Brownian Motion Products, Scaled Normals, and Martingale Increments
Summary
The document presents three probability questions with brief calculations. First, Itô's product rule shows that the product of two Brownian motions has a covariation drift; it is a martingale when that covariation vanishes, as with independent Brownian motions. Second, it rejects the claim that multiplying a standard normal variable by the square root of time automatically creates Brownian motion, using an increment-variance calculation to show the stated construction fails. The notation assumes a particular relationship between the normal variables at different times, which is not fully specified.
Third, it argues that nonoverlapping increments of a martingale are uncorrelated by conditioning the later increment on information available by the end of the earlier one. This result requires suitable integrability, such as square-integrable increments, for the expectations and products to be well-defined. The material is a set of proof sketches rather than a complete treatment of filtrations or regularity conditions.
Key ideas
- The product of two Brownian motions acquires a drift from their quadratic covariation.
- For independent Brownian motions, that covariation is zero and their product is a martingale.
- A time-scaled standard normal marginal alone does not establish the independent increments required of Brownian motion.
- Martingale increments over disjoint intervals are uncorrelated under appropriate integrability conditions.
- Proofs involving conditional expectations of products require assumptions ensuring those quantities exist.
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Full text
# Three proofs regarding brownian motions and martingales
# Three proofs regarding brownian motions and martingales
1. Let $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ be two standard Brownian motions and let $X_t := B_t W_t$. Is $(X_t)_{t \geq 0}$ a martingale?
The easiest way to proceed seems to be to apply Ito's Lemma, from which we get: \begin{align} dX_t = B_t dB_t + W_t dW_t + d<B, W>_t \end{align}
Hence, $X_t$ is a sum of two Ito integrals (which are martingales), a constant ($X_0$) and a last term, $d<B, W>_t$. $X$ should only be a martingale if the last term is null. In general, this term should be $d<B, W>_t = \rho dt$ and its magnitude depends on the correlation between the two Brownian motions. In general, the answer should thus be no, with the exception being two independent Brownian motions.
2. Let $Z_t \sim N(0,1)$ and $X_t = \sqrt{t} Z_t$. Is $X_t$ a standard Brownian motion?
We have:
\begin{align} E(X_{t+s} - X_t) = 0 \\ V(X_{t+s} - X_t) = E \left( (t+s)Z_{t+s}^2 + (t)Z_t^2 - \sqrt{t}\sqrt{t+s}Z_{t+s} Z_t \right) \\ \leftrightarrow V(X_{t+s} - X_t) = (t + s)(1) + t(1) + 0 = 2t + s \neq s. \end{align}
Hence, this isn't a standard Brownian motion.
3. Are disjoint increments of a martingale uncorrelated?
A martingale is a measurable and integrable process $(M_t)_{t \geq 0}$ with the following property $E(M_t | F_s) = M_s \; \forall t \geq s \geq 0$ from which we can get $E(M_t - M_s| F_s) = 0$. Let $0 < i < j < k < l$, then \begin{align} E((M_j - M_i)(M_l - M_k)) &= E\left( M_j (M_l - M_k) - M_i(M_l - M_k) \right) \\ &= E\left( M_j (M_l - M_k) \right) - E \left(M_i(M_l - M_k) \right) \\ &= E\left( M_j E(M_l - M_k | F_j) \right) - E \left(M_i E(M_l - M_k| F_i) \right) \\ &= 0 + 0 = 0 \end{align}
Hence, the answer is yes.
Am I correct or did I make a mistake somewhere?
## Answer by Daneel Olivaw (score 3)
https://quant.stackexchange.com/a/50143
The proofs seem good to me. An alternative answer to the first question is, with $s<t$:
$$\begin{align} E\left(B_tW_t|\mathscr{F}_s\right)&= E\left((B_t-B_s+B_s)(W_t-W_s+W_s)|\mathscr{F}_s\right) \\ &= E\left((B_t-B_s)(W_t-W_s)+(B_t-B_s)W_s+B_s(W_t-W_s)+B_sW_s|\mathscr{F}_s\right) \\ &= E((B_t-B_s)(W_t-W_s))+W_sE(B_t-B_s)+B_sE(W_t-W_s)+B_sW_s \\ &= E((B_t-B_s)(W_t-W_s))+B_sW_s \end{align}$$
Hence $(B_tW_t)_{t\geq0}$ is a martingale if the correlation between the two Brownian Motions is null.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.