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Solving integration for some parameters?

Posted 1 year ago

Hello everybody.

I wrote the attached code. this code gives a good result when "H" parameter is equal to 0.5. Even in this condition if I change other parameters, I can get a true result as well. But when I change "H" parameter to 0.6,0.7,0.8 or 0.9 the code can't be executed anymore and I can't get a result.

Based on the paper, for H=0.6,0.7,0.8 or 0.9. we have a result.

What's the problem?

Can anyone please help me to solve this problem?

thank you for your help in advance.

Attachments:
POSTED BY: M M
11 Replies

You may rewrite the nested integrals as double integrals on a two-variable domain. That way you can use a single NIntegrate.

POSTED BY: Gianluca Gorni
Posted 1 year ago

Excuse me, I didn't get your idea.

Is it possible to give me an example?

POSTED BY: M M
Posted 1 year ago

I could solve Integrals like this numerically. But there is a problem.

For example, I write q_1 enter image description here like this enter image description here

the result is close to real value but it is different a bit. For example for H=0.5, q_1 is computed 0.000195151 but it should be 0.0001943732.

why results are different a bit? this difference affects on the final result.

POSTED BY: M M

It is Fubini's theorem, hoping that your functions are bounded:

Integrate[f[s] Integrate[g[u], {u, 0, s}], {s, 0, T}] == 
 Integrate[Integrate[f[s] g[u], {u, 0, s}], {s, 0, T}] ==
 Integrate[f[s] g[u], {u, 0, s}, {s, 0, T}] ==
 Integrate[f[s] g[u], 
  Element[{u, s}, Triangle[{{0, 0}, {T, T}, {0, T}}]]]

It may be easier to set a precision target for a PrecisionGoal for a single NIntegrate than for a nested expression. However, I am no expert on this.

POSTED BY: Gianluca Gorni

This is what I was able to compute symbolically. The rest is to be done numerically:

\[Kappa] = 4; \[Gamma]0 = 1/10; \[Sigma]0 = 1/5; \[Theta] = 6/5;
H = 1/2 + 1/6; \[Rho] = -(3/4); K = 115;
Subscript[S, 0] = 100; r = 953/10000; T = 1/2;
F0[t_] = Subscript[S, 0]*E^(r*t);
Ktilde[t_] = 1 - K/F0[t];
Subscript[\[Lambda], 1][t_, s_] = (RealAbs[t - s])^(H - 1/2)/
  Gamma[H + 1/2];
E1[t_] = E^(\[Kappa]*t);
Ebar[t_] = E^(-\[Kappa]*t);
L0[t_] = \[Sigma]0*Ebar[t] + ((1 - Ebar[t])*\[Theta])/\[Kappa];
f0[t_] = L0[t];

Subscript[\[Lambda], 2][t_, s_] = 
 FullSimplify[\[Gamma]0*(Subscript[\[Lambda], 1][t, 
      s] - \[Kappa]^(1/2 - H) *E1[s]*Ebar[t]*
      (Integrate[x^(H - 1/2) E^x, {x, 0, (RealAbs[t - s]) \[Kappa]},
        Assumptions -> t > s > 0])),
  t > s > 0]
Vbar[s_] = FullSimplify[f0[s] +
   \[Rho]*Integrate[f0[u]  Subscript[\[Lambda], 2][s, u], {u, 0, s},
     Assumptions -> s > 0], s > 0]
POSTED BY: Gianluca Gorni
Posted 1 year ago

Thank you Gianluca for your time.

Actually, I reformed my code myself like this:

\[Kappa] = 4; \[Gamma]0 = 1/10; \[Sigma]0 = 1/5; \[Theta] = 6/5;
H = 1/2 + 1/6; \[Rho] = -(3/4); K = 115;
Subscript[S, 0] = 100; r = 953/10000; T = 1/2;
F0[t_] := Subscript[S, 0]*E^(r*t)
Ktilde[t_] := 1 - K/F0[t]
E1[t_] := E^(\[Kappa]*t)
Ebar[t_] := E^(-\[Kappa]*t)
L0[t_] := \[Sigma]0*Ebar[t] + ((1 - Ebar[t])*\[Theta])/\[Kappa]
f0[t_] := L0[t]
Subscript[\[Lambda], 1][t_, s_] := (t - s)^(H - 1/2)/Gamma[H + 1/2]
Subscript[\[Lambda], 2][t_, 
  s_] = \[Gamma]0*(Subscript[\[Lambda], 1][t, 
      s] - \[Kappa]^(1/2 - H) *E1[s]*Ebar[t]*
      Integrate [x^(H - 1/2) E^x, {x, 0, (t - s) \[Kappa]}, 
       Assumptions -> (t - s) \[Kappa] > 0]) // FullSimplify
Vbar[s_] = 
 f0[s] + \[Rho]*
    Integrate[f0[u] *Subscript[\[Lambda], 2][s, u], {u, 0, s}, 
     Assumptions -> s > 0] // FullSimplify
\[CapitalSigma][T] = NIntegrate[(Vbar[s])^2, {s, 0, T}]

when I reformed the code, I could compute Vbar[s_] and Subscript[[Lambda], 2][t, s] symbolically. Even I could compute [CapitalSigma][T] numerically. you can see our result in the below picture: enter image description here

It seems good. But now there is just another problem and it's when we want to compute q_1 enter image description here

. As you can see in q_1 when we want to compute each side of the plus sign numerically, there is an inner integral and it can't be computed. So outer integral can't be computed numerically as well.

I think the inner integral should be computed symbolically but it doesn't answer.

How can we compute q_1 numerically? any suggestions?

POSTED BY: M M
Posted 1 year ago

You have exponents of the form H-1/2, which are easy only when H=1/2. Otherwise all those nested integrals seem hopeless to evaluate symbolically in closed form. You may try numerical methods.

POSTED BY: Updating Name
Posted 1 year ago

Yes, exactly that's the problem. I've tried some numerical methods before, but unfortunately can't get a result. I don't know what I should do. I try different ways but none of them wasn't helpful.

do you know any special numerical methods or other suggestions that can be helpful in this condition?

POSTED BY: M M
Posted 1 year ago

Maybe for computing nested integrals, we can use numerical methods, but for example for computing q_1, first, we need to compute Inner integrals symbolically and after that use numerical methods. But as you can see in the picture, Inner integrals aren't computed.

enter image description here

POSTED BY: M M

Please don't unprotect Power, and please replace all those delayed definitions := with immediate definitions =. It does not make any sense to delay the calculation of all those integrals. This way I get CT = 3.53031

POSTED BY: Gianluca Gorni
Posted 1 year ago

Thanks Gianluca for your attention.

Yes, It's true. For H = 0.5, I got CT = 3.53031. In fact, for H=0.5 the code doesn't have any problems and can be executed well. but as I said before, the main problem appears when we change parameter H. For example, if you change parameter H to 0.6 or 0.7 the code can't be executed and this is my main problem.

how can I get true values for CT when parameter H is not equal to 0.5?

According to the paper, when H = 0.6 --> CT = 3.55 or when H = 0.7 --> CT = 3.58

POSTED BY: M M
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