what is the log of the PDF for a Normal Distribution?
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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
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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
$endgroup$
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
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– 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
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
$endgroup$
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
probability log
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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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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add a comment |
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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:
$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$
$endgroup$
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:
$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$
$endgroup$
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:
$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$
$endgroup$
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:
$$ell_mathbf{x}(mu,sigma^2) = sum_{i=1}^n ell_x(mu,sigma^2) = - n ln sigma - frac{n}{2} ln (2 pi) - frac{1}{2 sigma^2} sum_{i=1}^n (x_i-mu)^2.$$
answered 4 hours ago
BenBen
28.9k233129
28.9k233129
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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