If $X,Y$ are local martingales and $\lambda_n\to\lambda$, then $[X+\lambda_n Y]\to[X+\lambda Y]$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a complete filtration of $\mathcal A$
• $X$ and $Y$ be almost surely continuous local $\mathcal F$-martingales on $(\Omega,\mathcal A,\operatorname P)$

Let $$A_\lambda:=[X+\lambda Y]$$ and $$B_\lambda:=[X]+2\lambda[X,Y]+\lambda^2[Y]$$ for $\lambda\in\mathbb R$. By $(1)$, $$N:=\bigcup_{\lambda\in\mathbb Q}\left\{A_\lambda\ne B_\lambda\right\}$$ is a $\operatorname P$-null set.

Let $\lambda\in\mathbb R$. I want to prove that $$A_\lambda(\omega)=B_\lambda(\omega)\;\;\;\text{for all }\omega\in\Omega\setminus N\;.\tag2$$

Since $\mathbb Q$ is a dense subset of $\mathbb R$, $$\lambda_n\xrightarrow{n\to\infty}\lambda\tag3$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq\mathbb Q$. By definition, $$A_{\lambda_n}(\omega)=B_{\lambda_n}(\omega)\;\;\;\text{for all }\omega\in\Omega\setminus N\text{ and }n\in\mathbb N\;.\tag4$$ However, while it's obvious that $$B_{\lambda_n}\xrightarrow{n\to\infty}B_\lambda\tag5\;,$$ I fail to see why $$A_{\lambda_n}(\omega)\xrightarrow{n\to\infty}A_\lambda(\omega)\;\;\;\text{for all }\omega\in\Omega\setminus N\tag6\;.$$

So, how can we show $(2)$?

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$^1$ If $M$ is an almost surely continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$, then there is an $\mathcal F$-adapted stochastic process $[M]$ on $(\Omega,\mathcal A,\operatorname P)$ with

1. $[M]=0$
2. $[M]$ is continuous
3. $[M]$ is of locally bounded variation
4. $M^2-[M]$ is a local $\mathcal F$-martingale
5. $[M]$ is nondecreasing$^2$

By 1. - 4., $[M]$ is unique up to indistinguishability. If $N$ is an almost surely continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$, then $$[M+\lambda N]=[M]+2\lambda[M,N]+\lambda^2[N]\;\;\;\text{almost surely for all }\lambda\in\mathbb R\;.\tag1$$

Answer 1

It seems that claim (6) can never be established.

Reason:

Firstly, we need to clarify the notation $[X]$. By $[X]$, we mean a fixed but arbitrary choice of quadratic variation for the process $X$.

Fix a choice for each $[X]$, $[X,Y]$, and $[Y]$. Then for each $\lambda\in\mathbb{R}$, $B_{\lambda}$ is completely determined (not just up to indistinguishability).

For each $\lambda\in\mathbb{Q}$, fix a choice for $A_{\lambda}$. Then the set $N$ is completely determined and is independent from the choice of $A_{\lambda}$ for $\lambda\in\mathbb{Q}^{c}$.

Now let $\lambda_{0}\in\mathbb{Q}^{c}$. Choose a sequence $(\lambda_{n})$ in $\mathbb{Q}$ such that $\lambda_{n}\rightarrow\lambda_{0}$.

Fix a choice for $A_{\lambda_{0}}$.

Suppose that we have: For each $t\in[0,\infty)$ and $\omega\in N^{c}$, $$ \lim_{n\rightarrow\infty}A_{\lambda_{n}}(t,\omega)=A_{\lambda_{0}}(t,\omega). $$ Note that it is very likely (although I cannot prove it) to construct $N'\in\mathcal{A}$ with $P(N')=0$ and $N'\not\subseteq N$. For example, if there exists $\omega_{0}\in\Omega$ such that $\{\omega_{0}\}\in\mathcal{A}$, $P(\{\omega_{0}\})=0$, and $\omega_{0}\notin N$, we may define $N'=N\cup\{\omega_{0}\}$.

Now define another process $C$ by $$ C(t,\omega)=A_{\lambda_{0}}(t,\omega)+1_{[0,\infty)\times N'}(t,\omega). $$ Then $C$ and $A_{\lambda_{0}}$ are indistinguishable and hence $C$ is also a quadratic variation for $X+\lambda_{0}Y$.

However, the following is clearly false: For each $t\in[0,\infty)$ and $\omega\in N^{c}$, $$ \lim_{n\rightarrow\infty}A_{\lambda_{n}}(t,\omega)=C(t,\omega). $$

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For your original quesiton, it seems that we can prove in the following way:

Suppose that we are given a sequence $(\lambda_{n})$ of real numbers, $\lambda\in\mathbb{R}$ such that $\lambda_{n}\rightarrow\lambda$. Suppose that we have made a choice for $[X]$, $[X,Y]$, $[Y]$, and $[X+\lambda_nY]$.

For each $n$, there exists $\Omega_{n}\in\mathcal{A}$ with $P(\Omega_n)=1$ such that for any $t\in[0,\infty)$ and $\omega\in\Omega_{n}$ $$ [X+\lambda_{n}Y](t,\omega)=[X](t,\omega)+2\lambda_{n}[X,Y](t,\omega)+\lambda_{n}^{2}[Y](t,\omega). $$

Define $\Omega_{0}=\cap_{n}\Omega_{n}$. For each $t\in [0,\infty)$, $\omega\in\Omega_{0}$, clearly $$ \lim_{n\rightarrow\infty}[X+\lambda_{n}Y](t,\omega)=[X]+2\lambda[X,Y]+\lambda^{2}[Y](t,\omega). $$ It can be proved (I skip it) that $[X]+2\lambda[X,Y]+\lambda^{2}[Y]$ is a quadratic variation for the process $X+\lambda Y$. QED

Answer 2

What is true is that for each $\omega\notin N$ and each $\lambda$, the process $t\mapsto (X(t,\omega)+\lambda Y(t,\omega))^2 -B_\lambda(t,\omega)$ is a continuous local martingale, meaning that $B_\lambda$ is a version of $[X+\lambda Y]$.