Testing a Time-Scaled Brownian Integral for the Martingale Property
Summary
The document asks whether a stochastic integral with an integrand that depends on both the integration time and the current time is a martingale. It points out a limitation of a standard result for integrals with a deterministic integrand depending only on the integration variable: that result cannot be applied directly when the integrand changes with the endpoint time.
The answer rewrites the process as a time-scaled version of an auxiliary Brownian integral and applies Itô's formula. This yields a stochastic differential with a drift term proportional to the process itself, leading the answer to conclude that the martingale condition holds only when the scaling exponent is zero. The reasoning illustrates why the drift must be checked rather than inferred from the presence of a Brownian integral. The answer does not discuss technical conditions near time zero or distinguish in detail between local and true martingales, so those conditions require separate attention.
Key ideas
- A martingale result for integrands depending only on integration time does not directly cover endpoint-dependent integrands.
- Representing the process as a scaled auxiliary stochastic integral enables an Itô formula calculation.
- The derived drift term vanishes only when the scaling exponent is zero.
- The document's conclusion does not detail integrability or behavior near time zero.
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Full text
# Test if a process (with no drift) is a martingale
# Test if a process (with no drift) is a martingale
Consider the process $$Z(t)=\int_{0}^{t} \frac{u^a}{t^a}dW_u$$ for some real constant $a$ and $W_t$ is a wiener process. I want to check whether this process is a $F_t^W$-martingale. I noticed Lemma 4.9 in Bjork's book which states the following:
for $g\in L^2$ and $X_t$ defined as $$X_t=\int_{0}^{t} g(u) dW_u$$ is a $F_t^W$-martingale.
If we let $g(z,t)=z^at^{-a}$; Can we then use the lemma to conclude that $Z(t)=\int_{0}^{t} g(u,t)dW_u$ is $F_t^W$-martingale?
And that leads me another question: How do I test if a function $g \in L^2$
With Bjork I referto the textbook Bjork, Arbitrage Theory in Continous Time Finance (3rd Edition)
## Answer by Cettt (score 2, accepted)
https://quant.stackexchange.com/a/37842
the Lemma from Bjoerk's book is only valid for functions $g(u)$ not for function's $g(u,t)$.
What you can do instead is to use Ito's formula. First let's start with the process
$$ X_t = \int_{0}^t u^a \; dW_u. $$
This process can be written in differntial form as $$ dX_t = t^a dW_t. $$
Now we want to derive the differential form of $Z_t = \frac 1 {t^a}X_t$. Therefore we can use Ito's formula with $f(t,x) = \frac 1 {t^a}x$: $$ dZ_t = df(t,X_t) = f_t(t,X_t)dt + f_x(t,X_t)dX_t + \frac 12 f_{xx}(t,X_t) d[X]_t $$ Here $f_t$ resp. $f_x$ is the partial derivative of $f$ with respect to $t$ resp. $x$, and $f_{xx}$ is the second partial derivative of $f$ with respect to $x$. Note that we have $f_{xx}(t,x) = 0$, such that the differential form of $Z$ can be simplified to $$ dZ_t = - \frac a {t^{a+1}}X_t dt + \frac{1}{t^a} t^a dW_t = - \frac at Z_t dt + dW_t. $$ For a Ito process to be a (local) martingale the drift part has to be equal to zero. Therefore, $Z_t$ is a martingale if and only if $a = 0$ in which case $Z_t = W_t$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.