# A Gaussian martingale which is the sum of two independent Gaussian non-semimartingales

Marc Yor

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## To cite this version:

Marc Yor. A Gaussian martingale which is the sum of two independent Gaussian non-semimartingales. Electronic Communications in Probability, 2015, 20, pp.1-5. 10.1214/ECP.v20-4034. hal-01263074

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A Gaussian martingale which is the sum of two independent Gaussian non-semimartingales*

Marc Yor†

Abstract

In this paper two examples of two independent centered Gaussian processes are given such that at least one of them is not a semimartingale but their sum is a martingale.

**Keywords:** Martingales; semimartingales; Gaussian processes; Brownian bridges.

**AMS MSC 2010:** 60G15; 60G44.

Submitted to ECP on December 31, 2014, final version accepted on September 28, 2015.

# 1 Certain mixed Fractional Brownian motions are semimartingales

In his thesis, P. Cheridito [1, 2] obtained the following remarkable result: if $(B_t, t \ge 0)$ and $(B_t^{(H)}, t \ge 0)$ denote two independent Gaussian processes, the first one being a Brownian motion, and the second one a fractional Brownian motion with Hurst parameter $H \in ]3/4, 1]$, i.e.,

$$E[B_t^{(H)}] = 0 \quad \text{and} \quad E[(B_t^{(H)} - B_s^{(H)})^2] = |t-s|^{2H}, \quad s, t \ge 0,$$

then, for every $\alpha \in \mathbb{R}$, the sum:

$$\Sigma_t^{(H)} = B_t + \alpha B_t^{(H)}, \quad t \ge 0,$$

is a semimartingale with respect to its own natural filtration.

Notice that, for $H = 1$, one has: $B_t^{(1)} = t\xi$, where $\xi$ is a standard Gaussian variable, and consequently, $(\Sigma_t^{(1)}, t \ge 0)$ is a semimartingale in the filtration $B_t^{(\xi)} := \sigma\{B_s, s \le t; \xi\}$, made right continuous, hence, a fortiori, with respect to its own filtration. However, for $H \in ]3/4, 1]$, $(B_t^{(H)}, t \ge 0)$ has zero quadratic variation, but infinite variation on any time interval, hence it is not a semimartingale with respect to its own filtration, which makes Cheridito's result remarkable.

**Note:** Throughout the rest of this paper, when we say that a process $(\Pi_t, t \ge 0)$ is a semimartingale with no further qualification, we mean: semimartingale with respect to its own filtration made right continuous and P-complete.

<footnote>*Handwritten by Marc Yor in 2001. Typed and lightly edited by Patrick Cheridito in 2014.</footnote>

<footnote>†Université Paris VI, France.</footnote>
## 2 Some related questions

In the light of Cheridito's result, one may ask the following question:

(*) to give a "simpler" example of a pair of independent centered Gaussian processes, $(X_t, t \ge 0)$ and $(Y_t, t \ge 0)$, one of which at least is not a semimartingale, but such that the sum is a semimartingale.

In Section 3, we shall give an example where $(X_t, t \ge 0)$ is constructed from a Brownian bridge, and is not a semimartingale whereas $(Y_t, t \ge 0)$ has bounded variation. In Section 4, pushing the construction of Section 3 one step further, we shall give another example of (*), where neither $(X_t)$ nor $(Y_t)$ is a semimartingale. For the moment, we simply note that, in order to obtain some positive answer to (*), at least one of the Gaussian processes $(X_t)$ or $(Y_t)$ must have some non-zero quadratic variation, i.e., $\sum_{\tau_n} (\Delta X_{t_i})^2$ does not converge to 0, where $\tau_n = \{0 = t_0 < t_1 < \cdots < t_{p_n} = 1\}$, $\Delta X_{t_i} = X_{t_i} - X_{t_{i-1}}$, and $\sup_{\tau_n} (t_i - t_{i-1}) \xrightarrow{(n \to \infty)} 0$. This assertion follows from the

**Lemma 2.1.**

(i) Assume that $X$ and $Y$ are two independent centered Gaussian processes, and $\tau$ is a subdivision of $[0, 1]$. Then

$$ \max \left( E \left[ \sum_{\tau} |\Delta X_{t_i}| \right] ; E \left[ \sum_{\tau} |\Delta Y_{t_i}| \right] \right) \\ \le E \left[ \sum_{\tau} |\Delta(X + Y)_{t_i}| \right] \le E \left[ \sum_{\tau} |\Delta X_{t_i}| + \sum_{\tau} |\Delta Y_{t_i}| \right]. $$

(ii) If both, $X$ and $Y$, have zero quadratic variation and at least one of them has infinite variation on a set of positive probability, then $X + Y$ also enjoys these two properties.

