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what is the log of the PDF for a Normal Distribution?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to solve/compute for normal distribution and log-normal CDF inverse?Distribution of the convolution of squared normal and chi-squared variables?Cramer's theorem for a precise normal asymptotic distributionConditional Expected Value of Product of Normal and Log-Normal DistributionAsymptotic relation for a class of probability distribution functionsShow that $Y_1+Y_2$ have distribution skew-normalExpected Fisher's information matrix for Student's t-distribution?Expected Value of Maximum likelihood mean for Gaussian DistributionJoint density of the sum of a random and a non-random variable?Reversing conditional distribution
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
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
asked 2 hours ago
shi95shi95
83
$endgroup$
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
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:
$$
frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2
,-infty <x<infty
$$
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
asked 2 hours ago
shi95shi95
83
$endgroup$
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:
$$
frac 1sigma sqrt 2pi
e^-frac (x - mu)^22sigma ^2
,-infty <x<infty
$$
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
probability log
asked 2 hours ago
shi95shi95
83
asked 2 hours ago
shi95shi95
83
asked 2 hours ago
shi95shi95
83
asked 2 hours ago
shi95shi95
83
asked 2 hours ago
shi95shi95
83
83
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 |
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
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
2 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
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
$begingroup$
@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 Answer
1
active
oldest
votes
$begingroup$
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:
$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$
answered 1 hour ago
BenBen
28.9k233129
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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$begingroup$
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:
$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$
answered 1 hour ago
BenBen
28.9k233129
$endgroup$
add a comment |
$begingroup$
For a single observed value $x$ you have log-likelihood:
$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$
For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:
$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$
answered 1 hour ago
BenBen
28.9k233129
$endgroup$
add a comment |
$begingroup$
For a single observed value $x$ you have log-likelihood:
$$ell_x(mu,sigma^2) = - ln sigma - frac12 ln (2 pi) - frac12 Big( fracx-musigma Big)^2.$$
For a sample of observed values $mathbfx = (x_1,...,x_n)$ you then have:
$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$
answered 1 hour ago
BenBen
28.9k233129
$endgroup$
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:
$$ell_mathbfx(mu,sigma^2) = sum_i=1^n ell_x(mu,sigma^2) = - n ln sigma - fracn2 ln (2 pi) - frac12 sigma^2 sum_i=1^n (x_i-mu)^2.$$
answered 1 hour ago
BenBen
28.9k233129
answered 1 hour ago
BenBen
28.9k233129
answered 1 hour ago
BenBen
28.9k233129
answered 1 hour ago
BenBen
28.9k233129
28.9k233129
add a comment |
add a comment |
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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