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From the title I'm supposed to show $\bar{y} \rightarrow \mu$ (converges in probability) where $$y_t = \mu + u_t$$ $$ u_t = \rho u_{t-2} + \epsilon_t$$$$ E(\epsilon) = 0, E(\epsilon^2) = \sigma^2, E(\epsilon_t\epsilon_s) = 0, 0 \leq \rho \leq 1$$ So I'm not sure if my math is right, but If i show that $\bar{y} = \frac{1}{N}\sum (\mu + u_t) = \frac{1}{N}N\mu+\sum u_t $ and $\sum u_t $ goes to 0 will that be enough?

I can show each term of $\sum u_t$ will include a $\epsilon_t$ which the ExpValue is 0. But I'm not sure if that's enough.

I'm mildy mathematically mature finance grad student in my first econometrics class so any help would be really appreciated. Thanks.

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There are some misprints in your question: first, $u_t=\rho u_{t-1}+\epsilon_t$, not $u_t=\rho u_{t-2}+\epsilon_t$; second, one should assume that $\rho<1$, not that $\rho\le1$; third, one wants to show that $\frac1N\sum u_t$ goes to $0$, not that $\sum u_t$ goes to $0$. See below for the steps of a proof. –  Did Sep 14 '11 at 20:20
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Since $\bar y_N=\mu+\bar u_N$, it is sufficient to show that $\bar u_N=\dfrac1N\sum\limits_{t=1}^Nu_t$ converges to $0$ in $L^2$ (which implies the convergence in probability), in other words, that $\mathrm E((\bar u_N)^2)\to0$.

Here are the steps of the proof.

  1. Write every $u_t$ as a linear combination of the random variables $u_0$ and $\epsilon_s$ for $1\le s\le t$.
  2. Deduce from 1. an expression of $N\bar u_N$ as a linear combination of the random variables $u_0$ and $\epsilon_s$ for $1\le s\le N$.
  3. Use the independence of all these random variables and the fact that $\mathrm E(\epsilon)=0$ to deduce from 3. an expression of $N^2\mathrm E((\bar u_N)^2)$ in terms of the parameters $\rho$, $\sigma^2=\mathrm E(\epsilon^2)$ and $\mathrm E(u_0^2)$.
  4. Using the fact (which you shall prove) that $1+\rho+\rho^2+\cdots+\rho^t\le\dfrac1{1-\rho}$ for every nonnegative $t$, show that, for every $N$, $$ N^2\mathrm E((\bar u_N)^2)\le N\sigma^2+\mathrm E(u_0^2). $$
  5. Conclude that $\mathrm E((\bar u_N)^2)\to0$.
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