*Proof.* (i) Only the LHS inequality needs to be proven; but this follows from

$$ E [ |\Delta(X + Y)_{t_i}| ] = \sqrt{\frac{2}{\pi}} \| \Delta X_{t_i} + \Delta Y_{t_i} \|_2 \ge \sqrt{\frac{2}{\pi}} \| \Delta X_{t_i} \|_2 = E [ |\Delta X_{t_i}| ]. $$

(ii) It is clear that $X + Y$ has zero quadratic variation. On the other hand, it follows from

(i) and our hypothesis in (ii) that

$$ E \left[ \int_{0}^{1} |d(X_s + Y_s)| \right] = \infty. $$

Now it follows from Fernique's integrability result for the norms of Gaussian vectors that $\int_{0}^{1} |d(X_s + Y_s)|$ cannot be finite a.s. $\square$

## 3 Brownian bridges and a first solution to (*)

Let $u > 0$, and denote by $(\eta_u(t), t \le u)$ a Brownian bridge of length $u$, i.e., $(B_t, t \le u)$ conditioned to be equal to 0 at time $u$. Recall that it can be realized as $\eta_u(t) = B_t - \frac{t}{u} B_u$, $\eta_u$ is independent of $B_u$, and its canonical decomposition is

$$ \eta_u(t) = \beta_t - \int_0^t ds \frac{\eta_u(s)}{u-s}, \quad t \le u, \qquad (3.1) $$

where $(\beta_t, t \le u)$ is a Brownian motion in the filtration $(\mathcal{P}_t^{(u)}, t \le u)$ of $\eta_u$. Furthermore, there is the following
A Gaussian martingale which is the sum of two Gaussian non-semimartingales

**Proposition 3.1.** Let $f \in L^2([0, u])$. Then

(i) The process

$$ \int_{0}^{t} f(s) d\eta_{u}(s) = \int_{0}^{t} f(s) d\beta_{s} - \int_{0}^{t} ds f(s) \frac{\eta_{u}(s)}{u-s} $$

is well defined for any $t \le u$ with

$$ \int_{0}^{u} f(s) d\eta_{u}(s) = (L^2 \text{ and a.s.}) \lim_{t \uparrow u} \int_{0}^{t} f(s) d\eta_{u}(s). $$

(ii) $(\int_{0}^{t} f(s) d\eta_{u}(s), t \le u)$ is a semimartingale with respect to $(\mathcal{P}_{t}^{(u)}, t \le u)$ if and only if

$$ \int_{0}^{u} ds |f(s)| \frac{1}{\sqrt{u-s}} < \infty. $$

*Proof.* (i) The $L^2$ and a.s. convergence results are easily obtained from the representations of $\eta_u$ as $\eta_u(t) = B_t - \frac{t}{u}B_u$.

(ii) The semimartingale property of $(\int_{0}^{t} f(s) d\eta_{u}(s), t \le u)$ is clearly equivalent to

$$ \int_{0}^{u} ds |f(s)| \frac{|\eta_{u}(s)|}{u-s} < \infty. $$

The arguments developed in the proof of Theorem 3 in Jeulin and Yor [3] show that this is equivalent to

$$ \int_{0}^{u} ds |f(s)| \frac{1}{\sqrt{u-s}} < \infty. $$

□

In order to give explicit examples for (*) in the sequel of this paper, let us point out that for $u \in ]0, 1]$ and $\alpha \in ]1/2, 1]$, the function

$$ \psi(s) = \frac{1}{\sqrt{u-s}} |\log(u-s)|^{-\alpha} 1_{(u/2<s<u)} $$

satisfies

$$ \int_{0}^{u} ds \psi^{2}(s) < \infty \quad \text{but} \quad \int_{0}^{u} ds \psi(s) \frac{1}{\sqrt{u-s}} = \infty. $$

To obtain a solution to (*), we decompose a Brownian motion $(B_t, t \le u)$ as

$$ B_t = \eta_u(t) + \frac{t}{u} B_u, \quad t \le u, $$

and we consider $f_* \in L^2([0, u])$ such that

$$ \int_{0}^{u} ds |f_{*}(s)| \frac{1}{\sqrt{u-s}} = \infty \quad \text{and} \quad f_{*}(s) \neq 0 \text{ for every } s. $$

Then, taking

$$ X_t = \int_0^t f_*(s) d\eta_u(s) \quad \text{and} \quad Y_t = \frac{B_u}{u} \int_0^t f_*(s) ds, $$

we obtain a solution to (*) since X and Y are independent and $X_t + Y_t = \int_0^t f_*(s) dB_s$ is a martingale.
A Gaussian martingale which is the sum of two Gaussian non-semimartingales

