Why divide regularization factor by size of dataset?
$begingroup$
Suppose I'm trying to minimize a cost function:
$$ J(theta) = frac {1} {2m} sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 $$
Adding regularization, as seen here, we get:
$$ J(theta) = frac {1} {2m} [sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 + lambda sum_{i = 1}^n theta_i^2] $$
My question is: why is the regularization factor also divided by $2m$?
regression regularization linear-regression
$endgroup$
add a comment |
$begingroup$
Suppose I'm trying to minimize a cost function:
$$ J(theta) = frac {1} {2m} sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 $$
Adding regularization, as seen here, we get:
$$ J(theta) = frac {1} {2m} [sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 + lambda sum_{i = 1}^n theta_i^2] $$
My question is: why is the regularization factor also divided by $2m$?
regression regularization linear-regression
$endgroup$
1
$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08
add a comment |
$begingroup$
Suppose I'm trying to minimize a cost function:
$$ J(theta) = frac {1} {2m} sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 $$
Adding regularization, as seen here, we get:
$$ J(theta) = frac {1} {2m} [sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 + lambda sum_{i = 1}^n theta_i^2] $$
My question is: why is the regularization factor also divided by $2m$?
regression regularization linear-regression
$endgroup$
Suppose I'm trying to minimize a cost function:
$$ J(theta) = frac {1} {2m} sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 $$
Adding regularization, as seen here, we get:
$$ J(theta) = frac {1} {2m} [sum _{i = 1}^ m (h_theta (x^{(i)}) - y^{(i)})^2 + lambda sum_{i = 1}^n theta_i^2] $$
My question is: why is the regularization factor also divided by $2m$?
regression regularization linear-regression
regression regularization linear-regression
asked Dec 20 '15 at 7:59
CheshieCheshie
18118
18118
1
$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08
add a comment |
1
$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08
1
1
$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08
$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08
add a comment |
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$begingroup$
Such division is notational only. You can divide $lambda$ by $2m$ and make $lambda$ larger, or do not divide $lambda$ by $2m$ and make $lambda$ smaller. It doesn't matter
$endgroup$
– Yining Wang
Jun 20 '17 at 1:08