How to obtain expressions for coefficients from OLS formula?

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

1

Consider the standard linear regression model: $y_i = alpha + beta D_i + e_i$ where the coefficients are defined by linear projections and $D_i$ is a dummy variable. In the population, the coefficients are given by:

$$alpha = E[y_i mid D_i =0] textand beta = E[y_i mid D_i = 1] - E[y_i mid D_i =0]$$

Using OLS to estimate the coefficients, we get:

$$widehatalpha = frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

$$widehatbeta = frac1sum_i=1^N1(D_i=1)sum_i=1^N1(D_i=1)y_i-frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

In other words, $widehatalpha$ is just the sample mean of $y_i$ in the subsample with $D_i=0$.

My question is, how can we arrive at the above coefficient estimates by using the standard OLS formulas? That is:

$$widehatalpha = overliney - overlineDwidehatbeta textand widehatbeta = fracsum_i=1^N(D_i - overlineD)(y_i - overliney)sum_i=1^N(D_i - overlineD)^2$$ where the bars represent sample means.

statistics regression least-squares linear-regression

share|cite|improve this question

edited Jul 29 at 12:46

asked Jul 29 at 12:37

elbarto

1,519523

add a comment | 

up vote
1
down vote

favorite

1

Consider the standard linear regression model: $y_i = alpha + beta D_i + e_i$ where the coefficients are defined by linear projections and $D_i$ is a dummy variable. In the population, the coefficients are given by:

$$alpha = E[y_i mid D_i =0] textand beta = E[y_i mid D_i = 1] - E[y_i mid D_i =0]$$

Using OLS to estimate the coefficients, we get:

$$widehatalpha = frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

$$widehatbeta = frac1sum_i=1^N1(D_i=1)sum_i=1^N1(D_i=1)y_i-frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

In other words, $widehatalpha$ is just the sample mean of $y_i$ in the subsample with $D_i=0$.

My question is, how can we arrive at the above coefficient estimates by using the standard OLS formulas? That is:

$$widehatalpha = overliney - overlineDwidehatbeta textand widehatbeta = fracsum_i=1^N(D_i - overlineD)(y_i - overliney)sum_i=1^N(D_i - overlineD)^2$$ where the bars represent sample means.

statistics regression least-squares linear-regression

share|cite|improve this question

edited Jul 29 at 12:46

asked Jul 29 at 12:37

elbarto

1,519523

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Consider the standard linear regression model: $y_i = alpha + beta D_i + e_i$ where the coefficients are defined by linear projections and $D_i$ is a dummy variable. In the population, the coefficients are given by:

$$alpha = E[y_i mid D_i =0] textand beta = E[y_i mid D_i = 1] - E[y_i mid D_i =0]$$

Using OLS to estimate the coefficients, we get:

$$widehatalpha = frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

$$widehatbeta = frac1sum_i=1^N1(D_i=1)sum_i=1^N1(D_i=1)y_i-frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

In other words, $widehatalpha$ is just the sample mean of $y_i$ in the subsample with $D_i=0$.

My question is, how can we arrive at the above coefficient estimates by using the standard OLS formulas? That is:

$$widehatalpha = overliney - overlineDwidehatbeta textand widehatbeta = fracsum_i=1^N(D_i - overlineD)(y_i - overliney)sum_i=1^N(D_i - overlineD)^2$$ where the bars represent sample means.

statistics regression least-squares linear-regression

share|cite|improve this question

edited Jul 29 at 12:46

asked Jul 29 at 12:37

elbarto

1,519523

Consider the standard linear regression model: $y_i = alpha + beta D_i + e_i$ where the coefficients are defined by linear projections and $D_i$ is a dummy variable. In the population, the coefficients are given by:

$$alpha = E[y_i mid D_i =0] textand beta = E[y_i mid D_i = 1] - E[y_i mid D_i =0]$$

Using OLS to estimate the coefficients, we get:

$$widehatalpha = frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

$$widehatbeta = frac1sum_i=1^N1(D_i=1)sum_i=1^N1(D_i=1)y_i-frac1sum_i=1^N1(D_i=0)sum_i=1^N1(D_i=0)y_i $$

In other words, $widehatalpha$ is just the sample mean of $y_i$ in the subsample with $D_i=0$.

