Hello Math StackExchange!
I am very frustrated trying to find the definition of $\mathbb{E}_{\mu}$ (apparently it means expectation with initial distribution). I have tried to search on many different platforms and books , but am in vain.
Also, is it true that $\mathbb{E}_{\mu}[X] = \sum_{s} \mu(s)\hspace{1mm} \mathbb{E} [X_{n} | X_0 = s]$ ?
Best Answer
After conversation with @nejimbam, I used the definition from p. 13 of the notes I shared (https://www.math.bgu.ac.il/~yadina/RWnotes.pdf).
$\mathbb{E}_{\mu}[X] = \sum_{x} \mathbb{P}_{\mu}[X=x] \cdot x \hspace{45mm} \text{(By def. of $\mathbb{E}_{\mu}$)}\\ \hspace{11mm}= \sum_x \big[ \sum_s\mu(s) \hspace{1mm} \mathbb{P}[X=x | X_0 = s] \big] \cdot x \hspace{13mm} \text{(By def. of $\mathbb{P}_{\mu}$ as shared in the notes)}\\ \hspace{11mm}= \sum_x \sum_s\mu(s) \hspace{1mm} \mathbb{P}[X=x | X_0 = s]\cdot x \hspace{17mm} \text{(Distributivity of Summation)}\\ \hspace{11mm}= \sum_s \sum_x \mu(s)\hspace{1mm} \mathbb{P}[X=x | X_0 = s]\cdot x \hspace{17mm} \text{(Switching order of Sums)} \\ \hspace{11mm}= \sum_s \mu(s) \sum_x\mathbb{P}[X=x | X_0 = s]\cdot x \hspace{17mm} \text{(Rearranging)}\\ \hspace{11mm}=\sum_s \mu(s) \hspace{1mm} \mathbb{E}[X=x | X_0 = s] \hspace{29mm} \text{(By def. of $\mathbb{E}$)}$
I am sorry if this was trivial and “expected” to be known (pun is intended here), but I prefer clear, rigorous definitions / derivations.