Ridge Regression with Gradient Descent Converges to OLS estimatesSolving for regression parameters in closed-form vs gradient descentAssumptions of linear regression and gradient descentRidge regression using stochastic gradient descent in PythonModifying ordinary least squares (OLS) in ridge regression to perform transfer learningIs ridge regression useless in high dimensions ($n ll p$)? How can OLS fail to overfit?sklearn Linear Regression vs Batch Gradient DescentWhy is gradient descent so bad at optimizing polynomial regression?How To Choose Step Size (Learning Rate) in Batch Gradient Descent?stochastic gradient descent of ridge regression when regularization parameter is very bigInitialization for ridge regression
How will losing mobility of one hand affect my career as a programmer?
What linear sensor for a keyboard?
What is the grammatical term for “‑ed” words like these?
Constructing Group Divisible Designs - Algorithms?
Has Darkwing Duck ever met Scrooge McDuck?
Find last 3 digits of this monster number
Could the E-bike drivetrain wear down till needing replacement after 400 km?
Should I install hardwood flooring or cabinets first?
If a character with the Alert feat rolls a crit fail on their Perception check, are they surprised?
Two-sided logarithm inequality
Visiting the UK as unmarried couple
Query about absorption line spectra
Varistor? Purpose and principle
Why did the HMS Bounty go back to a time when whales are already rare?
Do Legal Documents Require Signing In Standard Pen Colors?
A Permanent Norse Presence in America
Does the Mind Blank spell prevent the target from being frightened?
Should I stop contributing to retirement accounts?
How do ground effect vehicles perform turns?
Can the Supreme Court overturn an impeachment?
Can I sign legal documents with a smiley face?
Could solar power be utilized and substitute coal in the 19th Century
Is possible to search in vim history?
Reply 'no position' while the job posting is still there
Ridge Regression with Gradient Descent Converges to OLS estimates
Solving for regression parameters in closed-form vs gradient descentAssumptions of linear regression and gradient descentRidge regression using stochastic gradient descent in PythonModifying ordinary least squares (OLS) in ridge regression to perform transfer learningIs ridge regression useless in high dimensions ($n ll p$)? How can OLS fail to overfit?sklearn Linear Regression vs Batch Gradient DescentWhy is gradient descent so bad at optimizing polynomial regression?How To Choose Step Size (Learning Rate) in Batch Gradient Descent?stochastic gradient descent of ridge regression when regularization parameter is very bigInitialization for ridge regression
$begingroup$
I'm implementing a homespun version of Ridge Regression with gradient descent, and to my surprise it always converges to the same answers as OLS, not the closed form of Ridge Regression.
This is true regardless of what size alpha I'm using. I don't know if this is due to something wrong with how I'm setting up the problem, or something about how gradient descent works on different types of data.
The dataset I'm using is the boston housing dataset, which is very small (506 rows), and has a number of correlated variables.
BASIC INTUITION:
I'd love to be corrected if my priors are inaccurate, but this is how I frame the problem.
The cost function for OLS regression:
Where output is the dot product of the feature matrix and the weights for each column + intercept.
The cost function for regression with L2 regularization (ie, Ridge Regression):
Where alpha is the tuning parameter and omega represents the regression coefficient, squared and summed together.
The derivative of the above statement can be written like this:
We then take the result of this expression and multiply it by the learning rate, and adjust each weight appropriately.
To test this out on my own, I made my own python class for Ridge Regression according to the above postulates, and it looks like so:
RIDGE REGRESSION CODE:
class RidgeRegression():
def __init__(self, alpha=1, eta =.0001, random_state=None, n_iter=10000):
self.eta = eta
self.random_state = random_state
self.n_iter = n_iter
self.alpha = alpha
self.w_ = []
def output(self, X):
return X.dot(self.w_[1:]) + self.w_[0]
def fit(self, X, y):
rgen = np.random.RandomState(self.random_state) # random number generator
self.w_ = rgen.normal(0, .01, size= 1 + X.shape[1]) # fill weights w/ near 0 values
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # alpha * sum of squared weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # alpha * sum of weights
for i in range(self.n_iter):
gradient = (y - self.output(X)) + self.l2_gradient # the gradient
self.w_[1:] += (X.T.dot(gradient) * (self.eta)) # update each weight w.r.t. its gradient
self.w_[0] += (y - self.output(X)).sum() * self.eta # update the intercept w.r.t. the gradient
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # adjust regularization term to reflect new values of weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # update derivate of regularization term to be used in gradient
This looks right to me, but I'm not sure how to interpret the fact that the results are always the same as OLS, vs what I'd get using the closed form of Ridge Regression.
