# Is $Y_n=S_n^3-3nS_n$ a martingale?

Let $$X_i$$ for $$i=1,\cdots,n$$ random variables i.i.d., such that $$P(X_1=1)=\frac{1}{2}$$ and $$P(X_1=-1)=\frac{1}{2}$$. Let $$S_n=X_1+\cdots+X_n$$ Let $$Y_n=S_n^3-3nS_n$$, is $$Y_n$$ a martingale or isn't?

Attempt: We have $$\mathbb{E}(X_1)= 1\cdot \frac{1}{2}-1\cdot\frac{1}{2}= 0$$. I also have verified that $$\mathbb{E}(|Y_n|)<\infty$$ and that taking $$h_n(X_1,\cdots,X_n)=(X_1+\cdots+X_n)^3-3n(X_1+\cdots+X_n)=Y_n$$ so is medible. If we can prove that $$\mathbb{E}(S_n^3-3nS_n|X_1,\cdots, X_{n-1})=Y_{n-1}=S_{n-1}^3-3(n-1)S_{n-1}$$ then we are done, but not sure how to proceed, because $$Y_n$$ is unknow at filtration at time $$X_{n-1}$$.

• What exactly is $S_{n}$?
– B_B
Commented May 8, 2021 at 17:03
• @B_B Is a random walk sum. Commented May 8, 2021 at 17:04
• OK, you have edited it.
– B_B
Commented May 8, 2021 at 17:05
• It might help to split $S_n$ into what is known at time $n-1$ and what is unknown. I.e. write $S_n = S_{n-1} + X_n$. Commented May 8, 2021 at 17:08
• Yes, Jamie is right.
– B_B
Commented May 8, 2021 at 17:36

If $$Y_{n}=S_{n}^{3}-3nS_{n}$$, then

$$Y_{n+1}=(S_{n}+X_{n+1})^{3}-3(n+1)(S_{n}+X_{n+1})$$

so after multiplication and re-arranging, we get

$$Y_{n+1}=Y_{n}+3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}$$

We need to check if

$$\mathbb{E}[Y_{n+1}|\mathcal{F}_{n}]=Y_{n}$$

so in our case, if

$$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}|\mathcal{F}_{n}\Big]=0$$

Note that:

1.

$$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)|\mathcal{F}_{n}\Big]=3S_{n}\Big(\mathbb{E}\Big[X_{n+1}^{2}|\mathcal{F}_{n}\Big]-1\Big)=3S_{n}\Big(\mathbb{E}\Big[X_{n+1}^{2}\Big]-1\Big)$$

2.

$$\mathbb{E}\Big[3S_{n}^{2}X_{n+1}|\mathcal{F}_{n}\Big]=3S_{n}^{2}\mathbb{E}\Big[X_{n+1}|\mathcal{F}_{n}\Big]=3S_{n}^{2}\mathbb{E}\Big[X_{n+1}\Big]$$

3.

$$\mathbb{E}\Big[3(n+1)X_{n+1}|\mathcal{F}_{n}\Big]=3(n+1)\mathbb{E}\Big[X_{n+1}|\mathcal{F}_{n}\Big]=3(n+1)\mathbb{E}\Big[X_{n+1}\Big]$$

4.

$$\mathbb{E}\Big[X_{n+1}^{3}|\mathcal{F}_{n}\Big]=\mathbb{E}\Big[X_{n+1}^{3}\Big]$$

Because $$X_{n+1}^{3}$$, $$X_{n+1}^{2}$$, $$X_{n+1}$$ and $$X_{1},\ldots,X_{n}$$ are independent random variables.

Now

$$\mathbb{E}\Big[X_{n+1}^{3}\Big]=\mathbb{E}\Big[X_{n+1}\Big]=0$$

$$\mathbb{E}\Big[X_{n+1}^{2}\Big]=1$$

so

$$\mathbb{E}\Big[3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}|\mathcal{F}_{n}\Big]=3S_{n}(1-1)+3S_{n}^{2}\cdot 0-3(n+1)\cdot 0+ 0=0$$

It means, that

$$\mathbb{E}[Y_{n+1}|\mathcal{F}_{n}]=Y_{n}$$

and $$Y_{n}=S_{n}^{3}-3nS_{n}$$ is a martingale.

• Alternatively, we can use $$Y_{n+1}=Y_{n}+3S_{n}(X_{n+1}^{2}-1)+3S_{n}^{2}X_{n+1}-3(n+1)X_{n+1}+X_{n+1}^{3}$$ and the distribution of $X_i$ to write $$Y_{n+1}=Y_{n}+(3S_{n}^{2}-3n-2)X_{n+1}$$ almost surely. Commented May 8, 2021 at 18:04