## 4 A "full" solution to (*)

Let $u \in ]0, 1[$. We shall use the same idea as in Section 3, but twice instead of once, by decomposing first $(B_t, t \le u)$ into $\eta_u(t) + \frac{t}{u}B_u$, and then

$$ (\hat{B}_t \equiv B_{t+u} - B_u, t \le 1-u) \quad \text{into} \quad \hat{\eta}_{1-u}(t) + \frac{t}{1-u} \hat{B}_{1-u}. \qquad (4.1) $$

Next, for $f \in L^2([0, 1])$, we write

$$ \begin{aligned} \int_0^t f(s)dB_s &= \int_0^t f(s)1_{(s \le u)}dB_s + 1_{(u < t)} \int_u^t f(s)dB_s \\ &= \int_0^t f(s)1_{(s \le u)}d\eta_u(s) + \frac{B_u}{u} \int_0^t f(s)1_{(s \le u)}ds \\ &\quad + 1_{(u < t)} \int_u^t f(s)d\hat{\eta}_{1-u}(s-u) + 1_{(u < t)} \frac{B_1 - B_u}{1-u} \int_u^t f(s)ds. \end{aligned} $$

We then choose $f_* \in L^2([0, 1])$ such that

$$ \int_0^u |f_*(s)| \frac{ds}{\sqrt{u-s}} = \infty, \quad \int_u^1 |f_*(s)| \frac{ds}{\sqrt{1-s}} = \infty \quad \text{and} \quad f_*(s) \neq 0 \text{ for all } s < 1. $$

Then

$$ X_t = \int_0^t f_*(s)1_{(s \le u)}d\eta_u(s) + 1_{(u < t)} \frac{B_1 - B_u}{1-u} \int_u^t f_*(s)ds $$

and

$$ Y_t = 1_{(u < t)} \int_u^t f_*(s) d\hat{\eta}_{1-u}(s-u) + \frac{B_u}{u} \int_0^t f_*(s) 1_{(s \le u)} ds $$

are two independent Gaussian processes such that $X_t + Y_t = \int_0^t f_*(s)dB_s$ is a martingale. Using the semimartingale characterization in part (ii) of Proposition 3.1, it is easily shown that neither $X$ nor $Y$ is a semimartingale. However, we give a few details:

Concerning $(X_t)$, we see that $X_t = \tilde{X}_t$ for $t \le u$, where $\tilde{X}_t = \int_0^t f_*(s)1_{(s \le u)}d\eta_u(s)$. Hence the non-semimartingale property of $X$ follows from that of $\tilde{X}$ as discussed in Section 3.

Concerning $(Y_t)$, we have

$$ Y_u = \frac{B_u}{u} \int_0^u f_*(s)ds \quad \text{and} \quad Y_t - Y_u = \int_u^t f_*(s)d\hat{\eta}_{1-u}(s-u), \quad t \in [u, 1]. $$

Now $Y$, being a Gaussian process, could only be a semimartingale if it were a quasi-martingale; see, e.g., Stricker [4]. If

$$ \mathcal{Y}_{u+t} = \sigma\{B_u, \hat{\eta}_{1-u}(s), s \le t\} $$

and $(\hat{\mathcal{P}}_t^{1-u})$ is the filtration of $\hat{\eta}_{1-u}$, it follows from the independence of $B_u$ and $\hat{\eta}_{1-u}$ that for $s < t$:

$$ E[Y_{u+t} - Y_{u+s} | \mathcal{Y}_{u+s}] = E[Y_{u+t} - Y_{u+s} | \hat{\mathcal{P}}_s^{1-u}]. $$

From Section 3 we know that $(Y_t - Y_u)$ is not a $\hat{\mathcal{P}}^{1-u}$-semimartingale. So it is not a $\hat{\mathcal{P}}^{1-u}$-quasimartingale. It follows that $(Y_t)$ is not a $\mathcal{Y}$-quasimartingale and therefore, also not a $\mathcal{Y}$-semimartingale.
A Gaussian martingale which is the sum of two Gaussian non-semimartingales

References

[1] Cheriditio P. (2001). Regularizing Fractional Brownian Motion with a view Towards Stock Price Modelling. PhD Thesis. ETH Zürich.

[2] Cheridito P. (2001). Mixed fractional Brownian motion. *Bernoulli* 7(6), 913–934. MR-1873835

[3] Jeulin, T. and Yor, M. (1979). Inégalité de Hardy, semimartingales, et faux-amix. *Séminaire de probabilités de Strasbourg* 13, *Lect. Notes in Math.* 721, 332–359. MR-0544805

[4] Stricker C. (1984). Quelques remarques sur les semimartingales Gaussiennes et le problème de l'innovation. *Lecture Notes in Control and Information Sciences* 61, 260–276. MR-0874835
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