EDIT:
Even if I set the value of alpha to something very large, like 1000, the results do not change.
CODE:
from sklearn.linear_model import Ridge
r = RidgeRegression(alpha=1000, n_iter=5000, eta=0.0001) # my algorithm
r.fit(X, y)
rreg = Ridge(alpha=1000) # scikit learn's ridge algorithm
rreg.fit(X, y)
ols_coeffs = np.linalg.inv(X.T @ X) @ (X.T @ y) # ols algorithm
If we compare these, we can see the ridge coefficients in my homespun version match the OLS estimates almost exactly:
pd.DataFrame(
'Variable' : X.columns,
'gd_Weight' : r.w_[1:], # coefficients for ridge regression w/ gd
'cf_Weight' : ridge_coeffs, # closed form version of ridge
'ols_weight' : coeffs # regular OLS, closed form
)
Which gives the following dataframe:
They all have roughly the same intercept of 22.5.
regression python regularization ridge-regression numpy
$endgroup$
add a comment |
$begingroup$
I'm implementing a homespun version of Ridge Regression with gradient descent, and to my surprise it always converges to the same answers as OLS, not the closed form of Ridge Regression.
This is true regardless of what size alpha I'm using. I don't know if this is due to something wrong with how I'm setting up the problem, or something about how gradient descent works on different types of data.
The dataset I'm using is the boston housing dataset, which is very small (506 rows), and has a number of correlated variables.
BASIC INTUITION:
I'd love to be corrected if my priors are inaccurate, but this is how I frame the problem.
The cost function for OLS regression:
Where output is the dot product of the feature matrix and the weights for each column + intercept.
The cost function for regression with L2 regularization (ie, Ridge Regression):
Where alpha is the tuning parameter and omega represents the regression coefficient, squared and summed together.
The derivative of the above statement can be written like this:
We then take the result of this expression and multiply it by the learning rate, and adjust each weight appropriately.
To test this out on my own, I made my own python class for Ridge Regression according to the above postulates, and it looks like so:
RIDGE REGRESSION CODE:
class RidgeRegression():
def __init__(self, alpha=1, eta =.0001, random_state=None, n_iter=10000):
self.eta = eta
self.random_state = random_state
self.n_iter = n_iter
self.alpha = alpha
self.w_ = []
def output(self, X):
return X.dot(self.w_[1:]) + self.w_[0]
def fit(self, X, y):
rgen = np.random.RandomState(self.random_state) # random number generator
self.w_ = rgen.normal(0, .01, size= 1 + X.shape[1]) # fill weights w/ near 0 values
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # alpha * sum of squared weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # alpha * sum of weights
for i in range(self.n_iter):
gradient = (y - self.output(X)) + self.l2_gradient # the gradient
self.w_[1:] += (X.T.dot(gradient) * (self.eta)) # update each weight w.r.t. its gradient
self.w_[0] += (y - self.output(X)).sum() * self.eta # update the intercept w.r.t. the gradient
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # adjust regularization term to reflect new values of weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # update derivate of regularization term to be used in gradient
This looks right to me, but I'm not sure how to interpret the fact that the results are always the same as OLS, vs what I'd get using the closed form of Ridge Regression.
EDIT:
Even if I set the value of alpha to something very large, like 1000, the results do not change.
CODE:
from sklearn.linear_model import Ridge
r = RidgeRegression(alpha=1000, n_iter=5000, eta=0.0001) # my algorithm
r.fit(X, y)
rreg = Ridge(alpha=1000) # scikit learn's ridge algorithm
rreg.fit(X, y)
ols_coeffs = np.linalg.inv(X.T @ X) @ (X.T @ y) # ols algorithm
If we compare these, we can see the ridge coefficients in my homespun version match the OLS estimates almost exactly:
pd.DataFrame(
'Variable' : X.columns,
'gd_Weight' : r.w_[1:], # coefficients for ridge regression w/ gd
'cf_Weight' : ridge_coeffs, # closed form version of ridge
'ols_weight' : coeffs # regular OLS, closed form
)
Which gives the following dataframe:
They all have roughly the same intercept of 22.5.
regression python regularization ridge-regression numpy
$endgroup$
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago
add a comment |
$begingroup$
I'm implementing a homespun version of Ridge Regression with gradient descent, and to my surprise it always converges to the same answers as OLS, not the closed form of Ridge Regression.
