Vjylku

Random body shuffle every night—can we still function?

How can I prevent/balance waiting and turtling as a response to cooldown mechanics

I can't produce songs

Resize vertical bars (absolute-value symbols)

Moving a wrapfig vertically to encroach partially on a subsection title

Test print coming out spongy

Can an iPhone 7 be made to function as a NFC Tag?

Tips to organize LaTeX presentations for a semester

What initially awakened the Balrog?

Is multiple magic items in one inherently imbalanced?

How to ternary Plot3D a function

How to change the tick of the color bar legend to black

What does it mean that physics no longer uses mechanical models to describe phenomena?

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

What are the main differences between Stargate SG-1 cuts?

Why is std::move not [[nodiscard]] in C++20?

What would you call this weird metallic apparatus that allows you to lift people?

Flight departed from the gate 5 min before scheduled departure time. Refund options

Can two person see the same photon?

Why not send Voyager 3 and 4 following up the paths taken by Voyager 1 and 2 to re-transmit signals of later as they fly away from Earth?

Did any compiler fully use 80-bit floating point?

Should a wizard buy fine inks every time he want to copy spells into his spellbook?

How does light 'choose' between wave and particle behaviour?

Mounting TV on a weird wall that has some material between the drywall and stud

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:

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

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 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:

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

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 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:

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

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 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:

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

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 | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f404191%2fwhat-is-the-log-of-the-pdf-for-a-normal-distribution%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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