For 4 coin tosses represented here as a moving average of my last 4 tosses. A win represented by the arbitrary side of the coin represented 1, a loss the other side represented by an outcome of 0.
Given $\epsilon_{t-3}\ =\ \epsilon_{t-2} = 1$
Mean: conditional on above information
$$\mathbb{E}(w_t\ | \epsilon_{t-3} = \epsilon_{t-2}= 1) = 1/4 \epsilon_t + 1/4\epsilon_{t-1} + 1/4\epsilon_{t-2} + 1/4 \epsilon_{t-3})$$
Since the 3rd and 2nd error are given a value of 1, the conditional expectation of $w_t$ is
$\mathbb{E}_{t-2}*w_t=\mathbb{E}(w_t\ | \epsilon_{t-3} = \epsilon_{t-2}= 1) = (1/4)(\mathbb{E}_{t-2}\epsilon_t + \mathbb{E}_{t-2}\epsilon_{t-1} + \mathbb{E}_{t-2}\epsilon_{t-2} + \mathbb{E}_{t-2}\epsilon_{t-3}$
$\mathbb{E}_{t-2} = 0.25(0 + 0+1+1) = 0.5$
I come about the same answer and it's an easy one. But I'm confused as the formal evaluation. In the example above copied from text, why is there an expectation at t-2? I'm not at all familiar with the notation $\mathbb{E}_{t-2}$and it seems everything in the final equation is converted to that. What does it mean? I'm coming about the same result so I'm not sure how much attention I should be giving this?