Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 29 / 2

The intended setting is $X=M+A$ with $A$ continuous and adapted. These conditions are needed for the requested continuous-semimartingale conclusion; finite variation alone does not supply them. For example, $M=0$, $A_t=1_{\{t\ge 1\}}$ satisfies the printed deterministic bound and starts at zero, but $f_n(X)$ has a jump for sufficiently large $n$. We use the continuous adapted interpretation throughout this question.

Define $f_n(x)={\int }_0^x\varphi (ny)dy$. Then $f_n\in C^2$, $|f_n^{\prime }|\le 1$ and $f_n^{\prime \prime }(x)=n{\varphi }^{\prime }(nx)\ge 0$. The Itô formula gives the canonical continuous semimartingale decompositionThe initial value is zero. The first term is a continuous martingale, since $M$ is bounded and the integrand is bounded. The second is continuous adapted finite variation, hence predictable. This decomposition is unique because a continuous finite-variation local martingale is constant. The phrase Doob-Meyer here must mean this semimartingale decomposition: $V^{(n)}$ need not be nondecreasing, and the hypotheses do not make $f_n(X)$ a submartingale as required by the Doob-Meyer decomposition theorem. Indeed $f_n^{\prime }(0)=-1$, so the bounded deterministic drift $A_t=t\wedge \delta$ makes $f_n(A_t)$ initially decrease, excluding the submartingale property.

The sign process is predictable because it is a Borel function of the continuous adapted, and therefore predictable, process $X$. Its paths need not themselves be left-continuous. Pointwise, including at $x=0$,For a fixed horizon $T$, $\mathbb{E}[M{]}_T=\mathbb{E}{M}_T^2\le K^2$. The dominated convergence theorem for the finite measure $\mathbb{E}d[M]$ on $[0,T]$ givesThe Itô isometry, extended from simple predictable integrands by their density, and the Doob L2 maximal inequality now implyHence

Fix a sample path on the probability-one event where $X$ is continuous and the total-variation process of $A$ satisfies $V_{\infty }\le K$. The signed measure $dA$ has absolute variation measure $dV$. ThusThe integrand tends to zero at every time, including times when $X_s=0$, and is bounded by $2$. Since $dV$ is a finite measure on this path, the dominated convergence theorem makes the right-hand side tend to zero. ThereforeIn fact the convergence is uniform over the entire time axis under the given total-variation bound. This pathwise estimate establishes one common full-probability event, rather than separate exceptional events for each time.

For $x\le 0$, $f_n(x)=|x|$; for $x>0$, only the interval $(0,1/n)$ contributes an error to the integral of the derivative. HenceBy the Itô formula, the continuous processesare nondecreasing, since $f_n^{\prime \prime }\ge 0$. Parts (a) and (b), together with the uniform approximation of absolute value, show that $C^{(n)}$ converges uniformly on compacts in probability toThe limit is continuous: the martingale integral is continuous, and the bounded-integrand finite-variation integral is continuous because $A$ is continuous. For any rational $s<t$, convergence in probability and $C_t^{(n)}-C_s^{(n)}\ge 0$ imply $Z_t-Z_s\ge 0$ almost surely. Intersect these countably many events and use continuity to extend to all real $s<t$. ThusSince $|X|={\mathrm{sgn}}_{-}(X)\cdot M+({\mathrm{sgn}}_{-}(X)\cdot A+Z)$ is a continuous local martingale plus continuous finite variation, $|X|$ is a continuous semimartingale. This is the smooth convex approximation proof of the Tanaka formula. With the specified sign value $-1$ at zero, $Z$ is the right local time at zero, rather than an unspecified symmetric normalization.

For an arbitrary continuous semimartingale starting at zero, use its continuous decomposition $X=M+A$, with both parts starting at zero, and letThe continuity and local finiteness of variation imply ${\rho }_k\uparrow \infty$ almost surely. Up to ${\rho }_k$ the required deterministic bound holds, and the stopped local martingale is bounded, hence a true $L^2$-bounded martingale. Apply the previous argument to $X^{{\rho }_k}$ for each integer $k$. Stopping the defining integrals gives precisely $Z_{t\wedge {\rho }_k}$. On the intersection of their full-probability events, each stopped $Z$ is nondecreasing; every finite interval is eventually contained before ${\rho }_k$, so $Z$ is nondecreasing globally. Therefore