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

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

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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:

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

taking log produces:

begin{align}
ln(frac {1}{sigma sqrt {2pi}}
e^{-frac {(x - mu)^2}{2sigma ^2}}) &=
ln(frac {1}{sigma sqrt {2pi}})+ln(e^{-frac {(x - mu)^2}{2sigma ^2}})\
&=-ln(sigma)-frac{1}{2} ln(2pi) - frac {(x - mu)^2}{2sigma ^2}
end{align}

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 2 hours 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
  2 hours 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:

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

taking log produces:

begin{align}
ln(frac {1}{sigma sqrt {2pi}}
e^{-frac {(x - mu)^2}{2sigma ^2}}) &=
ln(frac {1}{sigma sqrt {2pi}})+ln(e^{-frac {(x - mu)^2}{2sigma ^2}})\
&=-ln(sigma)-frac{1}{2} ln(2pi) - frac {(x - mu)^2}{2sigma ^2}
end{align}

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 2 hours 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
  2 hours 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:

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

taking log produces:

begin{align}
ln(frac {1}{sigma sqrt {2pi}}
e^{-frac {(x - mu)^2}{2sigma ^2}}) &=
ln(frac {1}{sigma sqrt {2pi}})+ln(e^{-frac {(x - mu)^2}{2sigma ^2}})\
&=-ln(sigma)-frac{1}{2} ln(2pi) - frac {(x - mu)^2}{2sigma ^2}
end{align}

which is very different from equation1.

is equation1 right? what am I missing?

probability log

share|cite|improve this question

asked 2 hours ago

shi95shi95

83

$endgroup$

share|cite|improve this question

asked 2 hours ago

shi95shi95

83

share|cite|improve this question

asked 2 hours ago

shi95shi95

83

asked 2 hours ago

shi95shi95

83

asked 2 hours ago

shi95shi95

83

1 Answer
1

active

oldest

votes

2

$begingroup$

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

$$ell_x(mu,sigma^2) = - ln sigma - frac{1}{2} ln (2 pi) - frac{1}{2} Big( frac{x-mu}{sigma} Big)^2.$$

For a sample of observed values $mathbf{x} = (x_1,...,x_n)$ you then have:

$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 1 hour ago

BenBen

28.9k233129

$endgroup$

add a comment | 

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2

$begingroup$

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

$$ell_x(mu,sigma^2) = - ln sigma - frac{1}{2} ln (2 pi) - frac{1}{2} Big( frac{x-mu}{sigma} Big)^2.$$

For a sample of observed values $mathbf{x} = (x_1,...,x_n)$ you then have:

$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 1 hour ago

BenBen

28.9k233129

$endgroup$

add a comment | 

2

$begingroup$

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

$$ell_x(mu,sigma^2) = - ln sigma - frac{1}{2} ln (2 pi) - frac{1}{2} Big( frac{x-mu}{sigma} Big)^2.$$

For a sample of observed values $mathbf{x} = (x_1,...,x_n)$ you then have:

$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 1 hour ago

BenBen

28.9k233129

$endgroup$

add a comment | 

$begingroup$

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

$$ell_x(mu,sigma^2) = - ln sigma - frac{1}{2} ln (2 pi) - frac{1}{2} Big( frac{x-mu}{sigma} Big)^2.$$

For a sample of observed values $mathbf{x} = (x_1,...,x_n)$ you then have:

$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$

share|cite|improve this answer

answered 1 hour ago

BenBen

28.9k233129

$endgroup$

share|cite|improve this answer

answered 1 hour ago

BenBen

28.9k233129

answered 1 hour ago

BenBen

28.9k233129

answered 1 hour ago

BenBen

28.9k233129