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Gradient decent optimization
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I am trying to understand gradient decent optimization in ML algorithms. I understand that there's a cost function - where the aim is to minimize the error y^-y. Now in a scenario where weights w1, w2 are being optimized to give the minimum error, When the optimization does occur through partial derivatives, in each turn does it change both w1 and w2 or is it a combination like in few iterations only w1 is changed and when w1 isn't reducing the error more, the derivative starts with w2 - to reach the local minima? The application can be a linear regression model or a logistic regression model or Boosting algorithms.
optimization gradient-descent
asked 1 hour ago
Pb89
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up vote
1
down vote
favorite
I am trying to understand gradient decent optimization in ML algorithms. I understand that there's a cost function - where the aim is to minimize the error y^-y. Now in a scenario where weights w1, w2 are being optimized to give the minimum error, When the optimization does occur through partial derivatives, in each turn does it change both w1 and w2 or is it a combination like in few iterations only w1 is changed and when w1 isn't reducing the error more, the derivative starts with w2 - to reach the local minima? The application can be a linear regression model or a logistic regression model or Boosting algorithms.
optimization gradient-descent
asked 1 hour ago
Pb89
539
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to understand gradient decent optimization in ML algorithms. I understand that there's a cost function - where the aim is to minimize the error y^-y. Now in a scenario where weights w1, w2 are being optimized to give the minimum error, When the optimization does occur through partial derivatives, in each turn does it change both w1 and w2 or is it a combination like in few iterations only w1 is changed and when w1 isn't reducing the error more, the derivative starts with w2 - to reach the local minima? The application can be a linear regression model or a logistic regression model or Boosting algorithms.
optimization gradient-descent
asked 1 hour ago
Pb89
539
I am trying to understand gradient decent optimization in ML algorithms. I understand that there's a cost function - where the aim is to minimize the error y^-y. Now in a scenario where weights w1, w2 are being optimized to give the minimum error, When the optimization does occur through partial derivatives, in each turn does it change both w1 and w2 or is it a combination like in few iterations only w1 is changed and when w1 isn't reducing the error more, the derivative starts with w2 - to reach the local minima? The application can be a linear regression model or a logistic regression model or Boosting algorithms.
optimization gradient-descent
optimization gradient-descent
asked 1 hour ago
Pb89
539
asked 1 hour ago
Pb89
539
asked 1 hour ago
Pb89
539
asked 1 hour ago
Pb89
539
asked 1 hour ago
Pb89
539
539
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
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Gradient decent is applied to both w1 and w2 for each iteration. During each iteration, the parameters updated according to the gradients. They would likely have different partial derivative.
Check https://spin.atomicobject.com/2014/06/24/gradient-descent-linear-regression.
answered 53 mins ago
SmallChess
5,44341837
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up vote
1
down vote
Gradient descent updates all parameters at each step. You can see this in the update rule:
$$
w^(t+1)=w^(t) - etanabla fleft(w^(t)right).
$$
Since the gradient of the loss function $nabla f(w)$ is vector-valued with dimension matching that of $w$, all parameters are updated at each iteration.
answered 39 mins ago
Sycorax
36k694180
So the algorithm may try different combinations like increasew1, decreasew2based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?
– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done tow1andw2or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?
– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Gradient decent is applied to both w1 and w2 for each iteration. During each iteration, the parameters updated according to the gradients. They would likely have different partial derivative.
Check https://spin.atomicobject.com/2014/06/24/gradient-descent-linear-regression.
answered 53 mins ago
SmallChess
5,44341837
add a comment |Â
up vote
1
down vote
Gradient decent is applied to both w1 and w2 for each iteration. During each iteration, the parameters updated according to the gradients. They would likely have different partial derivative.
Check https://spin.atomicobject.com/2014/06/24/gradient-descent-linear-regression.
answered 53 mins ago
SmallChess
5,44341837
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Gradient decent is applied to both w1 and w2 for each iteration. During each iteration, the parameters updated according to the gradients. They would likely have different partial derivative.
