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If we have a process defined as:

$Y_t = ε_t + Y_{t-1}$

How do you determine the expected value:

$E[Y_tY_{t-1}]$

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1  
What do you assume on $\varepsilon_t$? –  Davide Giraudo Mar 3 '12 at 15:02
    
You should state your question clearly, what's the realtion between $Y_{t}, \epsilon_t, Y_{t-1}$ –  yaoxiao Mar 3 '12 at 15:03
    
Epsilon is a standard normal random variable –  Soo Mar 3 '12 at 16:06
    
Do you know anything about, say, the distribution or expected value of $Y_0$? –  TMM Mar 3 '12 at 16:13
1  
Still not sufficient to solve the question. –  Did Mar 3 '12 at 16:48

2 Answers 2

up vote 4 down vote accepted

Since $Y_t = Y_0 + \sum_{i=1}^t \epsilon_i$, if $Y_0 = 0$ we get $Y_t = \sum_{i=1}^t \epsilon_i$. Furthermore you said $\epsilon_i \sim N(0,1)$, and since under the assumption of independence of the $\epsilon_i$ the sum of (standard) normal random variables is again a normal random variable (with the means and variances added) we get $Y_t = \sum_{i=1}^t \epsilon_i \sim N(0,t)$. Writing out the expected value a bit, we get

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[(Y_{t-1} + \epsilon_t)Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}].$$

From $Y_{t-1} \sim N(0,t-1)$ it follows that $\mathrm{Var}(Y_{t-1}) = \mathbb{E}[Y_{t-1}^2] - \mathbb{E}[Y_{t-1}]^2 = \mathbb{E}[Y_{t-1}^2] = t - 1$. Finally, since $\epsilon_t$ and $Y_{t-1}$ are independent and symmetric around $0$, it follows that

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}] = (t - 1) + 0 = t - 1.$$

Without the assumption of independence of the $\epsilon_i$, however, this does not work.

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+1. $ $ $ $ $ $ –  Did Mar 3 '12 at 16:59

$E[Y_{t}Y_{t-1}]=E[E[Y_{t}Y_{t-1}|\mathcal{F}_{t-1}]]=EY_{t-1}^{2}$

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It is usually not a good idea to simply write an equation, without an explanation of what your point is. That is why you are getting downvotes! Don't you want to explain a bit more? –  Mariano Suárez-Alvarez Mar 3 '12 at 19:31

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