How to compute the regression coefficient in Matlab with exp and ln?

3 次查看（过去 30 天）
san der
san der 2013-12-26
评论： dpb 2013-12-26
I would like to compute the regression coefficients `a` and `b` for my data using this equation:
y=exp(a * ln(1 - t / h) + b * ln(1 - t / t1))
and this data(example):
t = [1,2,5,4,8,7,5,1,2,5,4,1,2,1,5]
t1 = [1,2,4,4,5,3,7,5,6,8,7,1,2,1,5]
h = [1,2,3,2,9,6,8,3,6,7,4,5,2,1,5]
y = [1,2,1,4,4,6,5,8,5,7,3,1,4,1,5]
but I do not know how to include `exp` and `ln`. PLease help
  4 个评论
显示 2更早的评论
dpb
dpb 2013-12-26
a) Well, 'cuz it's a nonlinear equation, maybe? :)
b) The difficulty in estimating your a and b separately is that since they're an additive term in the exponential they combine as a single argument.
It would help if the data were collected for a design matrix that has those two terms evaluated independently altho I've not taken the time to try to work out a specific design for the case.
san der
san der 2013-12-26
in fact the original equation is like this {ln y = a ln(1-w) + b ln(1-x) where w=t/h and x=t/t1} so I need to find a and b based on initial values of y then I use a and b to find y. but then result will reflect ln(y) not y. what do you think?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Keith Dalbey
Keith Dalbey 2013-12-26
let t, h, t1, and y be column vectors then
if true
% code
G=[log(1-t./h) log(1-t./t1)]; ab=(G'*G)\(G'*log(y))
end
  4 个评论
显示 2更早的评论

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Linear Regression 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by