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Let $X = (X_t : t ∈ Z)$ be a $MA(1)$-Process. Define the time series $Y = (Y_t : t ∈ Z)$ with $Y_t = \mathbb 1_{(X_t > 0)}$. What is the Autocorrelation function of $Y_t$?

Anyone has any idea how to even start solving this?

There was a hint to show $P(U > 0,V > 0) = 1/ 4 + 1 /2π\cdot \arcsin(\rho)$ where $U$ annd $V$ are both normally distributed with mean $0$, same Variance $σ^2$ und correlation $\rho$. I have shown this equality but how does it help me further?

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  • $\begingroup$ Are we assuming that the white noise is normally distributed? $\endgroup$
    – Math1000
    Commented Jan 28, 2015 at 21:43
  • $\begingroup$ yes it is normally distributed. $\endgroup$
    – WaitForIt
    Commented Jan 28, 2015 at 21:47

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Let $t,s\in\mathbb Z$. Then the correlation of $Y_t$ with $Y_s$ is $$\rho_{t,s}=\frac{\operatorname{Cov}(Y_t,Y_s)}{\sqrt{\operatorname{Var}(Y_t)\operatorname{Var}(Y_s)}} = \frac{\mathbb E[Y_tY_s]-\mathbb E[Y_t]\mathbb E[Y_s]}{\sqrt{\operatorname{Var}(Y_t)\operatorname{Var}(Y_s)}}. $$ Now $$\mathbb E[Y_t]=\mathbb E\left[\mathbf 1_{(X_t>0)}\right] = \mathbb P(X_t>0), $$ $$\mathbb E[Y_tY_s] = \mathbb E\left[\mathbf 1_{(X_t>0)}\mathbf 1_{(X_s>0)}\right] =\mathbb P(X_t>0,X_s>0), $$ and $$\operatorname{Var}(Y_t)=\mathbb E[Y_t^2]- \mathbb E[Y_t]^2 = \mathbb P(X_t>0)(1-\mathbb P(X_t>0)). $$ So $$\rho_{t,s} = \frac{\mathbb P(X_t>0,X_s>0)-\mathbb P(X_t>0)\mathbb P(X_s>0)}{\mathbb P(X_t>0)(1-\mathbb P(X_t>0))\mathbb P(X_s>0)(1-\mathbb P(X_s>0))} .$$

Using the result from your hint it seems that you should be able to compute this quantity.

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  • $\begingroup$ I am getting $-1/4+\frac{1}{2\pi} \arcsin(\rho)$ for the numerator. But what about the denominator? $\endgroup$
    – WaitForIt
    Commented Jan 28, 2015 at 22:13
  • $\begingroup$ I am not sure exactly how you are defining the MA process, but if $X_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}$ then $\mathbb P(X_t>0) = \mathbb P(\varepsilon_t + \theta\varepsilon_{t-1}>-\mu)$. The sum of independent normal random variables is again a normal random variable, so... $\endgroup$
    – Math1000
    Commented Jan 28, 2015 at 22:18
  • $\begingroup$ Oh sorry I have $\theta=0.5$ and $\mu=0$. So I get 1/2 for each of the probabilities in the denominator since it's half of the bell curve right? Sorry if this sounds like a stupid question but I have been solving problems all day long and my brain doesn't wanna think propertly anymore, lol. $\endgroup$
    – WaitForIt
    Commented Jan 28, 2015 at 22:33
  • $\begingroup$ In that case $X_t=\varepsilon_t +\frac12\varepsilon_{t-1}$, so $X_t\sim\mathcal N\left(0,\frac54\sigma^2\right)$ (assuming $\varepsilon_t\sim\mathcal N(0,\sigma^2)$). $\endgroup$
    – Math1000
    Commented Jan 28, 2015 at 22:46
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    $\begingroup$ so the end result is $\rho_Y=\frac{2}{\pi}\arcsin(\rho_X)$. Thanks. $\endgroup$
    – WaitForIt
    Commented Jan 29, 2015 at 0:29

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