Karatzas, Ioannis & Shreve, Steven E. , "Methods of mathematical finance", Example 1.2.3, "Doubling Strategy"

Question

I'm reading the first chapter of Methods of mathematical finance from Karatzas as part of my B.Sc-thesis in mathematics and I don't understand 1 argument in example 2.3 of Chapter 1: The author defines the stochastic integral on [0,T),

$I(t) := $$\int_0^t \sqrt{\frac{1}{T-u}} d W(u) $

with W(.) is a 1-dimensional Brownian Motion and can (apparently) easily conclude that I(t) is a martingal on [0,T); but I cannot...

I can show that it is a local martingal on [0,T) (for example with : Philip E. Protter. Stochastic Integration and Differential Equations, III.5, Theo 17 : \sqrt{\frac{1}{T-u}} is continous and the Brownian Motion W is a local martingal and therefore I is a local martingale)

I've tried many things, we can't use:

1.)Bounded local martingals are martingals, because I isn't bounded.

2.)Stochastic integrals with bounded integrand are martingals, because the integrand $\sqrt{\frac{1}{T-u}}$ isn't bounded for $u\ge T$

Please help, Thank you in advance!

Answer 1

To say that $I$ is a martingale on $[0,T)$ is to say that $E[I(t)|\mathcal F_s]=I(s)$ for all $0\le s<t<T$; equivalently, $\{I(t):0\le t\le t_0\}$ is a martingale for each $t_0\in(0,T)$. But for fixed $t_0\in(0,T)$, the (deterministic) integrand $u\mapsto 1/\sqrt{T-u}$ is continuous and bounded on $[0,t_0]$. Consequently the stochastic integral $I(t)$, $0\le t\le t_0$, is a (square-integrable) martingale.

Answer 2

You can show that, for any $t_1$ and $t_2$, where $0 \le t_1 < t_2 < T$, \begin{align*} E\left(\int_{t_1}^{t_2}\sqrt{\frac{1}{T-u}} dW_u \right) =0. \end{align*}