Silly error somewhere in my code

Question

Mark Whirdy 2012-7-4

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/42747-silly-error-somewhere-in-my-code

Would appreciate someone's help in spotting a silly error (or bad math) somewhere here. Its difficult to explain precisely without being able to attach some equations but basically I'm trying to establish equivalents between q5 and q6 below. But instead I get 0.21 & 0.37 and can't see my own error.

q5 is the integral of a product of a quadratic polynomial and the norm pdf q6 is the normal moment-generating-function in t=2,1,0 where each moment is multiplied by its respective polynomial coefficient h1,h2,h3. I can't see why these are not equivalent. If the problem isn't clear, please email me at mark.whirdy@gmail.com and I will send the equations which should help(can't attach a doc here unfortunately).

mu = 0; sigma = 1;
h1 = -0.003;
h2 = 0.10;
h3 = 0.21;
a = -10^10; b = 10^10;  
g = @(mu,sigma,h1,h2,h3)(quad(@(x)((h1*x.^2 + h2*x.^1 + h3*x.^0).*normpdf(x,mu,sigma)), a, b));
q5 = g(mu,sigma,h1,h2,h3)
t = (2:-1:0)';
h = [h1, h2, h3];
q6 = h * exp(mu*t + 0.5.*t.*sigma^2) % moment generating function

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

AC 2012-7-4

在 MATLAB Online 中打开

Not the answer but a first error in the code is a square missing in q6. If I remember correctly, the moment generating function of a gaussian is

exp(mu*t + 0.5.*t.^2*sigma^2)

That doesn't solve your problem unfortunately, but it's a start!

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

AC 2012-7-4

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/42747-silly-error-somewhere-in-my-code#answer_52576

编辑：AC 2012-7-4

在 MATLAB Online 中打开

This time a real answer: it's a math problem. If that's ok, I'll answer with some latex code.

Let f be the norm(0,1) pdf

Q5= \int (h_1 x^2+h_2 x +h_3) f(x) dx
  = h_1 E(X^2) + h_2 E(X) + h_3

Editing: the following was not correct (wrong definition of moment generating function)

*********************************

Now, the moment generating function M(t) gives:

M(0)=1
M(1)=1+E(X)
M(2)=1+E(X)+2E(X^2)

So Q6 would look like this:

Q6=h_1 M(2) + h_2 M(1) + h_3 M(0)
  =h_1 (1+E(X)+2E(X^2)) + h_2 (1+E(X)) + h_3
************************************
And is therefore not equal to Q5.

Convinced?

3 个评论
显示 1更早的评论隐藏 1更早的评论

Jan 2012-7-5

Mark Whirdy has written:

Hi Anne-Claire

You're absolutely right - its not a coding error, its a much more basic error in reasoning the problem.

Thanks for your help here.

Mark

[EDITED, Jan, please post comments as comments, not as answers]

Mark Whirdy 2012-12-19

编辑：Mark Whirdy 2012-12-19

Jan,

It was posted as an Answer intentionally, to show that it was in fact answered (the answer is its a reasoning problem & not a coding problem), and that Anne-Claire provided that answer (by email). I accepted her answer above only to give her the stat. This is reasonable given the circumstances. Thanks

请先登录，再进行评论。