平方布朗運動的時間積分變異數
我想計算變異數
$$ I = \int_0^t W_s^2 ds $$ 我在想我可以定義函式 $ f(t,W_t) = tW_t^2 $ 然後應用 Ito 的引理，所以我得到
$$ f(t,W_t)-f(0,0) = \int_0^t \frac{\partial f}{\partial t}(s,W_s)ds + \int_0^t \frac{\partial f}{\partial x}(s,W_s)dW_s+ \frac{1}{2}\int_0^t \frac{\partial^2 f}{\partial x^2}(s,W_s)ds \= I + \int_0^t 2sW_sdW_s + \frac{t^2}{2} $$ 通過重新排列我得到
$$ I = tW_t^2 - \int_0^t 2sW_sdW_s - \frac{t^2}{2} $$ 然後我們得到了（我不確定，但我認為任何積分 wrt BM 的期望為零？）
$$ \mathbf{E}[I]=\frac{t^2}{2} $$ 和變異數
$$ \mathbf{V}[I] = \mathbf{V}[tW_t^2 - \int_0^t 2sW_sdW_s - \frac{t^2}{2}] = t^2\mathbf{V}[W_t^2]+\mathbf{E}[(\int_0^t 2sW_sdW_s)^2] \= 2t^4 + \mathbf{E}[\int_0^t 4s^2W_s^2ds]\quad\text{(Isometry property)} $$ Not sure if it is OK to change order of integration and expectation here, but if I do that, I get
$ \mathbf{V}[I]= 2t^4 + \int_0^t 4s^2\mathbf{E}[W_s^2]ds = 2t^4 + \int_0^t 4s^2\mathbf{E}[W_s^2]ds = 2t^4 + \int_0^t 4s^3ds=3t^4 $
However, the answer says the variance should be $ \frac{t^4}{3} $ , so I guess I do something wrong?
Other Way
By application of Ito’s lemma , we have $$ W^4_t=4\int_{0}^{t}W^3_sdW_s+6\int_{0}^{t}W^2_sds\tag 1 $$ We know
$$ \left{ \begin{align} &\mathbb{E}\left[ {{W}^{2n+1}}(t) \right]=0,,,,,,,,,,,,,, \ & \quad \mathbb{E}\left[ {{W}^{2n}}(t) \right]=\frac{(2n)!}{{{2}^{n}}n,!},{{t}^{n}} \ \end{align} \right. $$
therefore $$ \text{Var}(W^4_t)=\mathbb{E}[W^8_t]-\mathbb{E}[W^4_t]^2=105t^4-(3t^2)^2=96t^4\tag 2 $$ By application of Ito’s Isometry, we have $$ \text{Var}\left(4\int_{0}^{t}W^3_sdW_s\right)=16\int_{0}^{t}\mathbb{E}[W^6_s]ds=240\int_{0}^{t}s^3ds=60t^4\tag 3 $$ on the other hand $$ 2\text{Cov}\left(4\int_{0}^{t}W^3_sdW_s,,,6\int_{0}^{t}W^2_sds\right)=24t^4\quad\text{(Why?)}\tag 4 $$ Moreover $$ \text{Var}(W^4_t)=\text{Var}\left(4\int_{0}^{t}W^3_sdW_s+6\int_{0}^{t}W^2_sds\right)\tag 5 $$ thus $$ 96t^4=60t^4+36\text{Var}\left(\int_{0}^{t}W^2_sds\right)+24t^4 $$ i.e $$ \text{Var}\left(\int_{0}^{t}W^2_sds\right)=\frac{1}{3}t^4 $$