3
$\begingroup$

$\textbf{Question:}$

Let $B_t$ be a standard brownian motion started at $0$, $S_t=\max_{0\leq s\leq t} B_s$ and $F:\mathbb{R}_+\times\mathbb{R}\times\mathbb{R}_+ \to \mathbb{R}$ be a $C^{1,2,1}$ function.

(1) Apply Ito's formula to $F(t,B_t,S_t)$ for $t\geq 0$ and determine a continuous local martingale $M_t$ starting at $0$ and a continuous bounded variation process $A_t$ such that $F(t,B_t,S_t)=M_t+A_t$

(2) Show that if $F_t(t,x,s)+\frac{1}{2}F_{xx}(t,x,s)=0$ for all $(t,x,s)$ and $F_s(t,x,s)=0$ for $x=s$, then $F(t,B_t,S_t)$ is a continuous local martingale.

(3) Show that $(S_t-B_t)^6-15t(S_t-B_t)^4+45t^2(S_t-B_t)^2-15t^3$ is a martingale.

$\textbf{My attempt:}$

(1) $F(t,B_t,S_t) = F(0,B_0,S_0) + \int_0^t F_t(s,B_s,S_s) ds + \int_0^t F_x(s,B_s,S_s) dB_s + \int_0^t F_s(s,B_s,S_s) dS_s + \frac{1}{2} \int_0^t F_{xx}(s,B_s,S_s) ds$

So, $M_t = F(0,B_0,S_0) + \int_0^t F_x(s,B_s,S_s) dB_s$ and $A_t = \int_0^t [F_t(s,B_s,S_s) + \frac{1}{2} \int_0^t F_{xx}(s,B_s,S_s)] ds + \int_0^t F_s(s,B_s,S_s) dS_s$

(2) Skipping all the steps: Let $(t,x,s) = (s,B_s,S_s)$ and then only the $M_t$ terms would be left and so $F$ is a continuous local martingale.

Would this be enough or is more detail needed?

(3) Let $F(t,B_t,S_t) = (S_t-B_t)^6-15t(S_t-B_t)^4+45t^2(S_t-B_t)^2-15t^3$. Using Ito's formula, this simplifies to $$F(t,B_t,S_t) = \int_0^t [-6(S_s-B_s)^5 + 60s(S_s-B_s)^3 - 90s^2(S_s-B_s)] dB_s$$ $$+ \int_0^t 6(S_s-B_s)^5 - 60s(S_s-B_s)^3 + 90s^2(S_s-B_s) dS_s$$

I am not sure how I should proceed.

$\endgroup$

1 Answer 1

1
$\begingroup$

(2) You might add a little more detailed explanation why the $dS_s$-term vanishes (i.e. why it is enough to have $F_s(t,x,s)=0$ for $x=s$).

(3) Define $$F(t,x,s) := (s-x)^6 - 15 t (x-s)^4 + 45 t^2 (s-x)^2 - 15 t^3.$$ Then $$F_s(t,x,s) = 6 (s-x)^5-60 (x-s)^3 + 90 t(s-x)$$ which implies, in particular, $F_s(t,x,s) =0$ for $x=s$. Moreover, $$\begin{align*}&F_t(t,x,s) + \frac{1}{2} F_{xx}(t,x,s) \\ &= -15 (x-s)^4 + 90t (s-x)^2-45 t^2 + \frac{1}{2} \bigg(30 (x-s)^4 - 180 (x-s)^2 + 90t^2 \bigg) \\ &=0\end{align*}$$

Therefore it follows from step (2) and (3) that the process $M_t = F(t,B_t,S_t)$ is a local martingale and $$M_t = \int_0^t F_x(s,B_s,S_s) \, dB_s. \tag{1}$$ Because $(B_t)_{t \geq 0}$ is Gaussian (and therefore has finite moments of arbitrary order), it is not difficult to check that

$$\mathbb{E} \left( \int_0^T F_x(s,B_s,S_s)^2 \, ds \right)< \infty;$$

this implies that the local martingale $(1)$ is, in fact, a martingale.

$\endgroup$
3
  • $\begingroup$ what extra explanation should I add for Q2 that will justify my answer? $\endgroup$
    – user406357
    Commented Jan 13, 2017 at 17:16
  • 1
    $\begingroup$ @SC16 Well, what do you think why the term vanishes? In your solution, it sounds as you think that this is obvious. $\endgroup$
    – saz
    Commented Jan 13, 2017 at 17:26
  • $\begingroup$ Hello, sorry to ask, but I have already seen this question and because I do not know how to justify it vanishes (see: math.stackexchange.com/questions/3988046/…). Could you please give an intuition of this fact? $\endgroup$
    – R__
    Commented Jan 16, 2021 at 22:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .