Which likelihood function is used in linear regression?Maximizing: likelihood vs likelihood ratioWhen would maximum likelihood estimates equal least squares estimates?Comparing maximum likelihood estimation (MLE) and Bayes' TheoremMaximizing likelihood versus MCMC sampling: Comparing Parameters and DevianceLikelihood - Why multiply?AIC only applicable to maximum likelihood fit (not least squares)?Why does Maximum Likelihood estimation maximizes probability density instead of probabilitylinear regression with gaussian distributionUnderstand a statement about likelihood function
How many different ways are there to checkmate in the early game?
How does The Fools Guild make its money?
(11 of 11: Meta) What is Pyramid Cult's All-Time Favorite?
How do we avoid CI-driven development...?
How can I tell if a flight itinerary is fake?
Who are these characters/superheroes in the posters from Chris's room in Family Guy?
How would I as a DM create a smart phone-like spell/device my players could use?
Looking for a new job because of relocation - is it okay to tell the real reason?
Look mom! I made my own (Base 10) numeral system!
In a topological space if there exists a loop that cannot be contracted to a point does there exist a simple loop that cannot be contracted also?
Why "ch" pronunciation rule doesn't occur for words such as "durch", "manchmal"?
Is it really ~648.69 km/s delta-v to "land" on the surface of the Sun?
Performance of a branch and bound algorithm VS branch-cut-heuristics
Are there any financial disadvantages to living significantly "below your means"?
In reversi, can you overwrite two chips in one move?
What does "sardine box" mean?
Was the 2019 Lion King film made through motion capture?
What does Apple mean by "This may decrease battery life"?
Optimal way to extract "positive part" of a multivariate polynomial
Write an interpreter for *
Generator for parity?
Converting Piecewise function to C code
Ex-contractor published company source code and secrets online
Why couldn't soldiers sight their own weapons without officers' orders?
Which likelihood function is used in linear regression?
Maximizing: likelihood vs likelihood ratioWhen would maximum likelihood estimates equal least squares estimates?Comparing maximum likelihood estimation (MLE) and Bayes' TheoremMaximizing likelihood versus MCMC sampling: Comparing Parameters and DevianceLikelihood - Why multiply?AIC only applicable to maximum likelihood fit (not least squares)?Why does Maximum Likelihood estimation maximizes probability density instead of probabilitylinear regression with gaussian distributionUnderstand a statement about likelihood function
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
When trying to derive the maximum likelihood estimation for a linear regression, We start by a likelihood function. Does it matter if we use either of these 2 forms?
$P(y|x,w)$
$P(y,x|w)$
All pages that I read on the internet use the first one.
I found that $P(y,x|w)$ is equal to $P(y|x,w)*P(x)$
so maximizing $P(y,x|w)$ with respect to $w$ is the same as $P(y|x,w)$ because $x$ and $w$ are independent.
The second function looks better because it means "what is the probability of the parameter giving the data(x AND y)" but the first function doesn't show that.
Is my point correct or not?Is there any difference?
regression mathematical-statistics maximum-likelihood
$endgroup$
add a comment |
$begingroup$
When trying to derive the maximum likelihood estimation for a linear regression, We start by a likelihood function. Does it matter if we use either of these 2 forms?
$P(y|x,w)$
$P(y,x|w)$
All pages that I read on the internet use the first one.
I found that $P(y,x|w)$ is equal to $P(y|x,w)*P(x)$
so maximizing $P(y,x|w)$ with respect to $w$ is the same as $P(y|x,w)$ because $x$ and $w$ are independent.
The second function looks better because it means "what is the probability of the parameter giving the data(x AND y)" but the first function doesn't show that.
Is my point correct or not?Is there any difference?
regression mathematical-statistics maximum-likelihood
$endgroup$
add a comment |
$begingroup$
When trying to derive the maximum likelihood estimation for a linear regression, We start by a likelihood function. Does it matter if we use either of these 2 forms?
$P(y|x,w)$
$P(y,x|w)$
All pages that I read on the internet use the first one.
I found that $P(y,x|w)$ is equal to $P(y|x,w)*P(x)$
so maximizing $P(y,x|w)$ with respect to $w$ is the same as $P(y|x,w)$ because $x$ and $w$ are independent.
The second function looks better because it means "what is the probability of the parameter giving the data(x AND y)" but the first function doesn't show that.
Is my point correct or not?Is there any difference?
regression mathematical-statistics maximum-likelihood
$endgroup$
When trying to derive the maximum likelihood estimation for a linear regression, We start by a likelihood function. Does it matter if we use either of these 2 forms?
