Prove $X_n = \frac{1}{a_n} 1_{(0,a_n]}$ a martingale

Question

With the following premises, I want to prove, that the series of random variables $(X_n)_{n \in\mathbb{N}}$ is a Martingale:

Let $\Omega := (0, 1] \subset \mathbb{R}, \mathfrak{F}$ the Borel-$\sigma$-algebra on $(0, 1]$, P the Lebesgue-measure. Let $(a_n)_{n \in\mathbb{N}}$ be a monotonically decreasing series in $\Omega$.

$X_n := \dfrac{1_{(0, a_n]}}{a_n}$ where 1 is the indicator function.

Now I proved this arguing about the sets contained in the filtration I chose: $\mathfrak{F}_n := \sigma(X_0, ..., X_n)$ but I wonder whether there is a more formal and more direct way to get from

$\mathbb{E}[X_{n+1}|\mathfrak{F}] = \mathbb{E}[\dfrac{1_{(0, a_{n+1}]}}{a_{n+1}}|\mathfrak{F}] = \dfrac{1}{a_{n+1}} \mathbb{P}[X \in (0, a_{n+1}]|\mathfrak{F}]$ to ... $= X_n$

Thanks for any help!

Answer 1

Hints

1. Set $A_n := (0,a_n]$. Show that $$\sigma(X_1,\ldots,X_n) = \sigma(A_1,\ldots,A_n).$$
2. Recall that $\mathbb{E}(Y \mid \mathcal{F}) = X$ if, and only if, $X$ is $\mathcal{F}$-measurable and $$\int_G X \, d\mathbb{P} = \int_G Y \, d\mathbb{P} \tag{1}$$ for all $G \in \mathcal{G}$ where $\mathcal{G}$ is a $\cap$-stable generator of the $\sigma$-algebra $\mathcal{F}$. Apply this in the given setting, i.e. show $(1)$ for $\mathcal{G} := \{A_1,\ldots,A_n\}$, $X = X_n$, $Y=X_{n+1}$. (For this second part, we need that $(a_n)_n$ is decreasing.)

Answer 2

For every $k$, with the convention that $a_0=1$, consider the intervals $$I_k=(0,a_k],\qquad L_k=(a_k,a_{k-1}].$$ To compute $$Y_n=P(X \in I_{n+1}\mid\mathfrak{F}_n],$$ note first that $\mathfrak{F}_n$ is generated by a partition of $(0,1]$, namely, $$\{I_n\}\cup\{L_k\mid1\leqslant k\leqslant n\}.$$ Second, $I_{n+1}\subseteq I_n$ and $I_n$ is a member of the partition generating $\mathfrak{F}_n$ hence, for some suitable $x_n$, $$Y_n=x_n\mathbf 1_{I_n}.$$ Third, $$a_{n+1}=E(Y_n)=E(x_n\mathbf 1_{I_n})=x_nP(I_n)=x_na_n,$$ hence, $$\frac1{a_{n+1}}Y_n=\frac{x_n}{a_{n+1}}\mathbf 1_{I_n}=\frac1{a_n}\mathbf 1_{I_n}=X_n,$$ and finally, as desired, $$E\left(X_{n+1}\mid\mathfrak{F}_n\right)=\frac1{a_{n+1}}Y_n=X_n.$$

Answer 3

With the help given, I was able to solve the problem in a somewhat different way. Since we have a finite partition of $\Omega$ and (without resctriction $a_0 = 1$) $\mathfrak{F}_n = \sigma(\{(0, a_n], (a_n, a_{n-1}], ..., (a_1, a_0]\})$. Then

$\mathbb{E}[X_{n+1}|\mathfrak{F}_n] = 1_{(0, a_n]}\cdot\mathbb{E}[X_{n+1}|{(0, a_n]}] + \sum_{k=1}^n 1_{(a_k, a_{k-1}]}\cdot\mathbb{E}[X_{n+1}|{(a_k, a_{k-1}]}]$

Since $X_{n+1}\equiv 0$ in all terms of the sum, we have

$= 1_{(0, a_n]}\cdot\mathbb{E}[X_{n+1}|{(0, a_n]}] = 1_{(0, a_n]}\cdot\frac{\mathbb{E}[X_{n+1}\cdot 1_{(0, a_n]}]}{\mathbb{P}[(0, a_n]]} = X_n \cdot \mathbb{E}[X_{n+1}\cdot 1_{(0, a_n]}] = X_n$

Thanks for support!