Basic questions regarding Martingale and Conditonal Expectation

Question

I am given a sequence $(X_n)$ of independent Random Variables in $\mathfrak{L}^2$ (What is the right L in Latex?) on Some Probability Space $(\Omega,\mathfrak{F},\mathbb{P})$ with $\mathbb{E}(X_n)=a_n$ and $Var(X_n)=\sigma_n^2$. Further $S_n=\sum_{i=1}^n X_i$ and $\mathfrak{F}_n=\sigma(S_1,\dots,S_n)$ for all $n \in \mathbb{N}$. My task is a) to determine $\mathbb{E}(S_{n+1}|\mathfrak{F_n})$ and $\mathbb{E}(S_{n+1}^2|\mathfrak{F_n})$.

After that I should say how the $a_n$ has to be chosen in order that $(S_n)$ is a martingale relative to $(\mathfrak{F}_n)$.

Ad a) Since $S_{n+1}=S_n+X_{n+1}$ and $S_n$ is per definition $F_n$ measureable so that $$\mathbb{E}(S_{n+1}|F_n)=S_n\mathbb{E}(X_{n+1}|F_n).$$ However, I think that I am not done here. Is $\mathbb{E}(X_{n+1}|F_n)=a_{n+1}$?

For $\mathbb{E}(S_{n+1}^2|\mathfrak{F_n})$ I thought about using the variance since $\mathbb{E}(S_{n+1})^2=Var(S_{n+1})-\mathbb{E}(S_{n+1})^2$. Though, I also do not know how to go on with the restriction on $F_n$.

For the second part. I thought that $\mathbb{E}(|S_n|)<\infty$ is obvious and $S_n$ should be $F_n$ measureable per definition, too. So if my approach in a) is right, then $$\mathbb{E}(S_{n+1}|F_n)=S_n\mathbb{E}(X_{n+1}|F_n).$$ And if $\mathbb{E}(X_{n+1}|F_n)=a_{n+1}=1(?)$ then $$\mathbb{E}(S_{n+1}|F_n)=S_n$$ which is the definition for a martingale. I am new to Probability Theory and kind a confused with all the notation. I hope my problem is clear.

Answer 1

When dealing with conditional expectations you should always keep in mind the following important rules:

• Rule 1: Let $X \in L^1$ be a random variable which is measurable with respect to some $\sigma$-algebra $\mathcal{G}$. If $\mathcal{F}$ is another $\sigma$-algebra which is independent from $\mathcal{G}$, then $$\mathbb{E}(X \mid \mathcal{F}) = \mathbb{E}(X).$$
• Rule 2: If $Y \in L^1$ is a random variable which is measurable with respect to a $\sigma$-algebra $\mathcal{F}$, then $$\mathbb{E}(X Y \mid \mathcal{F}) = Y \mathbb{E}(X \mid \mathcal{F})$$ for any $X \in L^1$ such that $X \cdot Y \in L^1$.

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Now let's get back to your question. You already noted that $S_n$ is $\mathcal{F}_n$-measurable, and therefore (according to the 2nd rule above with $X:=1$) $$\mathbb{E}(S_n \mid \mathcal{F}_n) = S_n.$$ On the other hand, $X_{n+1}$ is independent from $\mathcal{F}_n$ (check it!) and therefore the first rule gives $$\mathbb{E}(X_{n+1} \mid \mathcal{F}_n) = \mathbb{E}(X_{n+1}) \stackrel{\text{def}}{=} a_{n+1}.$$ Combining both computations it follows from the linearity of the conditional expectation that $$\mathbb{E}(S_{n+1} \mid \mathcal{F}_n) = \mathbb{E}(S_n \mid \mathcal{F}_{n}) + \mathbb{E}(X_{n+1} \mid \mathcal{F}_n) = S_n+a_{n+1}. \tag{1}$$ I leave it to you to do a similar calculation for $S_n^2$, but I give you a hint: It is useful to write

$$S_{n+1}^2 = (S_n+X_{n+1})^2 = S_n^2 + 2 X_{n+1} S_n + X_{n+1}^2$$

in order to compute $\mathbb{E}(S_{n+1}^2 \mid \mathcal{F}_n)$.

Equation $(1)$ tells you how $(a_n)_{n \geq 1}$ has to be chosen such that $(S_n)_{n \geq 1}$ is a martingale.