how about find the joint density and then integrate

<t>the condition is n is big, p is small(say p less than 0.01) and n*p is not to big not too small QuoteOriginally posted by: freyziFasturtle, you are assuming that when n increases then p decrease. But that doesn't have to be the case. See for example the thread Probability of 8 Occurences?. There ...

yes I got it.Really Thanks!QuoteOriginally posted by: prosperosoE[X_u W^2_u] = int_0^u E[W^2_u W^2_s] ds = int_0^u (2s^2+us) ds = 7/6 u^3soE[X^2_t] = 2 int_0^t E[X_u W^2_u] du = 2 int_0^t 7/6 u^3 du = 7/12 t^4henceE[X^2_1] = 7/12.

<t>QuoteOriginally posted by: prosperosince X_t=int_0^t W^2_u du is a process of locally bouneded variation, by Itof(X_t)=f(x_0) + int_0^t f'(X_u) dX_u,for f(x)=x^2 you getX^2_t = 2 int_0^t X_u W^2_u du.Take expectation (and use Fubini)E[X^2_t] = 2 int_0^t E[X_u W^2_u] du.NextE[X_u W^2_u] = E[ W^2_u...

<t>B(t) is just standard brownian motion starting at 0, neither drifted BM nor diffusioncan I ask how you get the result?QuoteOriginally posted by: AaronIf the motion starts at B(0) with drift u and standard deviation s, I think the answer is [B(0) + u/2]^2 + s^2 for arithmetic Brownian motion. But ...

but X is the riemann integral of B(s)^2 over time interval [0,1].it seems ito's formula does not help a lot.QuoteOriginally posted by: prosperoExpand f(X_t), f(x)=x^2 by Ito's formula and take expectation.

sorry but the integrand is not square root of B(s) but square.

let B be a Brownian motion,define X=integral { sqr(B(s)) ds } 0<s<1,then what is the second moment of X?Thanks!

your guess is correct. the point here isp{B(t1)=x; B(t2)=y} = p(t1,0,x) p(t2-t1,x,y) =p(t2,0,y) p(t1(t2-t1)/t2, yt1/t2,x)here p(t,a,b) is the transition density: travelling from a to b in time t

hey, do you know the answer or not?

Lamberton and Lapeyre, orBjork,new edition