How to find expected value of normal cdf with a lognormal variable?

Hi. I am trying to find E[N((a+(si^2)/2)/si)] where a is a constant, si is lognormally distributed with mean m and variance v, (m and v are basically constants) and N is the standard normal cdf. Any help is greatly appreciated...

Answer 1

[Torsten]

MATHEMATICA cannot find an analytical expression for the corresponding integral.

So I guess you will have to assume numerical values for m and v.

Best wishes

Torsten.

[Richa]

Thanks for your reply. Supposing I do assume some value for m and v, even then how do i go about solving this expression? I have tried doing it using the definitions of both expected value and the cdf. but the integral is not solvable.

Answer 2

[Torsten]

Set values for s and m and enter the integral under

http://www.wolframalpha.com/input/?i=definite+integral

with

0.5*(1+Erf[x/Sqrt[2]])/(Sqrt[2*Pi]*s*x)*Exp[-(Log[x]-m)^2/(2*s^2)]

as integrand.

Best wishes

Torsten.

[Torsten]

Sorry, this is to calculate E[N(si)].

To get E[N((a+(si^2)/2)/si)], you will have to evaluate the integral with

0.5*(1+Erf[(a+x^2/2)/x/Sqrt[2]])/(Sqrt[2*Pi]*s*x)*Exp[-(Log[x]-m)^2/(2*s^2)]

as integrand.

Best wishes

Torsten.

[Richa]

Dear Torsten, thank you so much for your time. But how did you get this integral? I defined si as a lognormal object in MTLAB, but now it does not allow me to define (a+si^2/2)/si, saying that this is not allowed. How can I transform si to find its normal cdf value?

[Torsten]

I don't understand what you are trying to do.

If g: IR -> IR is measurable,the expected value of g(X) is given by integral_{x=-oo}^{x=+oo} g(x)*f(x) dx where f is the density function for the random variable X. In your case,

g(x)=0.5*(1+Erf((a+x^2/2)/x/sqrt(2))) and

f(x)=density function of the lognormal distribution.

(I assumed you mean the lognormal distribution with parameters m and v^2, not mean m and variance v^2. If you really mean mean m and variance v^2, you will have to calculate its parameters m' and v'^2 in advance.)

Best wishes

Torsten.