Check https://spin.atomicobject.com/2014/06/24/gradient-descent-linear-regression.
answered 53 mins ago
SmallChess
5,44341837
Gradient decent is applied to both w1 and w2 for each iteration. During each iteration, the parameters updated according to the gradients. They would likely have different partial derivative.
Check https://spin.atomicobject.com/2014/06/24/gradient-descent-linear-regression.
answered 53 mins ago
SmallChess
5,44341837
answered 53 mins ago
SmallChess
5,44341837
answered 53 mins ago
SmallChess
5,44341837
answered 53 mins ago
SmallChess
5,44341837
5,44341837
add a comment |Â
add a comment |Â
up vote
1
down vote
Gradient descent updates all parameters at each step. You can see this in the update rule:
$$
w^(t+1)=w^(t) - etanabla fleft(w^(t)right).
$$
Since the gradient of the loss function $nabla f(w)$ is vector-valued with dimension matching that of $w$, all parameters are updated at each iteration.
answered 39 mins ago
Sycorax
36k694180
So the algorithm may try different combinations like increasew1, decreasew2based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?
– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done tow1andw2or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?
– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
add a comment |Â
up vote
1
down vote
Gradient descent updates all parameters at each step. You can see this in the update rule:
$$
w^(t+1)=w^(t) - etanabla fleft(w^(t)right).
$$
Since the gradient of the loss function $nabla f(w)$ is vector-valued with dimension matching that of $w$, all parameters are updated at each iteration.
answered 39 mins ago
Sycorax
36k694180
So the algorithm may try different combinations like increasew1, decreasew2based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?
– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done tow1andw2or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?
– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Gradient descent updates all parameters at each step. You can see this in the update rule:
$$
w^(t+1)=w^(t) - etanabla fleft(w^(t)right).
$$
Since the gradient of the loss function $nabla f(w)$ is vector-valued with dimension matching that of $w$, all parameters are updated at each iteration.
answered 39 mins ago
Sycorax
36k694180
Gradient descent updates all parameters at each step. You can see this in the update rule:
$$
w^(t+1)=w^(t) - etanabla fleft(w^(t)right).
$$
Since the gradient of the loss function $nabla f(w)$ is vector-valued with dimension matching that of $w$, all parameters are updated at each iteration.
answered 39 mins ago
Sycorax
36k694180
answered 39 mins ago
Sycorax
36k694180
answered 39 mins ago
Sycorax
36k694180
answered 39 mins ago
Sycorax
36k694180
36k694180
So the algorithm may try different combinations like increasew1, decreasew2based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?
– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done tow1andw2or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?
– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
add a comment |Â
So the algorithm may try different combinations like increasew1, decreasew2based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?
– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done tow1andw2or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?
– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
So the algorithm may try different combinations like increase
w1 , decrease w2 based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?– Pb89
22 mins ago
So the algorithm may try different combinations like increase
w1 , decrease w2 based on the direction from partial derivative to reach local minima and just to confirm the algorithm will not necessarily give the global minima always?– Pb89
22 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done to
w1 and w2 or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?– Pb89
19 mins ago
and does the partial derivative also help to explain how much increase or decrease has to be done to
w1 and w2 or that is done by learning rate/shrinkage while partial derivative only provides direction of descent?– Pb89
19 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
The gradient is a vector, so it gives a direction and a magnitude. A vector can be arbitrarily rescaled by a positive scalar and it will have the same direction, but the rescaling will change its magnitude.
– Sycorax
9 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
If magnitude is also given by the gradient then what is the role of shrinkage or learning rate?
– Pb89
7 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
The learning rate rescales the gradient. Suppose $nabla f(x)$ has a large norm (length). Taking a large step will move you to a distant part of the loss surface (jumping from one mountain to another). The core justification of gradient descent is that it's a linear approximation in the vicinity of $w^(t)$. That approximation is always inexact, but it's probably worse the farther away you move -- hence, you want to take small steps, so you use some small $eta$, where 'small' is entirely problem-specific.
– Sycorax
4 mins ago
add a comment |Â
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