∂L/∂a = -2Σ(yi' - a*(1-exp(-bxi))) ∂L/∂b = -2Σ(yi' - a(1-exp(-bxi))) * (-xiexp(-b*xi)),求a,b

这是一个针对参数a和b的最小二乘法的损失函数的梯度。我们需要将损失函数设为：

L = Σ(yi' - a*(1-exp(-bxi)))^2

其中，yi'表示观测值，xi表示自变量，a和b是需要求解的参数。将L对a求偏导数：

∂L/∂a = -2Σ(yi' - a*(1-exp(-bxi)))

将L对b求偏导数：

∂L/∂b = -2Σ(yi' - a(1-exp(-bxi))) * (-xiexp(-b*xi))

接下来，我们可以使用最小二乘法来求解a和b。最小二乘法的目标是最小化L，即：

min(L) = Σ(yi' - a*(1-exp(-bxi)))^2

我们可以通过梯度下降法来求得a和b的解。梯度下降法的更新公式如下：

a = a - α * ∂L/∂a
b = b - α * ∂L/∂b

其中，α是学习率，控制参数更新的步长。

通过不断地使用上述更新公式，我们可以逐步接近最优解。需要注意的是，我们需要在每次更新参数之前对数据进行归一化处理，以避免参数更新过程中发生数值溢出的情况。

相关问题

用matlab编程Generate a candidate draw θ* from the normal proposal distribution θ ˜∗ = θ ˜(s−1) + z, z ∼ N (0, Σ) 。 where Σ a diagonal matrix, whose diagonal elements are the diagonal elements of the inverse of the negative Hessian matrix of ˜p(θ) evaluated at θ (0), i.e., Σ = diag  − ∂p˜(θ) 2 ∂θ∂θ′ θ=θ0 ! −1  

Sure, here's an example code in MATLAB:

% Define the proposal distribution
sigma = 2; % standard deviation
Sigma = diag([1/sigma^2 1/sigma^2]); % diagonal covariance matrix

% Define the negative Hessian matrix of the target distribution
H = [1 0.5; 0.5 2]; % example Hessian matrix
H_inv = inv(H); % inverse of the Hessian matrix

% Evaluate the diagonal elements of Sigma
Sigma_diag = diag(-H_inv);

% Generate a candidate draw theta_star
if s == 1 % if it's the first iteration, start from an initial value
    theta_s = [0; 0]; % initial value
else % if it's not the first iteration, use the previous value as the starting point
    theta_s = theta_star;
end
z = mvnrnd([0 0], Sigma); % generate z from the proposal distribution
theta_star = theta_s + z; % candidate draw

Note that you need to replace the example Hessian matrix (`H`) and standard deviation (`sigma`) with the actual values from your problem. Also, this code assumes that the target distribution is a bivariate normal distribution, so you may need to modify it if your target distribution is different.