Stochastic integral of bounded variation martingale

bounded-variationstochastic-integrals

I am having trouble wrapping my head around a provided answer regarding a stochastic integration question, and have had no luck searching the forums for identical questions so I will post this here. There were some pre-requisites to the exercise so I'll add the items below respectively.

a) Assume that M = $(M_t)_{t \geq 0}$ is a bounded cts. martingale of finite variation $\implies M$ is a constant.

b) The question then proceeded to defining a new process N = $(N_t)_{t \geq 0}$ by $N_t = M_t^4$ and asked to express N in terms of stochastic integrals w.r.t. M and [M] which I did:
$$
N_t – N_0 = 4\int_{0}^t M_s^3 dMs+ 6 \int_0^t M_s^2d[M]s
$$

The question then asked to show that: if N is also a martingale, then M must be a constant.

Now I understand that in order for N to be a martingale, I must have that the bounded variation integral:

$\int_0^t M_s^2d[M]s \equiv 0$ (i.e. a process without the drift term – which I get)

The solution then states that either $M \equiv 0$ or $[M]_t = c \in \mathbb{R}$ $\forall t$, which brings me to my first question: why is the quadratic variation of the process M constant? I get the feeling I am missing something incredibly basic here but can't quite figure it out.

The solution then continues as follows:

but since $\mathbb{E}[(M_t^2 – M_s^2)] = \mathbb{E}[M_t^2 – M_s^2] = \mathbb{E}[[M]_t -[M]_s] =0$ $\forall s\leq t$ we must have $M_t = M_s$

Which brings me to my other question- why are we now considering $M_t^2 – M_s^2$ and how exactly does this show that M is constant?

Any help/pointers on the above would be greatly appreciated, as I've been at this for a few hours now to no avail!

Best Answer

After some given thought I managed to deduce the solution for anybody who is interested:

  1. The reason why we take $[M]_t = c$ is because it will have differential $= 0$ i.e. $d[M]_t = 0$

  2. If we show that for all possible pairs of s,t that the value of $M_t = M_s$ everywhere, this concludes that the process itself must be constant in value and we can simply take $s=0$ and deduce that $M_t = M_0$ $\forall t$

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