I have $X_1,...,X_n \sim i.i.d.(0,1)$ and $S_n=X_1+...+X_n$,
I define trunctation $X_i^{(v)}=X_i\textbf1_{\{|X_i|\leq v\}}-E[X_i\textbf1_{\{|X_i|\leq v\}}]$ and $S_k^{(v)}=\Sigma_{i=1}^k X_i \textbf1_{\{|X_i|\leq v\}}-E \left[ \Sigma_{i=1}^k X_i \textbf1_{\{|X_i|\leq v\}} \right]$. So i obtain two random walks:
$\{ S_i^{(v)} \}_{i=1}^n$
$\left\{ S_n - S_n^{(v)} \right\}_{i=1}^n$
In the lecture is written that the random walks had mean 0.I currently know that they can be a martigales because of mean $0$ and P itegrability. However I'm not sure if I understand properly why they have the mean zero.
In trunctation we take only bounded random variables. So let's take $X_i^{(v)}$. Then we substract the mean of the $X_i^{(v)}$. Is that true $E[X_i\textbf1_{\{|X_i|\leq v\}}] = 0 $ because if we take the variable that is bounded it has a mean 0, on the other hand if we take variabe which isn't bounded by $v$ we get $0$ because of indicator function.
In conclusion $E[X_i^{(v)} = $ because we have mean $E[X_i]$ which is $0$ when $X_i$ is bounded and $0$ when $X_i$ isn't bounded because we have an indicator function.
So $E[S_n^{(v)}] = 0$ because of properties of mean value. $E[X+Y]=E[X]+E[Y]$
Is this way correct?
In general, if $Y$ is a random variable and $\mu = E[Y]$, then $$E[Y-\mu] = E[Y] - \mu = \mu - \mu = 0.$$ This is in particular true for $Y=X_i1_{\{|X_i| \leq \nu\}}$ also when $E[X_i1_{\{ |X_i| \leq \nu\}}]$ is not $0$.