OK, I think I have never seen this question before, its quite odd because this expectation is more like some given results since it is just an expectation of a lognormal variable.

If you really want to do it, I have an idea.

applying Ito formula to Z(s)

dZ(s)=sigma*Z(s)*dW(s)+sigma^2*Z(s)*ds

taking integral from t to T on both sides:

Z(T)-Z(t)=∫(t to T) sigma*Z(s)*dW(s) +∫(t to T) sigma^2*Z(s)*ds

taking E() conditional on time t on both sides, for convenience E(.|F(t)) is expressed as E(.)

E(Z(T))=Z(t)+∫(t to T) sigma^2*E(Z(s))*ds

note:

1. ∫(t to T) sigma*Z(s)*dW(s) is an Ito integral so that is it a martingale. Since the initial value of this integral is zero, its expectation at time T is zero
2. E() in most cases can be exchanged with ∫, since E() is basically another kind of integral.

So now you get:

E(Z(T))=Z(t)+∫(t to T) sigma^2*E(Z(s))*ds

X(T)=Z(t)+∫(t to T) sigma^2*X(s)*ds

take derivative respect to T on both sides:

dX(T)=sigma^2*X(T)

solve it you can get X(T)=Z(t)exp(0.5*sigma^2(T-t)) which is the result.

best,

如果Xt是一个鞅，Et[XT]很容易得到，Zt不是一个鞅，找一个函数f(t)，使f(t)*Zt是一个鞅，应用Ito公式可以得到f(t),则Et[ZT] = Et[f(T) * ZT] / f(T) = f(t) / f(T)  * Zt

Yes, this works in many cases.

But lognormal is a special case, it can not be solved like that. (at least in my memory, or maybe it can)

best,

2楼是我要的答案！
非常感谢版主给出详细推导过程，xuruilong100提供的思路也是一个启发，多谢！