Convex optimization and cost function

  • What is a convex function in cost function?

    The cost function is convex if its Second Order Derivative is positive semidefinite (i.e. ≥0 ).
    But this definition depends on the function with respect to which you take the derivative..

  • What is advantage of convex cost function over non convex function?

    A convex function: given any two points on the curve there will be no intersection with any other points, for non convex function there will be at least one intersection.
    In terms of cost function with a convex type you are always guaranteed to have a global minimum, whilst for a non convex only local minima..

  • What is cost function in optimization?

    The cost function is the sum of variables that drive the optimization of a schedule.
    For each activity in the schedule, the Optimizer minimizes the value of the cost function while optimizing the schedule.
    The cost function indirectly calculates the monetary cost of completing the activities in a schedule..

  • The cost function of neural network is J(W,b), and it is claimed to be non-convex.
  • The cost function of neural network is J(W,b), and it is claimed to be non-convex.Jul 9, 2014
Sep 9, 2020A function f is said to be a convex function if the seconder-order derivative of that function is greater than or equal to 0. Condition for 
Convexity in gradient descent optimization Our goal is to minimize this cost function in order to improve the accuracy of the model. MSE is a convex function (it is differentiable twice). This means there is no local minimum, but only the global minimum. Thus gradient descent would converge to the global minimum.

Is cost a convex function?

The need is for J(θ) to be convex (as a function of θ ), so you need Cost(hθ(x), y) to be a convex function of θ, not x

Note that the function inside the sigmoid is linear in θ

Consider a twice differentiable function of one variable f(z)

If the second derivative of f(z) is (always) non-negative, then f(z) is convex

So consider the function

Here I will prove the below loss function is a convex function. \begin{equation} L(\theta, \theta_0) = \sum_{i=1}^N \left( - y^i \log(\sigma(\thet...Best answer · 18

You are looking at the wrong variable. The need is for $J(\theta)$ to be convex (as a function of $\theta$), so you need $Cost(h_{\theta}(x), y)$ t...12

Consider a twice differentiable function of one variable $f(z)$. If the second derivative of $f(z)$ is (always) non-negative, then $f(z)$ is convex...2

,Cost function is also a kind of convex function. The best fit of the hypothesis appears on the minimal global optimum of the convex. Solving the global optimum in the convex function is called convex optimization.

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