ActiveUser

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6 years, 341 days
i would also not like to ask, but if not ask, what should i do?

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These are replies submitted by ActiveUser

sorry for misunderstanding, just consider all i in small letter or all in big letter

yes, by madan's result, in other words to see how integration of two things result in new distribution

i just want to produce madan's answer not only using gamma, i want to use all different kinds of distributions to do madan's result.

in the link of handbook of distribution's derivation of gamma, told the method

to find gamma that means using other process or your own generating function, you can also using the method and

then integrate with normal distribution, to find madan's result.

i want to investigate how these processes change the tail of normal distribution.

of course, furthermore not only normal distribution.

it is the fourier transform between characteristic function and density function

look at this definition

charc(t) = int( exp(i*t*x)*densityfunction, x=-infinity..infinity)

then

densityfunction = int(exp(-i*t*x)*charc(t), , x=-infinity..infinity)

i ask this before, someone reply me with this command.

n, p, y are only dummy variables

i is complex number sqrt(-1) i guess, most generating function are given in book, i also do not know these dummy

variables

use this official one, it evaluating a very long time, is it the substitution wrong?

restart;
with(Statistics):
X:=RandomVariable(Weibull(k-1,eta));
WeibullDensity := PDF(X,rho);
WeibullDensity := eta*rho^(-1+eta)*exp(-(rho/(k-1))^eta)/(k-1)^eta;
WeibullDensity := subs(k=a,WeibullDensity);
WeibullDensity := subs(rho=x,WeibullDensity);
WeibullDensity := subs(eta=b,WeibullDensity);
WeibullDensity := -b*((a-1)/x)^b*exp(-((a-1)/x)^b)/(x^2*(a-1));
WeibullDensity := subs(a=t/v, WeibullDensity); # do not know why subs this
WeibullDensity := subs(x=g, WeibullDensity); # do not know why subs this
WeibullDensity := subs(b=1/v, WeibullDensity); # do not know why subs this
WeibullDensity := simplify(WeibullDensity);
tm := simplify(int(expand(1/(rho*sqrt(2*Pi*g))*exp(-((X-theta*g)^2)/(2*g*rho^2))*WeibullDensity),g=0..infinity));
a := int(exp(i*X*u)*tm, X = -infinity .. infinity);
madan := simplify(a, power) assuming rho > 0, theta > 0, v > 0;

WeibullDist := int((η/z)*(((k-1)/z)^(η-1))*exp(-(((k-1)/z)^η)),z=0..rho);
WeibullDensity := diff(WeibullDist, rho) assuming rho >= 0;

WeibullDensity := simplify(simplify(simplify(expand(convert(simplify(WeibullDensity, size), exp)), size)));

above is steps, i follow previous post's link in derivation of gamma, you can start from below
WeibullDensity := eta*((k-1)/rho)^eta*exp(-((k-1)/rho)^eta)/(k-1);
WeibullDensity := subs(k=a,WeibullDensity);
WeibullDensity := subs(rho=x,WeibullDensity);
WeibullDensity := subs(eta=b,WeibullDensity);
WeibullDensity := -b*((a-1)/x)^b*exp(-((a-1)/x)^b)/(x^2*(a-1));
WeibullDensity := subs(a=t/v, WeibullDensity); # do not know why subs this
WeibullDensity := subs(x=g, WeibullDensity); # do not know why subs this
WeibullDensity := subs(b=1/v, WeibullDensity); # do not know why subs this
WeibullDensity := simplify(WeibullDensity);
tm := simplify(int(expand(1/(rho*sqrt(2*Pi*g))*exp(-((X-theta*g)^2)/(2*g*rho^2))*WeibullDensity),g=0..infinity));
a := int(exp(i*X*u)*tm, X = -infinity .. infinity);
madan := simplify(a, power) assuming rho > 0, theta > 0, v > 0;

when a comment, this site has error in firefox and i.e.

now become normal

when a comment, this site has error in firefox and i.e.

now become normal

@Preben Alsholm 

i find pochhammer can be converted to n! * binomial(n,k)

http://en.wikipedia.org/wiki/Pochhammer_symbol

i just close my computer,

if you tried, tell me whether the summation containing binomial be simplified to a correct generating function

 

if not, any other transformation formula can simplify to remove binomial in summation

@Preben Alsholm 

i find pochhammer can be converted to n! * binomial(n,k)

http://en.wikipedia.org/wiki/Pochhammer_symbol

i just close my computer,

if you tried, tell me whether the summation containing binomial be simplified to a correct generating function

 

if not, any other transformation formula can simplify to remove binomial in summation

@Preben Alsholm 

i use this to further calculate the generating function, it is not the official one.

Hermit

restart;
with(gfun):
test4 := diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x)=0;
sol := dsolve(test4);
series(rhs(sol),x,2);
sol2 := eval(rhs(sol), [_C2 = 0, _C1 = 1]);
convert(sol2, hypergeom);

restart;
with(SumTools):
An := 1/2-1/2*n;
Cn := 3/2;
An2 := 1/2-1/2*n+L;
Cn2 := 3/2+L;
gen := Summation(Product(An2, L=0..k-1)/Product(Cn2, L=0..k-1)/k!*z^k, k=0..infinity);
gen := Summation(pochhammer(An,k)/pochhammer(Cn,k)/k!*z^k, k=0..infinity);
genfun := simplify(gen);
correct_genfun := exp(2*x*z-z^2);

@Preben Alsholm 

i use this to further calculate the generating function, it is not the official one.

Hermit

restart;
with(gfun):
test4 := diff(P(x),x$2)-2*x*diff(P(x),x)+2*n*P(x)=0;
sol := dsolve(test4);
series(rhs(sol),x,2);
sol2 := eval(rhs(sol), [_C2 = 0, _C1 = 1]);
convert(sol2, hypergeom);

restart;
with(SumTools):
An := 1/2-1/2*n;
Cn := 3/2;
An2 := 1/2-1/2*n+L;
Cn2 := 3/2+L;
gen := Summation(Product(An2, L=0..k-1)/Product(Cn2, L=0..k-1)/k!*z^k, k=0..infinity);
gen := Summation(pochhammer(An,k)/pochhammer(Cn,k)/k!*z^k, k=0..infinity);
genfun := simplify(gen);
correct_genfun := exp(2*x*z-z^2);

@Carl Love 

http://en.wikipedia.org/wiki/Charlier_polynomials

compare with above, i guess a is mu, two structures of genfun are totally different.

even change to pochhammer, can not summation

@Markiyan Hirnyk 

not tired, want a smooth operation in this, i am not a physics graduate, not sure manually convert without maple, i do not have confidence to be sure whether is correct or not.

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