Well as the title suggests, I am looking for a book that covers the fore mention topics well for a potential quant.Cheers

Cheers guys for the hints. I have had a look at cdfs now, and looking at the problem again I think the "survival" test is the route forward in obtaining some limit to the conditional expectation. Once again thankyou for the help..

<t>Hi All,I am stuggling to understand what this question is trying to get at..Let X~N(0,1) a normally distributed random variable mu = 0, sigma = 1. suppose now x E R, x >0 find the upper and lower bound of the conditional expectation of X for E(X|X>=x)Here X and x are different and not a typo.Firs...

<t>Hi Caligula,If you can ask your friend, that would be great to get any aspect of this process for all job roles and candidates. Since searching the internet hasn't been fruitful, which I suspect is due to some kind of waiver for candidates to sign for secrecy.As for me personally I am testing for...

<t>Hi all,I am looking for some tips,hints, and advice, basically anything that can give me a little insight what to expect in my interview(on friday)..I have been given a sample test outlining similar questions to be expected in the screening test, of which i am good/excellent on some, poor at othe...

I get w(x) = sqrt(2)erf^(-1)[a*sqrt(2/pi)*x]??From w''=y(dy/dw) where y = w' w'' = w(w')^2=wy^2=y(dy/dw)Is this an equation from financial maths?

<t>I would personally transform with u = x^a V thus the equation becomes x^(a+2)V'' + (2a+1)x^(a+1)V'+[a^2+c]V=0choosing a^2=-c with c<0 ->a E RV''+(2a+1)x^(-1)V'=0V' = Bexp(mx^2) where m = (2a+1)/2V = int^x Bexp(mx^2)dx + Du = x^(sqrt(-c))VI forever working with equations, and finding a transform i...