Proving a Stochastic Integral’s Squared Value Is a Martingale
Summary
The document proves that the square of a stochastic integral, minus its accumulated squared integrand, is a martingale when the integrand is bounded and adapted. It defines the integral as a process, identifies its quadratic variation, and applies Itō’s lemma to its square. The quadratic variation term creates the drift that is canceled by subtracting the time integral of the squared integrand. The remaining differential is a stochastic integral, which is a true martingale under the boundedness condition.
A second proof uses conditional expectation and Itō isometry to show that the expected increment of the squared stochastic integral equals the expected increment of its quadratic variation. The arguments rely on the stated boundedness and filtration assumptions; the document does not discuss broader conditions under which the result might hold.
Key ideas
- The stochastic integral has quadratic variation equal to the time integral of the squared integrand.
- Applying Itō’s lemma to the squared integral produces a quadratic variation drift term.
- Subtracting accumulated quadratic variation removes the drift and leaves a martingale term.
- Boundedness of the adapted integrand supports the claim that the local martingale is a true martingale.
- Conditional expectation and Itō isometry provide an alternative proof.
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# Proving that a stochastic process is a martingale using Ito's Lemma
# Proving that a stochastic process is a martingale using Ito's Lemma
Assume a Wiener process W and a bounded F-adjusted stochastic process a. Show that the following process is a martingale on F
$$X(t)=(\int_{0}^{t}a(s)dW(s))^{2}-\int_{0}^{t}a^{2}(s)ds,\ t\geq0$$
Can someone help me on the above exercise? I tried to apply Ito's lemma but I got stuck
## Answer by siou0107 (score 6, accepted)
https://quant.stackexchange.com/a/63424
$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is bounded, a true martingale. Its quadratic variation is given by $$ \langle Y \left(\cdot\right)\rangle_t = \int_0^t{a^2 \left(s\right)\mathrm{d}s} $$ by definition of the stochastic integral with respect to the Wiener process.
Using Itō's lemma, $$ d \left[\left[Y \left(t\right)\right]^2\right] = 2 Y \left(t\right) d Y\left(t\right) + d \langle Y \left(\cdot\right)\rangle_t $$ Subtracting the differential of the time integral, i.e. $a^2 \left(t\right) \, dt$, removes the drift term due to $d \langle Y \left(\cdot\right)\rangle_t$ and you are done.
## Answer by ir7 (score 4)
https://quant.stackexchange.com/a/63437
Alternatively, we can use Ito isometry ($X$'s integrability and adaptability are assured by $a$'s boundness and adaptability, respectively):
$$E[X_t|{\cal F}_s] = E[X_s\big|{\cal F}_s] + E\left[\left(\int_s^t a_udW_u\right)^2 - \int_s^t a_u^2du \big|{\cal F}_s \right] $$
$$ = X_s + E\left[\left(\int_s^t a_udW_u\right)^2\big|{\cal F}_s\right] - E \left[ \int_s^t a_u^2du \big|{\cal F}_s \right] =X_s$$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.