdxx\[8pt]
= & -frac12int u^-1 , du quad text(ignore limits for now) \[8pt]
= & -frac12 log u \[8pt]
= & left. -frac12 log(1-log x) right|_a^c, ~~ textin lim a rightarrow 0\[8pt]
= & -frac12 lim_arightarrow 0 log(1-log c) -log(1-log a)
endaligned
And we see the limit doesn't exist. So, no, the entropy doesn't exist.

And I see that Martijn Weterings beat me to the punch 12 minutes ago! OK, he had it first. :)

share|cite|improve this answer

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I find the change of variables u = log(e/x) more elegant. But this is not really integration by parts.
  $endgroup$
  – Martijn Weterings
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MartijnWeterings. Quite right. It is u-sub not by-parts. Of course!
  $endgroup$
  – Peter Leopold
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Martijn Weterings and Peter Leopold, for your clearly explaining
  $endgroup$
  – mhmt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edits for proper MathJax usage.
  $endgroup$
  – Michael Hardy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelHardy, thank you for the edits. I'm a bit astonished that it is considered "proper MathJax usage" to left justify a set of sequentially-derived equations on the '=' sign. Aligning equations on the central '=' sign is considered proper $AMS ~LaTeX$ style, I believe. (See 117 of ams.org/publications/authors/AMS-StyleGuide-online.pdf.) But I guess your point is that MathJax is not $LaTeX$. OK!
  $endgroup$
  – Peter Leopold
  5 hours ago
  
  

  
  
  
  

add a comment
 | 

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2

$begingroup$

$$int_0^cfrac1xlogfracexfrac12dx $$

substitute $t=e/x$ (and use $dt/dx=-e/x^2$)

$$int_e/c^inftyfrace^2tlog tfrac12dt = log(log(t)) bigvert_e/c^infty$$

which diverges because $log(log(t))$ becomes infinite as $t to infty$

share|cite|improve this answer

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

$$int_0^cfrac1xlogfracexfrac12dx $$

substitute $t=e/x$ (and use $dt/dx=-e/x^2$)

$$int_e/c^inftyfrace^2tlog tfrac12dt = log(log(t)) bigvert_e/c^infty$$

which diverges because $log(log(t))$ becomes infinite as $t to infty$

share|cite|improve this answer

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

$$int_0^cfrac1xlogfracexfrac12dx $$

substitute $t=e/x$ (and use $dt/dx=-e/x^2$)

$$int_e/c^inftyfrace^2tlog tfrac12dt = log(log(t)) bigvert_e/c^infty$$

which diverges because $log(log(t))$ becomes infinite as $t to infty$

share|cite|improve this answer

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

answered 8 hours ago

Martijn WeteringsMartijn Weterings

16k2424 silver badges7070 bronze badges

2

$begingroup$

By u-substitution,

beginaligned
& int_0^c frac1xlog^2 fracex frac12 logfracex , dx
= int_0^c frac12 left(logfracex right)^-1 frac1x dx \[8pt]
& u=logfracex=1-log(x), qquad du=-fracdxx\[8pt]
= & -frac12int u^-1 , du quad text(ignore limits for now) \[8pt]
= & -frac12 log u \[8pt]
= & left. -frac12 log(1-log x) right|_a^c, ~~ textin lim a rightarrow 0\[8pt]
= & -frac12 lim_arightarrow 0 log(1-log c) -log(1-log a)
endaligned
And we see the limit doesn't exist. So, no, the entropy doesn't exist.

And I see that Martijn Weterings beat me to the punch 12 minutes ago! OK, he had it first. :)

share|cite|improve this answer

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I find the change of variables u = log(e/x) more elegant. But this is not really integration by parts.
  $endgroup$
  – Martijn Weterings
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MartijnWeterings. Quite right. It is u-sub not by-parts. Of course!
  $endgroup$
  – Peter Leopold
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Martijn Weterings and Peter Leopold, for your clearly explaining
  $endgroup$
  – mhmt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edits for proper MathJax usage.
  $endgroup$
  – Michael Hardy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelHardy, thank you for the edits. I'm a bit astonished that it is considered "proper MathJax usage" to left justify a set of sequentially-derived equations on the '=' sign. Aligning equations on the central '=' sign is considered proper $AMS ~LaTeX$ style, I believe. (See 117 of ams.org/publications/authors/AMS-StyleGuide-online.pdf.) But I guess your point is that MathJax is not $LaTeX$. OK!
  $endgroup$
  – Peter Leopold
  5 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

By u-substitution,

beginaligned
& int_0^c frac1xlog^2 fracex frac12 logfracex , dx
= int_0^c frac12 left(logfracex right)^-1 frac1x dx \[8pt]
& u=logfracex=1-log(x), qquad du=-fracdxx\[8pt]
= & -frac12int u^-1 , du quad text(ignore limits for now) \[8pt]
= & -frac12 log u \[8pt]
= & left. -frac12 log(1-log x) right|_a^c, ~~ textin lim a rightarrow 0\[8pt]
= & -frac12 lim_arightarrow 0 log(1-log c) -log(1-log a)
endaligned
And we see the limit doesn't exist. So, no, the entropy doesn't exist.

And I see that Martijn Weterings beat me to the punch 12 minutes ago! OK, he had it first. :)

share|cite|improve this answer

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I find the change of variables u = log(e/x) more elegant. But this is not really integration by parts.
  $endgroup$
  – Martijn Weterings
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MartijnWeterings. Quite right. It is u-sub not by-parts. Of course!
  $endgroup$
  – Peter Leopold
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Martijn Weterings and Peter Leopold, for your clearly explaining
  $endgroup$
  – mhmt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edits for proper MathJax usage.
  $endgroup$
  – Michael Hardy
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @MichaelHardy, thank you for the edits. I'm a bit astonished that it is considered "proper MathJax usage" to left justify a set of sequentially-derived equations on the '=' sign. Aligning equations on the central '=' sign is considered proper $AMS ~LaTeX$ style, I believe. (See 117 of ams.org/publications/authors/AMS-StyleGuide-online.pdf.) But I guess your point is that MathJax is not $LaTeX$. OK!
  $endgroup$
  – Peter Leopold
  5 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

By u-substitution,

beginaligned
& int_0^c frac1xlog^2 fracex frac12 logfracex , dx
= int_0^c frac12 left(logfracex right)^-1 frac1x dx \[8pt]
& u=logfracex=1-log(x), qquad du=-fracdxx\[8pt]
= & -frac12int u^-1 , du quad text(ignore limits for now) \[8pt]
= & -frac12 log u \[8pt]
= & left. -frac12 log(1-log x) right|_a^c, ~~ textin lim a rightarrow 0\[8pt]
= & -frac12 lim_arightarrow 0 log(1-log c) -log(1-log a)
endaligned
And we see the limit doesn't exist. So, no, the entropy doesn't exist.

And I see that Martijn Weterings beat me to the punch 12 minutes ago! OK, he had it first. :)

share|cite|improve this answer

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

edited 6 hours ago

Michael Hardy

4,8241515 silver badges3131 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges

answered 8 hours ago

Peter LeopoldPeter Leopold

1,33433 silver badges1818 bronze badges