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G'day, I am able to derive the answer, but unsure about the given argumentation of the prof.

Problem: Show that $M_t = W^3_{t} - 3 \int_0^{t}W_sds$ is a martingale. Let $W_t$ be Brownian motion.

$W^3_{t} = 3 \int_0^{t}W^2_{s}dW_s + 3 \int_0^{t}W_sds$

1) Substituting $W_t^{3}$ into the equation $M_t$

$M_t = 3 \int_0^{t}W^2_{s}dW_s + 3 \int_0^{t}W_sds - 3 \int_0^{t}W_sds $

2) Terms of $+ 3 \int_0^{t}W_sds - 3 \int_0^{t}W_sds$ cancels out leaving us with the answer

$M_t = 3 \int_0^{t}W^2_{s}dW_s$ and hence a martingale. However am I allowed to do step 2) just like that?

The prof gives the following reasoning which I don't understand: $E(\int_0^{t}(W_s^{2})^2ds) = E(\int_0^{t}W_s^{4}ds) = \int_0^{t}3s^{2}ds < ∞$ for all t>= 0.

Can someone explain the reasoning?

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  • $\begingroup$ Which book are you reading? $\endgroup$
    – user9464
    Nov 24, 2019 at 14:45
  • $\begingroup$ Stochastic Calculus for Finance and Arbitrage in Continuous Time $\endgroup$
    – simsalabim
    Nov 24, 2019 at 14:46
  • $\begingroup$ Can't find this book. Who's the author of the book? $\endgroup$
    – user9464
    Nov 24, 2019 at 14:48
  • $\begingroup$ First one is Shreve, the second one Thomas Bjork.. $\endgroup$
    – simsalabim
    Nov 24, 2019 at 14:49
  • $\begingroup$ In the step 1), we used the Ito formula for the function $f(x) = x^3$ with the process $W$. Then in the step 2), we showed that the process $M$ is actually a local martingale. In fact, we can even show that is a real (square) martingale by showing that the $L^2$-norm is finite. This is done by applying the Ito isometry. $\endgroup$
    – Sesame
    Nov 25, 2019 at 10:43

1 Answer 1

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In order to make sense of a stochastic integral of the form $\int_0^t f(B_s) \, dB_s$ you need to verify two properties:

  • $f$ is suitably measurable, e.g. progressively measurable
  • $f$ is suitably integrable, e.g. $\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$ or $\mathbb{E}(\int_0^{t \wedge \tau_n} f(s)^2 \, ds)<\infty$ for a sequence of stopping times $\tau_n \uparrow \infty$

Under the above conditions, the stochastic integral $M_t := \int_0^t f(B_s) \, dB_s$ is a local martingale but, in general, it might fail to be a martingale. In order to ensure that $(M_t)_{t \geq 0}$ is a martingale (not only a local one), $f$ has to satisfy the stronger integrability condition of the above-mentioned conditions, that is, $\mathbb{E}(\int_0^t f(s)^2 \, ds)<\infty$ for all $t\geq 0$.

This is exactly the condition which the authors of the solution verified for the particular case $f(s):=B_s^2$.

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