why gradient descent when we can solve linear regression analytically

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走了就别回头了
走了就别回头了 2021-01-29 22:16

what is the benefit of using Gradient Descent in the linear regression space? looks like the we can solve the problem (finding theta0-n that minimum the cost func) with analytic

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  • 2021-01-29 22:46

    You should provide more details about yout problem - what exactly are you asking about - are we talking about linear regression in one or many dimensions? Simple or generalized ones?

    In general, why do people use the GD?

    • it is easy to implement
    • it is very generic optimization technique - even if you change your model to the more general one, you can stil use it

    So what about analytical solutions? Well, we do use them, your claim is simply false here (if we are talking in general), for example the OLS method is a closed form, analytical solution, which is widely used. If you can use the analytical solution, it is affordable computationaly (as sometimes GD is simply cheapier or faster) then you can, and even should - use it.

    Neverlethles this is always a matter of some pros and cons - analytical solutions are strongly connected to the model, so implementing them can be inefficient if you plan to generalize/change your models in the future. They are sometimes less efficient then their numerical approximations, and sometimes there are simply harder to implement. If none of above is true - you should use the analytical solution, and people do it, really.

    To sum up, you rather use GD over analytical solution if:

    • you are considering changes in the model, generalizations, adding some more complex terms/regularization/modifications
    • you need a generic method because you do not know much about the future of the code and the model (you are only one of the developers)
    • analytical solution is more expensive computationaly, and you need efficiency
    • analytical solution requires more memory, which you do not have
    • analytical solution is hard to implement and you need easy, simple code
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  • 2021-01-29 23:02

    I saw a very good answer from https://stats.stackexchange.com/questions/23128/solving-for-regression-parameters-in-closed-form-vs-gradient-descent

    Basically, the reasons are:

    1.For most nonlinear regression problems there is no closed form solution.

    2.Even in linear regression (one of the few cases where a closed form solution is available), it may be impractical to use the formula. The following example shows one way in which this can happen.

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  • 2021-01-29 23:11

    When you use the normal equations for solving the cost function analytically you have to compute:

    enter image description here

    Where X is your matrix of input observations and y your output vector. The problem with this operation is the time complexity of calculating the inverse of a nxn matrix which is O(n^3) and as n increases it can take a very long time to finish.

    When n is low (n < 1000 or n < 10000) you can think of normal equations as the better option for calculation theta, however for greater values Gradient Descent is much more faster, so the only reason is the time :)

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  • 2021-01-29 23:13

    Other reason is that gradient descent is immediately useful when you generalize linear regression, especially if the problem doesn't have a closed-form solution, like for example in Lasso (which adds regularization term consisting on sum of absolute values of weight vector).

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