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3

Note that by Itô's formula (applied for $f(t,x) := t^2 \cdot x$), we have $$s^2 B_s = \int_0^s r^2 \, dB_r + 2 \int_0^s r B_r \, dr.$$ This is equivalent to $$Y_s = s^2 B_s - 2 \int_0^s r B_r \, dr$$ and that's exactly the identity which your professor used.


2

Hint: Use the summation by parts formula (also known as Abel's summation formula): $$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_{n+1} g_{n+1}-f_m g_m) - \sum_{k=m}^n g_{k+1} (f_{k+1}-f_k).$$


1

Using summation by parts, as suggested, it actually becomes quite clear. $$\sum_{n=1}^{N}t_n(W(t_{n+1})-W(t_n))=t_N W(t_N)-t_1W(t_1)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ With $t_N=T$ and $t_1=0$. This gives: $$\int_0^TtdW(t)= TW(T)-\sum_{n=1}^N W(t_n)(t_{n+1}-t_n)$$ Taking $n \rightarrow \infty$ gives the Riemann integral. Thus the results ends up being: ...


1

So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here : Lemma 3 : Let $X$ be a càdlàg square integrable martingale and ${\xi}$ be a bounded predictable process. Then, $\int\xi\,dX$ is a square integrable martingale. In your case $W$ is a Brownian motion, so you fit the conditions for the lemma as a ...



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