$P(y|x,w)$
$P(y,x|w)$
All pages that I read on the internet use the first one.
I found that $P(y,x|w)$ is equal to $P(y|x,w)*P(x)$
so maximizing $P(y,x|w)$ with respect to $w$ is the same as $P(y|x,w)$ because $x$ and $w$ are independent.
The second function looks better because it means "what is the probability of the parameter giving the data(x AND y)" but the first function doesn't show that.
Is my point correct or not?Is there any difference?
regression mathematical-statistics maximum-likelihood
regression mathematical-statistics maximum-likelihood
edited 7 hours ago
floyd
asked 8 hours ago
floydfloyd
3445 silver badges20 bronze badges
3445 silver badges20 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the details of what $p(x)$ may look like. This is because for the model $Y = f(X) + varepsilon$, where $varepsilon sim mathcalN(0, sigma^2)$, we want the estimate function $hatf$ we produce to be as close to $f$ as possible on average, measured by, say, least squares. This is the same thing as pulling $hatf$ close to $f$ on all its input values, individually---it doesn't matter the distribution $p(x)$ of how the inputs are scattered, as long as $hatf$ is close to $f$ on those values. We only want to model how $y$ responds to $x$, and we don't care about how $x$ itself.
More mathematically, from an MLE standpoint the goal is to get our likelihoods $P(x,y|w)$ as large as possible, but as you say, we are not modeling $P(x)$ through $w$, so $P(x)$ factors out and doesn't matter. From a fitting standpoint, if we want to minimize expected prediction error over $f$,
$$min_f textEPE(f) = min_f mathbbE(Y - f(X))^2$$
omitting some computation we obtain the minimizing $f$ to be
$$f(x) = mathbbE(Y | X = x)$$
so for least squares loss, the best possible $f$ depends only on the conditional distribution $p(y | x)$, and estimating it should not require any additional information.
However, in classification problems (logistic regression, linear discriminant analysis, Naive Bayes) there is a difference between a "generative" and a "discriminative" model; generative models do not condition on $x$ and discriminative models do. For instance, in linear discriminant analysis we do model $P(x)$ through $w$---we also do not use least squares loss in those.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
add a comment |
$begingroup$
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want to determine the probability of Y conditioned on X and w (usually denoted as $beta$ in statistical settings).
Usually, in simple linear regression,
$$Y = beta_0 + beta_1 X + epsilon$$
you model X as being fixed and independent from $epsilon$ which follows a $sim N(0,sigma^2)$ distribution. That is, Y is modeled as a linear function of some fixed input (X) with some random noise ($epsilon$). Therefore, it makes sense to model it as $p(y|X,w)$ in this setting as X is fixed or modeled as constant, which is usually denoted explicitly as $X=x$.
Mathematically, when X is fixed, p(X) =1 as it is constant. Therefore
$$p(y,X|w) = p(y|X,w)p(X) implies p(y,X|w)= p(y|X,w)$$
If X is not constant and given, then p(X) is no longer 1, and the above doesn't hold, so it all comes down to whether you model X as being random or fixed.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f421598%2fwhich-likelihood-function-is-used-in-linear-regression%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the details of what $p(x)$ may look like. This is because for the model $Y = f(X) + varepsilon$, where $varepsilon sim mathcalN(0, sigma^2)$, we want the estimate function $hatf$ we produce to be as close to $f$ as possible on average, measured by, say, least squares. This is the same thing as pulling $hatf$ close to $f$ on all its input values, individually---it doesn't matter the distribution $p(x)$ of how the inputs are scattered, as long as $hatf$ is close to $f$ on those values. We only want to model how $y$ responds to $x$, and we don't care about how $x$ itself.
More mathematically, from an MLE standpoint the goal is to get our likelihoods $P(x,y|w)$ as large as possible, but as you say, we are not modeling $P(x)$ through $w$, so $P(x)$ factors out and doesn't matter. From a fitting standpoint, if we want to minimize expected prediction error over $f$,
$$min_f textEPE(f) = min_f mathbbE(Y - f(X))^2$$
omitting some computation we obtain the minimizing $f$ to be
$$f(x) = mathbbE(Y | X = x)$$
so for least squares loss, the best possible $f$ depends only on the conditional distribution $p(y | x)$, and estimating it should not require any additional information.
