I would like to know if in some book or how could I compute the quadratic variation of the supremum of the bronian motion $S_t=\sup_{s\in[0,t]}W_s$ where $W$ is a Brownian motion. I was thinking about using the fact that $S_t\overset{d}{=}|W_t|$ but also I don't know if two martingales with the same distribution has the same quadratic variation (I know that the converse is not true). Anyone could recomend some lectures about these subjects please, or give some advice about how to start the proofs I'll really appreciate it. Thanks.
Quadratic Variation of Supremum of Brownian Motion – Probability
brownian motionpr.probability
Related Solutions
Assuming that $\mu$ and $S_0$ are positive, the process stays almost surely non-negative. This is easily seen as when $S$ hits zero, it has a deterministic drift upwards. However, the process does not necessarily stay strictly positive, the hitting probability of zero can be positive.
This model is well studied in the financial mathematics literature and goes there under the name CEV model. The generic reference on it and the close relationship to Bessel processes is the following paper by Delbaen and Shirakawa: A Note of Option Pricing for Constant Elasticity of Variance Model, https://people.math.ethz.ch/~delbaen/ftp/preprints/CEV.pdf
First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtration of the local martingale (i.e., the smallest filtration w.r.t. which $X$ is a local martingale).
Second, your prove/disprove statement does not consist of one, but of multiple claims. I will try to disentangle them first before discussing their correctness. I will refer throughout to the the text Continuous Martingales and Brownian Motion which you mention, too.
(1) Let $X$ be a continuous local martingale, then there exists a Brownian motion $B$ and a predictable process $\xi$ such that $X_t = B_{\int_0^t \xi_s^2 \, ds}$ for all $t \in [0, \infty)$.
(2) Let $X$ be a continuous local martingale, then there exists a Brownian motion $\tilde{B}$ and a predictable process $\tilde{\xi}$ such that $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$ for all $t \in [0, \infty)$.
(3) $\xi$ and $\tilde{\xi}$ as well as $B$ and $\tilde{B}$ agree in an appropriate sense, say, are versions of each other.
(4) The quadratic variation of $\langle X \rangle_t$ is then given by $\int_0^t \xi^2_s \, ds$ (resp. $\int_0^t \tilde{\xi}^2_s \, ds$).
(5) It holds true that $\xi$ is increasing and of finite variation.
(6) The statement (3) holds true at least in the case when $\xi$ is is deterministic.
Lets now asses the claims in detail:
(1) This statement holds true as long as $\langle X \rangle_\infty = \infty$, it is called Dambis -- Dubins-Schwarz theorem (cf. R-Y, Thm. V.1.6). Without the assumption this is not necessarily true in the natural filtration (as it might be not rich enough to support a Brownian motion). However, it is possible to enlarge the probability space resp. the filtration such that this statement is true in the larger space (cf. R-Y, Thm V.1.7).
(2) This is the \textit{martingale representation theorem}. It is usually formulated in the context of Brownian filtrations, i.e., it is a priori assumed that $X$ is a local martingale with respect to a filtration generated by a Brownian motion. However, it holds true without this assumption as long as the measure generated by the quadratic variation process, $d\langle X \rangle_t$, is equivalent to the Lebesgue measure $dt$. If there is no equivalence between the measures but at least $d\langle X\rangle_t$ is absolutely continuous with respect to the Lebesgue measure, one can save again the situation by enlarging the probability space (cf. R-Y, V.3.8 and discussion thereafter).
Without absolute continuity the statement is false as can be seen on the following example. Let $C_t^i$ be copies of the Cantor function on the unit interval $[0,1)$ and define the extension to the positive halfline by
$$ K_t = \sum_{i=0}^\infty i + C_{t-i}^i {{\mathchoice{1\mskip-4mu\mathrm l}{1\mskip-4mu\mathrm l} {1\mskip-4.5mu\mathrm l}{1\mskip-5mu\mathrm l}}}_{[i,i+1)}(t) $$
Let $W$ be a Brownian motion and define $X$ by $X_t = W_{K_t}$. This is clearly a martingale with quadratic variation $\langle X \rangle_t = K_t$. However, if $X$ could be written $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$, then we would have by (4) $\langle X \rangle_t = \int_0^t \tilde{\xi}^2_s \, ds$. However, this is always an absolutely continuous function whereas $K$ is not, therefore they cannot agree.
(3) From the restrictions above it is clear that this cannot hold true in general. But you can even find very simple counterexamples, e.g., set $\tilde{\xi} =1$. Then you have
$$ \int_0^t (-1) \, d\tilde{B}_s = - \tilde{B}_t = B_t = B_{\int_0^t 1^2 \, ds} = B_{\int_0^t (-1)^2 \, ds}$$
So in this case $\tilde{B} = - B$ whereas $\xi$ is not uniquely determined (e.g., it could be $1$ or $-1$). This can be of course made worse using processes jumping between $1$ and $-1$. (4) This is true and in both cases (direct calculation using properties of Brownian motion and R-Y, Prop. IV.2.7 resp.).
(5) Every increasing process is of finite variation by definition. However, the claimed result is clearly not true, just think about an arbitrary integral with deterministic non- monotone integrand, i.e. $X_t = \int_0^t \sin{(s)} \, \, dW_s$ for some Brownian motion $W$. I assume this is simply a typo and you intended to claim that the quadratic variation itself is increasing (or rather: non-decreasing) and thus of bounded variation (which is evidently true, you just add more squares).
(6) No, the example from (3) or the Cantor function example from (2) are counterexamples.
Let me conclude with two remarks about the context:
(a) The idea to place stochastic calculus on change of time instead of stochastic integration has some interesting history. Actually, it predates Ito calculus and was the first form of stochastic calculus, developed by Wolfgang Döblin (Vincent Doblin) in his famous lettre scellée to the French Academy of Science, https://mathoverflow.net/a/100040/20026.
(b) I disagree with your statement that local martingale is a misnomer. Yes, a local martingale is much more general than a martingale. But it is intentionally not called generalized martingale but local martingale, as it is locally (i.e. up to an increasing sequence of stopping times) a martingales. Generalizations that resemble locally the original object are very often far more general than the original object. The gap between manifolds and Euclidean spaces is for sure much larger than that between true and local martingales.
Best Answer
The quadratic variation is identically $0$, i.e.
$$\langle S, S \rangle_t = 0$$
for all $t$, almost surely.
To see this, note that $S$ is almost surely increasing, hence has bounded variation almost surely. Thus the quadratic variation is simply equal to the sum of the squared jumps of the process. However, $S$ is also almost surely continuous, so the quadratic variation is identically $0$.
Some comments are in order:
The formula you gave for $S$ in terms of $|W|$ does not hold. The correct equivalence is $D_t := S_t - W_t \overset{d}{=} |W_t|$. For $S$ itself, we have that $S_t \overset{d}{=} 2 L_t(0)$, where $L_t (0)$ denotes the local time of $W$ at $0$.
Neither $S$ nor $D$ are martingales. However, $S$ is almost surely increasing, hence trivially a submartingale. On the other hand $D$ is a semimartingale, with semimartingale decomposition given by Tanaka’s formula applied to $|W_t|$. Some justification is required as to why two processes with the same law admit the same (in law) semimartingale decomposition.