This is true regardless of what size alpha I'm using. I don't know if this is due to something wrong with how I'm setting up the problem, or something about how gradient descent works on different types of data.
The dataset I'm using is the boston housing dataset, which is very small (506 rows), and has a number of correlated variables.
BASIC INTUITION:
I'd love to be corrected if my priors are inaccurate, but this is how I frame the problem.
The cost function for OLS regression:
Where output is the dot product of the feature matrix and the weights for each column + intercept.
The cost function for regression with L2 regularization (ie, Ridge Regression):
Where alpha is the tuning parameter and omega represents the regression coefficient, squared and summed together.
The derivative of the above statement can be written like this:
We then take the result of this expression and multiply it by the learning rate, and adjust each weight appropriately.
To test this out on my own, I made my own python class for Ridge Regression according to the above postulates, and it looks like so:
RIDGE REGRESSION CODE:
class RidgeRegression():
def __init__(self, alpha=1, eta =.0001, random_state=None, n_iter=10000):
self.eta = eta
self.random_state = random_state
self.n_iter = n_iter
self.alpha = alpha
self.w_ = []
def output(self, X):
return X.dot(self.w_[1:]) + self.w_[0]
def fit(self, X, y):
rgen = np.random.RandomState(self.random_state) # random number generator
self.w_ = rgen.normal(0, .01, size= 1 + X.shape[1]) # fill weights w/ near 0 values
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # alpha * sum of squared weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # alpha * sum of weights
for i in range(self.n_iter):
gradient = (y - self.output(X)) + self.l2_gradient # the gradient
self.w_[1:] += (X.T.dot(gradient) * (self.eta)) # update each weight w.r.t. its gradient
self.w_[0] += (y - self.output(X)).sum() * self.eta # update the intercept w.r.t. the gradient
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # adjust regularization term to reflect new values of weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # update derivate of regularization term to be used in gradient
This looks right to me, but I'm not sure how to interpret the fact that the results are always the same as OLS, vs what I'd get using the closed form of Ridge Regression.
EDIT:
Even if I set the value of alpha to something very large, like 1000, the results do not change.
CODE:
from sklearn.linear_model import Ridge
r = RidgeRegression(alpha=1000, n_iter=5000, eta=0.0001) # my algorithm
r.fit(X, y)
rreg = Ridge(alpha=1000) # scikit learn's ridge algorithm
rreg.fit(X, y)
ols_coeffs = np.linalg.inv(X.T @ X) @ (X.T @ y) # ols algorithm
If we compare these, we can see the ridge coefficients in my homespun version match the OLS estimates almost exactly:
pd.DataFrame(
'Variable' : X.columns,
'gd_Weight' : r.w_[1:], # coefficients for ridge regression w/ gd
'cf_Weight' : ridge_coeffs, # closed form version of ridge
'ols_weight' : coeffs # regular OLS, closed form
)
Which gives the following dataframe:
They all have roughly the same intercept of 22.5.
regression python regularization ridge-regression numpy
$endgroup$
I'm implementing a homespun version of Ridge Regression with gradient descent, and to my surprise it always converges to the same answers as OLS, not the closed form of Ridge Regression.
This is true regardless of what size alpha I'm using. I don't know if this is due to something wrong with how I'm setting up the problem, or something about how gradient descent works on different types of data.
The dataset I'm using is the boston housing dataset, which is very small (506 rows), and has a number of correlated variables.
BASIC INTUITION:
I'd love to be corrected if my priors are inaccurate, but this is how I frame the problem.
The cost function for OLS regression:
Where output is the dot product of the feature matrix and the weights for each column + intercept.
The cost function for regression with L2 regularization (ie, Ridge Regression):
Where alpha is the tuning parameter and omega represents the regression coefficient, squared and summed together.
The derivative of the above statement can be written like this:
We then take the result of this expression and multiply it by the learning rate, and adjust each weight appropriately.
