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





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$begingroup$


I am learning Maximum Likelihood Estimation.



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



enter image description here



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









$endgroup$








  • 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
    4 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
    4 hours ago


















1












$begingroup$


I am learning Maximum Likelihood Estimation.



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



enter image description here



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









$endgroup$








  • 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
    4 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
    4 hours ago














1












1








1





$begingroup$


I am learning Maximum Likelihood Estimation.



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



enter image description here



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









$endgroup$




I am learning Maximum Likelihood Estimation.



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



enter image description here



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




probability log






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









shi95shi95

103




103








  • 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
    4 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
    4 hours ago














  • 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
    4 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
    4 hours ago








3




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
4 hours ago






$begingroup$
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
4 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
4 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
4 hours ago










1 Answer
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$begingroup$

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:



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






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    active

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    $begingroup$

    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









    $endgroup$


















      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









      $endgroup$
















        2












        2








        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









        $endgroup$



        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        BenBen

        28.9k233129




        28.9k233129






























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