However, in classification problems (logistic regression, linear discriminant analysis, Naive Bayes) there is a difference between a "generative" and a "discriminative" model; generative models do not condition on $x$ and discriminative models do. For instance, in linear discriminant analysis we do model $P(x)$ through $w$---we also do not use least squares loss in those.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
add a comment |
$begingroup$
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the details of what $p(x)$ may look like. This is because for the model $Y = f(X) + varepsilon$, where $varepsilon sim mathcalN(0, sigma^2)$, we want the estimate function $hatf$ we produce to be as close to $f$ as possible on average, measured by, say, least squares. This is the same thing as pulling $hatf$ close to $f$ on all its input values, individually---it doesn't matter the distribution $p(x)$ of how the inputs are scattered, as long as $hatf$ is close to $f$ on those values. We only want to model how $y$ responds to $x$, and we don't care about how $x$ itself.
More mathematically, from an MLE standpoint the goal is to get our likelihoods $P(x,y|w)$ as large as possible, but as you say, we are not modeling $P(x)$ through $w$, so $P(x)$ factors out and doesn't matter. From a fitting standpoint, if we want to minimize expected prediction error over $f$,
$$min_f textEPE(f) = min_f mathbbE(Y - f(X))^2$$
omitting some computation we obtain the minimizing $f$ to be
$$f(x) = mathbbE(Y | X = x)$$
so for least squares loss, the best possible $f$ depends only on the conditional distribution $p(y | x)$, and estimating it should not require any additional information.
However, in classification problems (logistic regression, linear discriminant analysis, Naive Bayes) there is a difference between a "generative" and a "discriminative" model; generative models do not condition on $x$ and discriminative models do. For instance, in linear discriminant analysis we do model $P(x)$ through $w$---we also do not use least squares loss in those.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
add a comment |
$begingroup$
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the details of what $p(x)$ may look like. This is because for the model $Y = f(X) + varepsilon$, where $varepsilon sim mathcalN(0, sigma^2)$, we want the estimate function $hatf$ we produce to be as close to $f$ as possible on average, measured by, say, least squares. This is the same thing as pulling $hatf$ close to $f$ on all its input values, individually---it doesn't matter the distribution $p(x)$ of how the inputs are scattered, as long as $hatf$ is close to $f$ on those values. We only want to model how $y$ responds to $x$, and we don't care about how $x$ itself.
More mathematically, from an MLE standpoint the goal is to get our likelihoods $P(x,y|w)$ as large as possible, but as you say, we are not modeling $P(x)$ through $w$, so $P(x)$ factors out and doesn't matter. From a fitting standpoint, if we want to minimize expected prediction error over $f$,
$$min_f textEPE(f) = min_f mathbbE(Y - f(X))^2$$
omitting some computation we obtain the minimizing $f$ to be
$$f(x) = mathbbE(Y | X = x)$$
so for least squares loss, the best possible $f$ depends only on the conditional distribution $p(y | x)$, and estimating it should not require any additional information.
However, in classification problems (logistic regression, linear discriminant analysis, Naive Bayes) there is a difference between a "generative" and a "discriminative" model; generative models do not condition on $x$ and discriminative models do. For instance, in linear discriminant analysis we do model $P(x)$ through $w$---we also do not use least squares loss in those.
$endgroup$
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the details of what $p(x)$ may look like. This is because for the model $Y = f(X) + varepsilon$, where $varepsilon sim mathcalN(0, sigma^2)$, we want the estimate function $hatf$ we produce to be as close to $f$ as possible on average, measured by, say, least squares. This is the same thing as pulling $hatf$ close to $f$ on all its input values, individually---it doesn't matter the distribution $p(x)$ of how the inputs are scattered, as long as $hatf$ is close to $f$ on those values. We only want to model how $y$ responds to $x$, and we don't care about how $x$ itself.
More mathematically, from an MLE standpoint the goal is to get our likelihoods $P(x,y|w)$ as large as possible, but as you say, we are not modeling $P(x)$ through $w$, so $P(x)$ factors out and doesn't matter. From a fitting standpoint, if we want to minimize expected prediction error over $f$,
$$min_f textEPE(f) = min_f mathbbE(Y - f(X))^2$$
omitting some computation we obtain the minimizing $f$ to be
$$f(x) = mathbbE(Y | X = x)$$
so for least squares loss, the best possible $f$ depends only on the conditional distribution $p(y | x)$, and estimating it should not require any additional information.
However, in classification problems (logistic regression, linear discriminant analysis, Naive Bayes) there is a difference between a "generative" and a "discriminative" model; generative models do not condition on $x$ and discriminative models do. For instance, in linear discriminant analysis we do model $P(x)$ through $w$---we also do not use least squares loss in those.
answered 6 hours ago
Drew N Drew N
3462 silver badges8 bronze badges
3462 silver badges8 bronze badges
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
add a comment |
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
1
1
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
$begingroup$
I wouldn't call $X$ the model. When they write $m$ that way, they're conditioning on a probabilistic model for the data in addition to the data $x$; when they write just $x$ instead of $m$ the model is assumed implicitly.
