Scimann

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16 years, 7 days

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These are questions asked by Scimann

Is there a way to obtain the code for BrownianMotion function in Maple15?

I tried

interface(verboseproc = 2);
readlib(Finance:-BrownianMotion);
print(Finance:-BrownianMotion);

but these only give

module () local ModuleApply, BrownianMotion, BrownianMotion1D; end module

Where do I go from here? Thanks.

With Maple15, how do I get this

sigma^2*(1+(lambda[0]+lambda[1]-2)^2+(1-lambda[0])^2)

from

with(Statistics):
R := 'RandomVariable(Normal(0, sigma))'; a := [`$`('R', 5)]
Variance((1-lambda[0])*a[3]+(lambda[0]+lambda[1]-2)*a[4]+a[5]);
simplify(%, size); collect(%, {sigma, lambda[0], lambda[1]})

In general, how do I simplify an expression when it can be stated as a sum of squares? I suspect I am missing a general simplification technique. Thanks.

I'd like to numerically compute integrals of the forms:

int(exp(-t)/((-3+2*exp(-(1/10)*t))*(-3+2*exp(-(9/10)*t))), t = 0 .. infinity); evalf(%)

where there are much more terms in the denominator, e.g.,

Int(e^(-t)/(product(-3+2*exp(-.1*(1-.1)^i*t), i = 0 .. A)), t = 0 .. infinity); evalf(%)

for some positive integer A. Even when I have three terms it takes so long... So how do I compute such integrals numerically with maple; what's the efficient...

With f := i->`if`(i = 0, 1, alpha), I'd like to evaluate expression A below entirely symbolically, i.e. with l=1..3 being replaced with l=1..h in

sum('sum(sum(f(i)*f(i+j), i = 0 .. l-1), j = 1 .. h-l)', l = 1 .. 3); subs(alpha = .3, h = 3, %);#A

but using  A with l= 1 .. h provides a different number, the one one obtains with

sum(sum(sum(f(i)*f(i+j), i = 0 .. l-1), j = 1 .. h-l), l = 1 .. h); subs(alpha = .3, h = 3, %); #B

Expressions...

How do I evaluate moments of Q4 in a reasonable time with a reasonable accuracy, i.e. to a few decimal points:
   
with(Statistics): RV := 'RandomVariable(Geometric(1/3))+1'; R := `$`('RV', 5);

Q4 := 1/(alpha*R[1]+alpha^2*R[2]+alpha^3*R[3]+(1-alpha-alpha^2-alpha^3)*R[4]);

Moment(Q4, 1);

E.g., can I tell maple that replacing infinite series underlying this calculation with sufficiently long partial sums is okay with me? Or maybe there...

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