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Two ways:

1. Integrate it directly. This can be done, trust me. Just calculate the expectation with respect to a normal density. ∫z^6 f(z) dz. Of course you need many steps. The good thing is that normal density is an exponential function which you can put into the differential operator directly

I don't recommend the first method.

2. Using Ito Formula.

You know E(B(t)^2) which is t.

Applying Ito lemma to B(t)^4, dB(t)^4=4B^3dB+12B^2*dt integrate and take expectation. dB term drops ( because Ito integral is a martingale) so you get something like E(B(t)^4)=E∫12B^2dt, E() is a linear operator. In most of the cases, it can be moved into the integral.

We get: E(B(t)^4)=∫12E(B^2)dt, put E(B^2)=t into it, it is a simple Riemann integral. Now you get E(B^4). Similarly you can get E(B^6) from E(B^4). This is a clever way to solve this problem.

Best,

非常非常感谢！！ 我试试。。。刚学习这些，感觉自己没有理解了内涵。。。

全英文，看不懂呢

全英文，看不懂呢