Show that $I(t),0\leq t\leq T$ is a martingale

martingalesprobabilityself-learningstochastic-calculus

Exercise 4.1 Suppose $M(t),0\leq t \leq T$ is a martingale with respect to some filtration $\mathcal{F}(t),0\leq t, \leq T$. Let $\Delta(t),0\leq t \leq T$, be a simple process adapted to $\mathcal{F}$ (i.e., there is a partition $\mathcal{\Pi}=\{t_0,t_1,\dots\,t_n\}$ of $[0,T]$ such that, for every $j$, $\Delta(t_j)$ is $\mathcal{F}(t_j)$-measurable and $\Delta(t)$ is a constant in $t$ on each subinterval $[t_j,t_j+1))$. For $t\in[t_k,t_{k+1})$, define the stochastic integral
$$I(t)=\sum^{k-1}_{j=0}\Delta(t_j)[M(t_{j+1})-M(t_j)]+\Delta(t_k)[M(t)-M(t_k)].$$
We think of $M(t)$ as the price of an asset at time $t$ and $\Delta(t_j)$ as the number of shares of the asset held by an investor between times $t_j$ and $t_{j+1}$. Then $I(t)$ is the capital gains that accrue to the investor between times $0$ and $t$. Show that $I(t),0\leq t\leq T$, is a martingale.

Source: Stochastic Calculus for Finance II by Steve Shreve, pg. 189

My Work So Far

Proof. Set $t$ for any $s\lt t$ and we assume $s\in[t_m,t_{m-1}]$

  • Case 1:$\ m=k$, thus
    $$I(t)-I(s)=\Delta_{t_k}(M_t-M_{t_k})-\Delta_{t_k}(M_s-M_{t_k})=\Delta_{t_k}(M_t-M_s)$$
    So,
    $$\mathbb{E}[I(t)-I(s)|\mathcal{F}_t]=\Delta_{t_k}\mathbb{E}[M_t-M_s|\mathcal{F}_s]=0$$
  • Case 2:$\ m\lt k$, thus
    $$t_m\lt s \lt t_{m+1}\leq t_k\leq t \leq t_{k+1}$$
    So,
    \begin{align*}
    I(t)-I(s)&=\sum^{k-1}_{j=m}\Delta_{t_j}(M_{t_{j+1}}-M_{t_j})+\Delta_{t_k}(M_s-M_{t_k})-\Delta_{t_m}(M_s-M_{t_m})\\
    &=\sum^{k-1}_{j=m+1}\Delta_{t_j}(M_{t{_j+1}}-M_{t_j})+\Delta_{t_k}(M_t-M_{t_k})+\Delta_{t_m}(M_{t_{m+1}}-M_s).\\\\
    \end{align*}
    Therefore, we note that
    \begin{align*}
    &\mathbb{E}[I(t)-I(s)|\mathcal{F}_s]\\
    &=\sum^{k-1}_{j=m+1}\mathbb{E}[\Delta_{t_j}\mathbb{E}[M_{t_{j+1}}-M_{t_j}|\mathcal{F}_{t_j}]|\mathcal{F}_s]+\mathbb{E}[\Delta_{t_k}\mathbb{E}[M_t-M_{t_k}|\mathcal{F}_{t_k}]|\mathcal{F}_s]+\Delta_{t_m}\mathbb{E}[M_{t_{m+1}}-M_s|\mathcal{F}_s]\\
    &=0.\\
    \end{align*}
    When combined, we can conclude that $I(t)$ is a martingale. $\square$

Would this be correct? Is there a better way to prove this?

Best Answer

You have a few typos (for example, in case 1 in the second line of equalities the first conditioning should be on $\mathcal{F}_s$; in case 2 you on the contrary use index $s$ instead of $t$ in the expression for $I(t)$ in the first line of equalities), but other than that everything is correct. I also don't think there is a better way to show it. If you are interested, the same way (though with omitting some details) is used to prove this statement in another book of Steven Shreve, namely "Brownian Motion and Stochastic Calculus" by I.Karatzas and S.Shreve (formula (2.13) on page 137 and its proof on page 138).

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