Inverse Cumulative distribution function

Hi,

if i have the following CDF :

F(X)=1/1+b(1+b-(1+b+qx/b))(1+x/b)^-q

How can I find x=inverse(F(x))

with my best reguards

Answer 1

[Jeff Miller]

Suppose you have a function called thisCDF which computes the CDF for your distribution. Then define

function x = inverseF(p)
    inverseFerr = @(x) thisCDF(x) - p;
    x = fzero(inverseFerr,0.5);  % replace 0.5 with a starting guess for x
end

Now

x = inverseF(0.5)  % compute the median, etc

[Bashar AlHalaq]

thank you for replay,

but i want generate x's data vector as x=[x1,x2,x3,.....,xn] as sample sizes n for exapmle

if i have F(x)=u=1-exp(-ax)

then x=-ln(1-u)/a

where u is rand number

with my respect

[Jeff Miller]

I don't see the problem. Generate as many u's as you would like, uniformly between 0 and 1, and then call inverseF for each value of u to get the corresponding value of x. In the end you will have your sample of x's.

[Bashar AlHalaq]

thank you very mach

[Bashar AlHalaq]

I wrote the following code for estimate the parameters b and q:

clc;

clear;

b=0.6;

q=1.5;

T=1;

n=150;

for t=1:T

for i=1:n

x(i)=inverseF(b,q);

xx=sort(x);

pdf=q.*(b.^2+q*xx-xx)./((b.^3+b.^2).*(1+(xx./b)).^(1+q));

F=((1+b-(1+b+q.*xx/b).*(1+xx/b).^-q))/(1+b);

R=1-((1+b-(1+b+(q.*xx./b)).*(1+(xx./b)).^-q))/(1+b);

end

%% 1-MLE

[par1 f]=fsolve(@(S) MLE(xx,S),[1 1]);

b_mle(t)=par1(1); q_mle(t)=par1(2);

f_mle=q_mle.*(b_mle.^2+q_mle*xx-xx)./((b_mle.^3+b_mle.^2).*(1+(xx./b_mle)).^(1+q_mle));

F_mle=((1+b_mle-(1+b_mle+q_mle.*xx/b_mle).*(1+xx/b_mle).^-q_mle))/(1+b_mle);

R_mle=1-(((1+b_mle-(1+b_mle+q_mle.*xx/b_mle).*(1+xx/b_mle).^-q_mle))/(1+b_mle));

%% 2-MOM

[par2 f]=fsolve(@(S) MOM_Method(xx,S),[b q]);

b_mom(t)=par2(1); q_mom(t)=par2(2);

f_mom=q_mom.*(b_mom.^2+q_mom*xx-xx)./((b_mom.^3+b_mom.^2).*(1+(xx./b_mom)).^(1+q_mom));

F_mom=((1+b_mom-(1+b_mom+q_mom.*xx/b_mom).*(1+xx/b_mom).^-q_mom))/(1+b_mom);

R_mom=1-(((1+b_mom-(1+b_mom+q_mom.*xx/b_mom).*(1+xx/b_mom).^-q_mom))/(1+b_mom));

end

%% Plot PDF

figure(1)

plot(pdf,'linewidth',2)

hold on;

plot(f_mle,'linewidth',2)

plot(f_mom,'linewidth',2)

legend('f','f_mle','f_mom')

xlabel('x')

ylabel('f(x)')

title('PDF for DTWPD distrbution')

%% Plot CDF

figure(2)

plot(F,'linewidth',2)

hold on;

plot(F_mle,'linewidth',2)

plot(F_mom,'linewidth',2)

legend('F','F_mle','F_mom')

xlabel('x')

ylabel('F(x)')

title('CDF for DTWPD distrbution')

%% Plot R

figure(3)

plot(R,'linewidth',2)

hold on;

plot(R_mle,'linewidth',2)

plot(R_mom,'linewidth',2)

legend('R','R_mle','R_mom')

xlabel('t')

ylabel('R(t)')

title('Reliability for mixlo distrbution')

with sub function:

1) for generate data:

function x = inverseF(b,q)

u=rand;

inverseFerr = @(x) ((1+b-(1+b+q*x/b)*(1+x/b)^-q))/(1+b);

x = inverseFerr(u);

end

2) mle

function [F]=MLE(xx,S)

n=length(xx);

b=S(1);

q=S(2);

k1=(n*(3*b^2+2*b));

k2=(1+q);

k3=(n/q);

k4=(2*b);

s1=0; s2=0; s3=0; s4=0;

for i=1:n

s1=s1+(1/(b^2+q*xx(i)-xx(i)));

s2=s2+((b^-2)/((1+xx(i))/b));

s3=s3+xx(i)/(b^2+q*xx(i)-xx(i));

s4=s4+log((1+xx(i))/b);

end

F=[k4*s1-k1+k2*s2 k3+s3-s4];

3)mom

function [F]=MOM_Method(xx,S)

m1=mean(xx);

b=S(1);

q=S(2);

n=length(xx);

s1=0;

for i=1:n

s1=s1+xx(i)^2;

end

m2=s1/n;

M1=b^2/(1+b)*(q-1)+2*b/(1+b)*(q-2);

M2=2*b^3/(1+b)*(q-1)*(q-2)+6*b^2/(1+b)*(q-2)*(q-3);

F=[sqrt((M1-m1)^2) sqrt((M2-m2)^2)];

please are these code correct or not لاecause I noticed the curve of the cumulative distribution function stabilizes at the value 0.7 and does not reach (1) . Is the data generation correct or not?

thank you for your attention

[Jeff Miller]

I don't know what you are trying to compute so I can't say for sure whether your code is correct, but a few things make me suspect it is not correct:

• The CDF should reach 1 and not stabilize at 0.7. Maybe your CDF function is not correct (e.g., why does it start with 1/1+... which is the same as 1+....?). What distribution are you actually trying to work with?
• Your inverseF function does not call fzero so it is not getting x values with the cdf of u. Instead, it is computing the CDF of u.

Answer 2

[Riley]

You can use function ksdensity.