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what is the log of the PDF for a Normal Distribution?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

I am learning Maximum Likelihood Estimation.

per this post, the log of the PDF for a Normal Distribution looks like this.

let's call this equation1.

according to any probability theory textbook the formula of the PDF for a Normal Distribution:

$$
frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2
,-infty <x<infty
$$

taking log produces:

beginalign
ln(frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2) &=
ln(frac 1sigma sqrt 2pi)+ln(e^-frac (x - mu)^22sigma ^2)\
&=-ln(sigma)-frac12 ln(2pi) - frac (x - mu)^22sigma ^2
endalign

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 1 hour ago

shi95shi95

83

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your first equation is the joint log-pdf of a sample of n iid normal random variables (AKA the log-likelihood of that sample). The second equation is the the log-pdf of a single normal random variable
  $endgroup$
  – Artem Mavrin
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArtemMavrin, I think your comment would be a perfectly good answer if you expanded on just a bit to make it slightly more clear.
  $endgroup$
  – StatsStudent
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I am learning Maximum Likelihood Estimation.

per this post, the log of the PDF for a Normal Distribution looks like this.

let's call this equation1.

according to any probability theory textbook the formula of the PDF for a Normal Distribution:

$$
frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2
,-infty <x<infty
$$

taking log produces:

beginalign
ln(frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2) &=
ln(frac 1sigma sqrt 2pi)+ln(e^-frac (x - mu)^22sigma ^2)\
&=-ln(sigma)-frac12 ln(2pi) - frac (x - mu)^22sigma ^2
endalign

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 1 hour ago

shi95shi95

83

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your first equation is the joint log-pdf of a sample of n iid normal random variables (AKA the log-likelihood of that sample). The second equation is the the log-pdf of a single normal random variable
  $endgroup$
  – Artem Mavrin
  1 hour ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArtemMavrin, I think your comment would be a perfectly good answer if you expanded on just a bit to make it slightly more clear.
  $endgroup$
  – StatsStudent
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

I am learning Maximum Likelihood Estimation.

per this post, the log of the PDF for a Normal Distribution looks like this.

let's call this equation1.

according to any probability theory textbook the formula of the PDF for a Normal Distribution:

$$
frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2
,-infty <x<infty
$$

taking log produces:

beginalign
ln(frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2) &=
ln(frac 1sigma sqrt 2pi)+ln(e^-frac (x - mu)^22sigma ^2)\
&=-ln(sigma)-frac12 ln(2pi) - frac (x - mu)^22sigma ^2
endalign

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 1 hour ago

shi95shi95

83

$endgroup$

share|cite|improve this question

asked 1 hour ago

shi95shi95

83

share|cite|improve this question

asked 1 hour ago

shi95shi95

83

asked 1 hour ago

shi95shi95

83

asked 1 hour ago

shi95shi95

83

1 Answer
1

active

oldest

votes

2

$begingroup$

For a single observed value $x$ you have log-likelihood:

$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$

For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:

$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 56 mins ago

BenBen

28.9k233129

$endgroup$

add a comment | 

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2

$begingroup$

For a single observed value $x$ you have log-likelihood:

$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$

For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:

$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 56 mins ago

BenBen

28.9k233129

$endgroup$

add a comment | 

2

$begingroup$

For a single observed value $x$ you have log-likelihood:

$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$

For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:

$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 56 mins ago

BenBen

28.9k233129

$endgroup$

add a comment | 

$begingroup$

For a single observed value $x$ you have log-likelihood:

$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$

For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:

$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 56 mins ago

BenBen

28.9k233129

$endgroup$

share|cite|improve this answer

answered 56 mins ago

BenBen

28.9k233129

answered 56 mins ago

BenBen

28.9k233129

answered 56 mins ago

BenBen

28.9k233129