My question is, how can we arrive at the above coefficient estimates by using the standard OLS formulas? That is:

$$widehatalpha = overliney - overlineDwidehatbeta textand widehatbeta = fracsum_i=1^N(D_i - overlineD)(y_i - overliney)sum_i=1^N(D_i - overlineD)^2$$ where the bars represent sample means.

statistics regression least-squares linear-regression

share|cite|improve this question

edited Jul 29 at 12:46

asked Jul 29 at 12:37

elbarto

1,519523

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 12:46

edited Jul 29 at 12:46

edited Jul 29 at 12:46

asked Jul 29 at 12:37

elbarto

1,519523

asked Jul 29 at 12:37

elbarto

1,519523

asked Jul 29 at 12:37

elbarto

1,519523

1,519523

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Denote by $n_0$ the number of zeroes (of $D$) and by $n_1$ the number of ones, such that the total number of observation $n$ is $n_0 + n_1$. For $beta$ you can see here full derivation, that is compactly can be written as
$$
hatbeta = bary_1 - bary_0.
$$
For $alpha$ you can just plug in the result, namely,
beginalign
hatalpha &= bary_n - barDhatbeta\
&=fracn_!bary_1 + n_0bary_0 n_0 + n_1 - frac n_1 n_0 + n_1 ( bary_1 - bary_0) \
& = frac n_0 + n_1n_0 + n_1 bary_0 \
&=bary_0.
endalign

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: 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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866041%2fhow-to-obtain-expressions-for-coefficients-from-ols-formula%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Denote by $n_0$ the number of zeroes (of $D$) and by $n_1$ the number of ones, such that the total number of observation $n$ is $n_0 + n_1$. For $beta$ you can see here full derivation, that is compactly can be written as
$$
hatbeta = bary_1 - bary_0.
$$
For $alpha$ you can just plug in the result, namely,
beginalign
hatalpha &= bary_n - barDhatbeta\
&=fracn_!bary_1 + n_0bary_0 n_0 + n_1 - frac n_1 n_0 + n_1 ( bary_1 - bary_0) \
& = frac n_0 + n_1n_0 + n_1 bary_0 \
&=bary_0.
endalign

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

add a comment | 

up vote
1
down vote



accepted

Denote by $n_0$ the number of zeroes (of $D$) and by $n_1$ the number of ones, such that the total number of observation $n$ is $n_0 + n_1$. For $beta$ you can see here full derivation, that is compactly can be written as
$$
hatbeta = bary_1 - bary_0.
$$
For $alpha$ you can just plug in the result, namely,
beginalign
hatalpha &= bary_n - barDhatbeta\
&=fracn_!bary_1 + n_0bary_0 n_0 + n_1 - frac n_1 n_0 + n_1 ( bary_1 - bary_0) \
& = frac n_0 + n_1n_0 + n_1 bary_0 \
&=bary_0.
endalign

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Denote by $n_0$ the number of zeroes (of $D$) and by $n_1$ the number of ones, such that the total number of observation $n$ is $n_0 + n_1$. For $beta$ you can see here full derivation, that is compactly can be written as
$$
hatbeta = bary_1 - bary_0.
$$
For $alpha$ you can just plug in the result, namely,
beginalign
hatalpha &= bary_n - barDhatbeta\
&=fracn_!bary_1 + n_0bary_0 n_0 + n_1 - frac n_1 n_0 + n_1 ( bary_1 - bary_0) \
& = frac n_0 + n_1n_0 + n_1 bary_0 \
&=bary_0.
endalign

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

Denote by $n_0$ the number of zeroes (of $D$) and by $n_1$ the number of ones, such that the total number of observation $n$ is $n_0 + n_1$. For $beta$ you can see here full derivation, that is compactly can be written as
$$
hatbeta = bary_1 - bary_0.
$$
For $alpha$ you can just plug in the result, namely,
beginalign
hatalpha &= bary_n - barDhatbeta\
&=fracn_!bary_1 + n_0bary_0 n_0 + n_1 - frac n_1 n_0 + n_1 ( bary_1 - bary_0) \
& = frac n_0 + n_1n_0 + n_1 bary_0 \
&=bary_0.
endalign

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 21:33

V. Vancak

9,7802926

answered Aug 1 at 21:33

V. Vancak

9,7802926

answered Aug 1 at 21:33

V. Vancak

9,7802926

9,7802926

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866041%2fhow-to-obtain-expressions-for-coefficients-from-ols-formula%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email