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I'm working through an econometrics textbook and came upon this supposedly simple problem early on.

Suppose you win $\$1$ if a fair coin shows heads and lose $\$1$ if it shows tails. Denote the outcome on toss $t$ by $ε_t$. Assume you want to calculate your average winnings on the last four tosses. For each coin toss $t$, your average payoff on the last four tosses is $w_t= 0.25ε_t + 0.25ε_{t-1} + 0.25ε_{t-2} + 0.25ε_{t-3}$

Find var($w_t$), and find var($w_t$) conditional on $ε_{t-3} = ε_{t-2} = 1$

For the first part of the question (finding the variance without any conditions on $ε$), I know that:

Var($w_t$) = Var(0.25ε_t + 0.25ε_{t-1} + 0.25ε_{t-2} + 0.25ε_{t-3}

= Var(0.25ε_t) + Var(0.25ε_{t-1}) + Var(0.25ε_{t-2}) + Var(0.25ε_{t-3})

=0.0625 Var($ε_t$) + 0.0625 Var($ε_{t-1}$) + 0.0625 Var($ε_{t-2}$) + 0.0625 Var($ε_{t-3}$)

This is the point in the problem that confuses me. I know the variance for the entire time series is 1, because each possible result of the coin flip (-1 or 1) is $\$1$ away from series' expected value of 0. Does this mean that Var($ε_t$) = 1, $∀ t$ as well? Based on that logic, I could expect Var($w_t$) = 0.0625 (1+1+1+1) = 0.25, but this textbook does not provide any solutions, so I have no way of checking my logic.

I'm confident that once I figure out this part of the question, the part of the question that assumes $ε_{t-3} = ε_{t-2} = 1$, but I added that part to my initial question in case there are other nuances I'm missing, in which case I can use said nuances to form a new question.

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The variance of w_t is 0.25, for the reasons you explained. – Did Oct 7 '12 at 20:27
@did Then the condition that $ε_{t-3} = ε_{t-2} = 1$ doesn't change the problem in the slightest, correct? – Ricardo Altamirano Oct 7 '12 at 20:28
It does change the problem since now, $w_t=0.25ε_t+0.25ε_{t−1}+$some constant, hence the conditional variance is only 0.125. – Did Oct 7 '12 at 20:33
@did Ah, ok. Post your comments as an answer and I'll accept it immediately. Thank you for the help. – Ricardo Altamirano Oct 7 '12 at 20:39
up vote 2 down vote accepted

The variance of $w_t$ is 0.25, for the reasons you explained. But conditioning on $ε_{t−3}=ε_{t−2}=1$ changes the problem since now, $w_t=$ 0.25$ε_t+$ 0.25$ε_{t−1}+$ some constant, hence the conditional variance is only 0.125.

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