I am new to Stochastic Calculus.

$$ \ I = \int_0^t W(s)dW(s)\,. $$

So I want to find out E(I) and Var(I). This is the answer I am coming up with, is this the correct way?

$$ I = 0.5(W(t)^2 - t) $$ $$ E[I] = 0.5(E[W(t)^2] - t) = 0.5(E[(W(t)-W(0))^2] - t) = 0.5(t-t) = 0 $$ $$ Var(I) = 0.25(Var[W(t)^2]) = 0.25(E[W(t)^4] - (E[W(t)^2])^2) = 0.25(3t^2 - t^2) = 0.5t^2 $$


What you did is correct. I think it's useful to point out that $\forall X,\ Y$ $\in M^2[0,T]$ the following properties hold: $$E[\int_a^b X_sdW_s]=0, \ \ \ \ \ 0\leq a<b\leq T$$

$$Cov(\int_a^b X_sdW_s,\int_a^b Y_sdW_s)=E[\int_a^b X_sY_s ds], \ \ \ \ \ 0\leq a<b\leq T.$$ In your case $$E[\int_0^t W_sdW_s]=0$$ and $$Var(\int_0^t W_sdW_s)=E[\int_0^tW_s^2ds]=\int_0^tE[W_s^2]ds=\int_0^tsds=\frac{t^2}{2}$$ which is exactly your result.

  • $\begingroup$ This is great, without actually solving the Integral we can find the moments. Thanks @QWERTZ $\endgroup$ – Deb Mar 13 '17 at 11:43
  • $\begingroup$ You're welcome. This is indeed a very useful result; I think you can find a reference for it in any book about stochastic calculus, since it follows from the definition of the stochastic integral in $M^2[0,T]$ and the Ito's isometry. $\endgroup$ – Uskebasi Mar 13 '17 at 12:40

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