How to define a cumulative distribution function with variable in it
13 ビュー (過去 30 日間)
古いコメントを表示
Hi everyone, I'm having problem trying to solve for the integral of the cumulative distribution function for my thesis. I want to hold x as variable for F(x) and solve it later to optimize it. The problem is the function normcdf(x,mu,sigma) doesn't allow me to define x as symbol, it require x to be a double value instead. So my question is, is there anyway to define x as "syms x" and then use normcdf(x,mu,sigma)?
With the uniform distribution I write the function straightforward in the program like this:
gamma = 0; n = 150; syms e Q_sc Q_r X Pi_sc Pi_r S_Q S_Q2 x h t; S_Q = Q_sc - int(((e-gamma+n)/(2*n)),e,0,Q_sc - D_r); S_Q2 = Q_r - int(((e-gamma+n)/(2*n)),e,0,Q_r - D_r); Pi_sc = (S_Q*(p-c_m-s+g_r+g_m+(G1*(s_return-l_m-l_r-r)))) + ((s-c_r)*Q_sc) - ((g_r+g_m)*D_r); Diff_Pi_sc = diff(Pi_sc,'Q_sc'); Q_sc = solve(Diff_Pi_sc); Q_sc = vpa(Q_sc);
So instead of writing unifcdf(...), I use ((e-gamma+n)/(2*n)) instead, because I want to keep "e" as variable, I don't want the program to calculate in right away. Is there anyway I can achieve this task using unifcdf function provided by the program?
Thank you very much in advance
0 件のコメント
採用された回答
bym
2013 年 2 月 3 日
f = @(x) normcdf(x,mu,sigma);
2 件のコメント
Tom Lane
2013 年 2 月 4 日
It's not clear to me what you tried. Your post mentions integrating the cdf. I don't know the integral of the cdf, but here's the cdf itself:
>> mu = 0; sigma = 1; f = @(x) normcdf(x,mu,sigma);
I would expect the cdf to be close to 1 over this interval of length 5, so I'd expect the integral to be about 5:
>> integral(f,10,15)
ans =
5
I'd expect the cdf to average to 0.5 over this symmetric interval of length 2, so I'd expect the answer to be about 1:
>> integral(f,-1,1)
ans =
1
その他の回答 (3 件)
Tom Lane
2013 年 2 月 4 日
If you need to use Symbolic Toolbox sym variables, then you may want to use the erfc function in place of normcdf. Consider this:
>> mu = 10; sigma = 2;
>> normcdf(10:13,mu,sigma)
ans =
0.5000 0.6915 0.8413 0.9332
Here's how to get to get that answer using a sym:
>> syms x
>> y = .5*erfc(-(x-mu)/(sigma*sqrt(2)));
>> subs(y,x,10:13)
ans =
[ 1/2, 1 - erfc(2^(1/2)/4)/2, 1 - erfc(2^(1/2)/2)/2, 1 - erfc((3*2^(1/2))/4)/2]
>> double(ans)
ans =
0.5000 0.6915 0.8413 0.9332
Pritee Ray
2015 年 3 月 20 日
Dear Tom Lane, I want to integrate a cdf function, however getting error "??? Undefined function or method 'erfc' for input arguments of type 'sym'." The Matlab code is given below.Please suggest me what to do? sym x f1(i) =int(normcdf(x,mu,sigma),0,1000); f2(i) =int(normcdf(x,mu,sigma),1600,2000); G(i) = double(f1(i)+f2(i));
0 件のコメント
Mohammad Wamique
2020 年 2 月 21 日
編集済み: Mohammad Wamique
2020 年 2 月 21 日
Use an equivalent 'complementary error function' instead. For example:
''normcdf(x)'' is equivalent to "0.5*erfc(-x/(sqrt(2)))''.
Try: [normcdf(.5); 0.5*erfc(-.5/(sqrt(2)))] in command window
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Number Theory についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!