Htykuut

I have to show that the following process $(X_t)_{tin [0,infty)}$ is no martingale.

Let $Y_n$ be a sequence of independent random variables with
$$P(Y_n=n)=frac{1}{2n^2},quad P(Y_n=-n)=frac{1}{2n^2},quad P(Y_n=0)=1-frac{1}{n^2}.$$
We put
$$X_t=sum_{left{n,;, 1-tleq frac 1{n}right}} Y_nquadtext{for all } tgeq 0.$$

An ansatz would be to assume that is a martingale and then, by the fundamental theorem of local martingales, you can decompose $X$ as
$$X=A+B$$
where $A$ is a local martingale with bounded jumps and $B$ is a process of locally integrable variation.
But we have
$$sum_{{sleq t,:,Delta X_sgeq 1}}|Delta X_s|=infty,$$
which contradicts $B$ having integrable variation. But it does not necessarily contradict $B$ having locally integrable variation.

Any help is appreciated!

Set

$$S_k := sum_{n=1}^k Y_n.$$

Since

$$sum_{n geq 1} mathbb{P}(Y_n neq 0) = sum_{n geq 1} frac{1}{n^2} < infty,$$

it follows from the Borel Cantelli lemma that for almost all $omega in Omega$ we have $Y_n(omega)=0$ for all $n geq N(omega)$ sufficiently large. This implies, in particular, that

$$S_{infty} := lim_{k to infty} S_k = sum_{n geq 1} Y_n$$

exists almost surely (as a pointwise limit).

Claim: $S_{infty}$ is not integrable. In particular, $(S_n)_{n in mathbb{N} cup {infty}}$ is not a martingale.

Proof: For $n in mathbb{N}$ define $$A_n := {Y_n = n} cap bigcap_{k neq n} {Y_k=0}.$$ Since $$a:=prod_{k geq 1} left(1- frac{1}{k^2} right) > 0,$$ it follows from the independence of the random variables $(Y_k)_{k in mathbb{N}}$ that $$ mathbb{P}(A_n) = mathbb{P}(Y_n=n) prod_{k neq n} mathbb{P}(Y_k=0) geq mathbb{P}(Y_n=n) prod_{k geq 1} mathbb{P}(Y_k=0)= a frac{1}{2n^2}$$ for all $n in mathbb{N}$. As the sets $A_n$ are pairwise disjoint, we find from the monotone converence theorem that $$begin{align*} mathbb{E}(|S_{infty}|) geq mathbb{E} left(|S_{infty}| sum_{n in mathbb{N}} 1_{A_n} right) &= sum_{n in mathbb{N}} mathbb{E}(|S_{infty}| 1_{A_n}) \ &= sum_{n in mathbb{N}} n mathbb{P}(A_n) \ &geq a sum_{n in mathbb{N}} n frac{1}{2n^2} = infty end{align*}$$ which proves the assertion.

Finally, we note that by the very definition of $(X_t)_{t geq 0}$, it holds that

$$X_1 = sum_{n in mathbb{N}} Y_n = S_{infty};$$

since we have shown that $X_1$ fails to be integrable, the process $(X_t)_{t geq 0}$ is not a martingale.