To test this out on my own, I made my own python class for Ridge Regression according to the above postulates, and it looks like so:
RIDGE REGRESSION CODE:
class RidgeRegression():
def __init__(self, alpha=1, eta =.0001, random_state=None, n_iter=10000):
self.eta = eta
self.random_state = random_state
self.n_iter = n_iter
self.alpha = alpha
self.w_ = []
def output(self, X):
return X.dot(self.w_[1:]) + self.w_[0]
def fit(self, X, y):
rgen = np.random.RandomState(self.random_state) # random number generator
self.w_ = rgen.normal(0, .01, size= 1 + X.shape[1]) # fill weights w/ near 0 values
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # alpha * sum of squared weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # alpha * sum of weights
for i in range(self.n_iter):
gradient = (y - self.output(X)) + self.l2_gradient # the gradient
self.w_[1:] += (X.T.dot(gradient) * (self.eta)) # update each weight w.r.t. its gradient
self.w_[0] += (y - self.output(X)).sum() * self.eta # update the intercept w.r.t. the gradient
self.l2_regularization = self.alpha * self.w_[1:].dot(self.w_[1:]) # adjust regularization term to reflect new values of weights
self.l2_gradient = self.alpha * np.sum(self.w_[1:]) # update derivate of regularization term to be used in gradient
This looks right to me, but I'm not sure how to interpret the fact that the results are always the same as OLS, vs what I'd get using the closed form of Ridge Regression.
EDIT:
Even if I set the value of alpha to something very large, like 1000, the results do not change.
CODE:
from sklearn.linear_model import Ridge
r = RidgeRegression(alpha=1000, n_iter=5000, eta=0.0001) # my algorithm
r.fit(X, y)
rreg = Ridge(alpha=1000) # scikit learn's ridge algorithm
rreg.fit(X, y)
ols_coeffs = np.linalg.inv(X.T @ X) @ (X.T @ y) # ols algorithm
If we compare these, we can see the ridge coefficients in my homespun version match the OLS estimates almost exactly:
pd.DataFrame(
'Variable' : X.columns,
'gd_Weight' : r.w_[1:], # coefficients for ridge regression w/ gd
'cf_Weight' : ridge_coeffs, # closed form version of ridge
'ols_weight' : coeffs # regular OLS, closed form
)
Which gives the following dataframe:
They all have roughly the same intercept of 22.5.
regression python regularization ridge-regression numpy
regression python regularization ridge-regression numpy
edited 2 hours ago
Jonathan Bechtel
asked 3 hours ago
Jonathan BechtelJonathan Bechtel
1234
1234
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago
add a comment |
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not very well versed in Python, but if I interpret your code correctly then X.T.dot(gradient) takes the gradient variable you computed on the previous line, and computes its dot product with $X$. This doesn't seem right since 'gradient' includes the "L2-gradient" which shouldn't be multiplied by $X$. Only the residual (y-self.output(X)) should be in that product. You want to add the L2-gradient only afterwards, and then multiply the result by eta.
Also, the L2-gradient shouldn't sum over the $w$'s (remember the gradient should be vector-valued since you're taking the derivative of a scalar-valued function w.r.t a vector). Together these errors probably explain the confusing results you're getting, although I would have expected with an incorrect gradient like that the output would actually just be garbage rather than the OLS-solution, so there is probably something subtle happening that I'm not seeing.
(Note that your mathematical expressions for the gradient also aren't quite right, but there you actually omitted the pre-multiplication of the residuals with $X^T$. And you also again have a sum over $w$'s in the L2-gradient which shouldn't be there.)
Going forward I would recommend first checking your implementation of each individual part of the cost function and gradient and satisfy yourself that they are correct, before running gradient descent.
$endgroup$
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: "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%2f399230%2fridge-regression-with-gradient-descent-converges-to-ols-estimates%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm not very well versed in Python, but if I interpret your code correctly then X.T.dot(gradient) takes the gradient variable you computed on the previous line, and computes its dot product with $X$. This doesn't seem right since 'gradient' includes the "L2-gradient" which shouldn't be multiplied by $X$. Only the residual (y-self.output(X)) should be in that product. You want to add the L2-gradient only afterwards, and then multiply the result by eta.
Also, the L2-gradient shouldn't sum over the $w$'s (remember the gradient should be vector-valued since you're taking the derivative of a scalar-valued function w.r.t a vector). Together these errors probably explain the confusing results you're getting, although I would have expected with an incorrect gradient like that the output would actually just be garbage rather than the OLS-solution, so there is probably something subtle happening that I'm not seeing.
(Note that your mathematical expressions for the gradient also aren't quite right, but there you actually omitted the pre-multiplication of the residuals with $X^T$. And you also again have a sum over $w$'s in the L2-gradient which shouldn't be there.)
