The problem is to find a closed form for the expectation (mu) and standard deviation (sigma) of the probability density function underneath (Maple notation):(adsbygoogle = window.adsbygoogle || []).push({});

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/

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# Finding an expression for the standard deviation.

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