$endgroup$
– Drew N
5 hours ago
add a comment |
$begingroup$
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want to determine the probability of Y conditioned on X and w (usually denoted as $beta$ in statistical settings).
Usually, in simple linear regression,
$$Y = beta_0 + beta_1 X + epsilon$$
you model X as being fixed and independent from $epsilon$ which follows a $sim N(0,sigma^2)$ distribution. That is, Y is modeled as a linear function of some fixed input (X) with some random noise ($epsilon$). Therefore, it makes sense to model it as $p(y|X,w)$ in this setting as X is fixed or modeled as constant, which is usually denoted explicitly as $X=x$.
Mathematically, when X is fixed, p(X) =1 as it is constant. Therefore
$$p(y,X|w) = p(y|X,w)p(X) implies p(y,X|w)= p(y|X,w)$$
If X is not constant and given, then p(X) is no longer 1, and the above doesn't hold, so it all comes down to whether you model X as being random or fixed.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
add a comment |
$begingroup$
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want to determine the probability of Y conditioned on X and w (usually denoted as $beta$ in statistical settings).
Usually, in simple linear regression,
$$Y = beta_0 + beta_1 X + epsilon$$
you model X as being fixed and independent from $epsilon$ which follows a $sim N(0,sigma^2)$ distribution. That is, Y is modeled as a linear function of some fixed input (X) with some random noise ($epsilon$). Therefore, it makes sense to model it as $p(y|X,w)$ in this setting as X is fixed or modeled as constant, which is usually denoted explicitly as $X=x$.
Mathematically, when X is fixed, p(X) =1 as it is constant. Therefore
$$p(y,X|w) = p(y|X,w)p(X) implies p(y,X|w)= p(y|X,w)$$
If X is not constant and given, then p(X) is no longer 1, and the above doesn't hold, so it all comes down to whether you model X as being random or fixed.
$endgroup$
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
add a comment |
$begingroup$
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want to determine the probability of Y conditioned on X and w (usually denoted as $beta$ in statistical settings).
Usually, in simple linear regression,
$$Y = beta_0 + beta_1 X + epsilon$$
you model X as being fixed and independent from $epsilon$ which follows a $sim N(0,sigma^2)$ distribution. That is, Y is modeled as a linear function of some fixed input (X) with some random noise ($epsilon$). Therefore, it makes sense to model it as $p(y|X,w)$ in this setting as X is fixed or modeled as constant, which is usually denoted explicitly as $X=x$.
Mathematically, when X is fixed, p(X) =1 as it is constant. Therefore
$$p(y,X|w) = p(y|X,w)p(X) implies p(y,X|w)= p(y|X,w)$$
If X is not constant and given, then p(X) is no longer 1, and the above doesn't hold, so it all comes down to whether you model X as being random or fixed.
$endgroup$
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want to determine the probability of Y conditioned on X and w (usually denoted as $beta$ in statistical settings).
Usually, in simple linear regression,
$$Y = beta_0 + beta_1 X + epsilon$$
you model X as being fixed and independent from $epsilon$ which follows a $sim N(0,sigma^2)$ distribution. That is, Y is modeled as a linear function of some fixed input (X) with some random noise ($epsilon$). Therefore, it makes sense to model it as $p(y|X,w)$ in this setting as X is fixed or modeled as constant, which is usually denoted explicitly as $X=x$.
Mathematically, when X is fixed, p(X) =1 as it is constant. Therefore
$$p(y,X|w) = p(y|X,w)p(X) implies p(y,X|w)= p(y|X,w)$$
If X is not constant and given, then p(X) is no longer 1, and the above doesn't hold, so it all comes down to whether you model X as being random or fixed.
answered 6 hours ago
Samir Rachid ZaimSamir Rachid Zaim
864 bronze badges
864 bronze badges
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
add a comment |
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
$begingroup$
Thank you for the explanation. I saw some pages on the internet call $X$ "the model" and $y$ "the sample/data" and the likelihood is $P(data|m, w)$ where $m$ stands for "model". So they write the likelihood as $P(y|x,w)$. In my question I called both $x$ AND $y$ "data", Am I right? or should I call $x$ "the model"?
$endgroup$
– floyd
5 hours ago
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f421598%2fwhich-likelihood-function-is-used-in-linear-regression%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown