I am sorry that I did something wrong, since there is some very complicated issue when change the order of the Ito integral with a Riemann integral. I am so lazy that I try to find some short cut :-)

So now some "standard" way to solve this problem. Of course it is Ito Lemma

Consider 1/2*s^2*w(s)

applying Ito lemma:

d(1/2*s^2*w(s))=1/2*s^2*dw(s)+s*w(s)ds (the second order term is zero)

taking integral from 0 to t at both sides.

∫(0 to t)d(1/2*s^2*w(s))=∫(0 to t)1/2*s^2*dw(s)+∫(0 to t)s*w(s)ds

∫(0 to t)s*w(s)ds which is A(t) is thus equal to:

A(t)=∫(0 to t)d(1/2*s^2*w(s))-∫(0 to t)1/2*s^2*dw(s)

=1/2*t^2*w(t)-∫(0 to t)1/2*s^2*dw(s)

set 1/2*t^2*w(t)=x and ∫(0 to t)1/2*s^2*dw(s)=y

so now it comes to find the variance of A(t)

the variance of A(t) is equal to var(x)+var(y)-2cov(x,y), since x, y are overlapped on 0 to t, they are correlated.

var(x)=1/4*t^4*var(w(t))=1/4*t^5

var(y)=∫(0 to t)1/4*s^4*ds=1/20*t^5 (Ito isometry as mentioned by the floor above)

cov(x,y)=E(xy) (since E(x)=0 and E(y)=0)

we can rewrite x into: ∫(0 to t)1/2*t^2*dw(s)=x

so that cov(x,y) is equal to E(∫(0 to t)1/2*t^2*dw(s)∫(0 to t)1/2*s^2*dw(s))

again applying Ito isometry:

cov(x,y)=E(1/4*t^2*∫(0 to t)*s^2*ds)=1/12*t^5

so var(A(t))=1/4*t^5+1/20*t^5-2*1/12*t^5=2/15*t^5

It should be correct since I test the result via simulation with matlab.

best,

Thank you very much, Chemist_MZ!You just made my day!

I was stuck with the E[Wt ∫(0 to t)1/2*s^2*dw(s)]!
I learned something new!

Also, I did not know about the Ito's isometry in the form of E[∫(0 to t) g(s)dWs ∫(0 to t) h(s)dWs ] = E[∫(0 to t) g(s)h(s)ds]


Thank you again!

Thank you for educating me on this form of Ito's isometry.

another approach to solve it
A^2 = 2int_0^t A_s dA
       = 2int_0^t A_s s W_s ds
       = 2int_0^t (int_0^s r W_r dr) s W_s ds
       = 2int_0^t int_0^s rW_r sW_s drds(关于r和s的双重积分)
计算E(A^2)时利用积分交换，计算E(rW_r sW_s) = r^2 * s
剩下的就是简单的双重积分
2int_0^t int_0^s r^2*s drds
=2/15*t^5

You can solve it without Ito isometry

Wow, you killed it nicely!

ok, that's cool.

A colleague of mine showed me another approach, which is another cool way to kill it.

E(A^2) = E(int_0^t u W_u du int_0^t s W_s ds)
           = E(int_0^t   int_0^t  u W_u s W_s du ds) --- now double integral on a square where 0 < u, s < t
we can break the square integral into two triangle integrals, where u > s and s < u, respectively, that is,
E(A^2) = E(int_0^t   int_0^s u W_u s W_s du ds) + E(int_0^t   int_0^u u W_u s W_s ds du)
           = int_0^t   int_0^s u s E(W_u W_s) du ds + int_0^t   int_0^u u s E(W_u W_s) ds du
           = int_0^t   int_0^s u s u du ds + int_0^t   int_0^u u s s ds du
           = int_0^t s int_0^s u^2 du ds  + int_0^t u int_0^u s^2 ds du
           = int_0^t s s^3/3 ds + int_0^t u u^3/3 ds
           = t^5/15 + t^5/15
           = 2 t^5/15
Similar to xuruilong100' approach in a way

I am glad to see this heat discussion.