- #1
AdVen
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The problem is to find a closed form for the expectation (mu) and standard deviation (sigma) of the probability density function underneath (Maple notation):
f(t) = a*exp(a*t)*BesselK(0, 2*sqrt(exp(a*t)))/BesselK(1, 2)
I was able to find an expression for mu::
mu = Int(t*a*exp(a*t)*BesselK(0, 2*sqrt(exp(a*t)))/BesselK(1, 2), t = 0 .. infinity)
Substitution with x = exp(a*t) leads to:
mu = Int((1/(x*a))*ln(x)*x*BesselK(0, 2*sqrt(x))/BesselK(1, 2),x=1..infinity)
Partial integration integration leads to:
-ln(x)*sqrt(x)*BesselK(1, 2*sqrt(x))/(a*BesselK(1, 2))-Int((-sqrt(x)*BesselK(1,
2*sqrt(x))/(a*BesselK(1, 2)))*(1/x),x)
Integration leads to:
-(ln(x)*sqrt(x)*BesselK(1, 2*sqrt(x))+BesselK(0, 2*sqrt(x)))/(a*BesselK(1, 2))
Finally the following closed form can be obtained:
mu = BesselK(0, 2)/(a*BesselK(1, 2))
I was not able to find a closed form for sigma. I suspect, however, that the solution must be proportional to: BesselK(0, 2)/(a*BesselK(1, 2)).
Who can help me?
Ad van der Ven.
PS. Mijnhomepage: http://www.socsci.ru.nl/~advdv/
f(t) = a*exp(a*t)*BesselK(0, 2*sqrt(exp(a*t)))/BesselK(1, 2)
I was able to find an expression for mu::
mu = Int(t*a*exp(a*t)*BesselK(0, 2*sqrt(exp(a*t)))/BesselK(1, 2), t = 0 .. infinity)
Substitution with x = exp(a*t) leads to:
mu = Int((1/(x*a))*ln(x)*x*BesselK(0, 2*sqrt(x))/BesselK(1, 2),x=1..infinity)
Partial integration integration leads to:
-ln(x)*sqrt(x)*BesselK(1, 2*sqrt(x))/(a*BesselK(1, 2))-Int((-sqrt(x)*BesselK(1,
2*sqrt(x))/(a*BesselK(1, 2)))*(1/x),x)
Integration leads to:
-(ln(x)*sqrt(x)*BesselK(1, 2*sqrt(x))+BesselK(0, 2*sqrt(x)))/(a*BesselK(1, 2))
Finally the following closed form can be obtained:
mu = BesselK(0, 2)/(a*BesselK(1, 2))
I was not able to find a closed form for sigma. I suspect, however, that the solution must be proportional to: BesselK(0, 2)/(a*BesselK(1, 2)).
Who can help me?
Ad van der Ven.
PS. Mijnhomepage: http://www.socsci.ru.nl/~advdv/