Going forward I would recommend first checking your implementation of each individual part of the cost function and gradient and satisfy yourself that they are correct, before running gradient descent.
$endgroup$
add a comment |
$begingroup$
I'm not very well versed in Python, but if I interpret your code correctly then X.T.dot(gradient) takes the gradient variable you computed on the previous line, and computes its dot product with $X$. This doesn't seem right since 'gradient' includes the "L2-gradient" which shouldn't be multiplied by $X$. Only the residual (y-self.output(X)) should be in that product. You want to add the L2-gradient only afterwards, and then multiply the result by eta.
Also, the L2-gradient shouldn't sum over the $w$'s (remember the gradient should be vector-valued since you're taking the derivative of a scalar-valued function w.r.t a vector). Together these errors probably explain the confusing results you're getting, although I would have expected with an incorrect gradient like that the output would actually just be garbage rather than the OLS-solution, so there is probably something subtle happening that I'm not seeing.
(Note that your mathematical expressions for the gradient also aren't quite right, but there you actually omitted the pre-multiplication of the residuals with $X^T$. And you also again have a sum over $w$'s in the L2-gradient which shouldn't be there.)
Going forward I would recommend first checking your implementation of each individual part of the cost function and gradient and satisfy yourself that they are correct, before running gradient descent.
$endgroup$
add a comment |
$begingroup$
I'm not very well versed in Python, but if I interpret your code correctly then X.T.dot(gradient) takes the gradient variable you computed on the previous line, and computes its dot product with $X$. This doesn't seem right since 'gradient' includes the "L2-gradient" which shouldn't be multiplied by $X$. Only the residual (y-self.output(X)) should be in that product. You want to add the L2-gradient only afterwards, and then multiply the result by eta.
Also, the L2-gradient shouldn't sum over the $w$'s (remember the gradient should be vector-valued since you're taking the derivative of a scalar-valued function w.r.t a vector). Together these errors probably explain the confusing results you're getting, although I would have expected with an incorrect gradient like that the output would actually just be garbage rather than the OLS-solution, so there is probably something subtle happening that I'm not seeing.
(Note that your mathematical expressions for the gradient also aren't quite right, but there you actually omitted the pre-multiplication of the residuals with $X^T$. And you also again have a sum over $w$'s in the L2-gradient which shouldn't be there.)
Going forward I would recommend first checking your implementation of each individual part of the cost function and gradient and satisfy yourself that they are correct, before running gradient descent.
$endgroup$
I'm not very well versed in Python, but if I interpret your code correctly then X.T.dot(gradient) takes the gradient variable you computed on the previous line, and computes its dot product with $X$. This doesn't seem right since 'gradient' includes the "L2-gradient" which shouldn't be multiplied by $X$. Only the residual (y-self.output(X)) should be in that product. You want to add the L2-gradient only afterwards, and then multiply the result by eta.
Also, the L2-gradient shouldn't sum over the $w$'s (remember the gradient should be vector-valued since you're taking the derivative of a scalar-valued function w.r.t a vector). Together these errors probably explain the confusing results you're getting, although I would have expected with an incorrect gradient like that the output would actually just be garbage rather than the OLS-solution, so there is probably something subtle happening that I'm not seeing.
(Note that your mathematical expressions for the gradient also aren't quite right, but there you actually omitted the pre-multiplication of the residuals with $X^T$. And you also again have a sum over $w$'s in the L2-gradient which shouldn't be there.)
Going forward I would recommend first checking your implementation of each individual part of the cost function and gradient and satisfy yourself that they are correct, before running gradient descent.
answered 1 hour ago
Ruben van BergenRuben van Bergen
3,8141923
3,8141923
add a comment |
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%2f399230%2fridge-regression-with-gradient-descent-converges-to-ols-estimates%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
$begingroup$
By "the weight of each variable in the dataset", do you mean the regression coefficient?
$endgroup$
– beta1_equals_beta2
3 hours ago
$begingroup$
Are you sure alpha is not zero? Your code doesn't show what it is.
$endgroup$
– Mark L. Stone
2 hours ago
$begingroup$
@beta1_equals_beta2 - yes. Changed to make it more clear. I also added some sample code to demonstrate what happens when alpha is set to 1000.
$endgroup$
– Jonathan Bechtel
2 hours ago