Solution. (i) \(E(Y_t) = \alpha \sum _{j\geq 0}\beta ^j = \alpha /(1 - \beta )\), which exists if and only if \(|\beta | < 1\), the condition for the geometric series to converge.

(ii) The \(U\)’s are uncorrelated with common variance, so \(\Var (Y_t) = \sigma ^2(1 + \beta ^2 + \beta ^4 + \cdots ) = \sigma ^2/(1 - \beta ^2)\).

(iii) \(Y_{t+1} - \mu = U_{t+1} + \beta (Y_t - \mu )\) and \(U_{t+1}\) is uncorrelated with \(Y_t\), so \(\gamma _1 = \beta \Var (Y_t) = \beta \sigma ^2/(1 - \beta ^2)\). Similarly \(Y_{t+2} - \mu = U_{t+2} + \beta U_{t+1} + \beta ^2(Y_t - \mu )\) gives \(\gamma _2 = \beta ^2\sigma ^2/(1 - \beta ^2)\).

(iv) The same argument with \(k\) steps gives \(\gamma _k = \beta ^k\sigma ^2/(1 - \beta ^2)\), \(k = 0, 1, 2, \dots \) The mean, the variance and every autocovariance are free of \(t\), and \(\gamma _k\) depends only on the lag \(k\), so the scheme is covariance stationary (Theorem 4.5.2).

(v) \(|\beta | = 0.95 < 1\). Its mean is \(0.05/(1 - 0.95) = 1\) and its variance \(\sigma ^2/(1 - 0.9025) = 10.26\sigma ^2\). With \(\beta \) this close to one the process is highly persistent (\(\rho _{10} = 0.95^{10} = 0.60\)), but it does return to its mean.

Solution. This is the AR(1) scheme with \(\alpha = 10\) and \(\beta = \frac 34\). By Theorem 4.5.2, \[ E(Y_t) = \frac {10}{1 - \frac 34} = 40, \qquad \Var (Y_t) = \frac {\sigma ^2}{1 - \frac {9}{16}} = \frac {16}{7}\sigma ^2 = 2.286\sigma ^2 , \] \[ \gamma _1 = \tfrac 34\cdot \tfrac {16}{7}\sigma ^2 = \tfrac {12}{7}\sigma ^2, \qquad \gamma _2 = \tfrac {9}{16}\cdot \tfrac {16}{7}\sigma ^2 = \tfrac {9}{7}\sigma ^2, \qquad \gamma _3 = \tfrac {27}{64}\cdot \tfrac {16}{7}\sigma ^2 = \tfrac {27}{28}\sigma ^2 . \] The autocorrelations are \(\rho _1 = 0.75\), \(\rho _2 = 0.5625\), \(\rho _3 = 0.4219\): each is \(\frac 34\) of the one before. They decay geometrically, the signature of a stationary AR(1).

Solution. (a) As in Example 4.2.6, \(\Var (\bar Y) = \theta (1 - \theta )/n \to 0\) with \(E(\bar Y) = \theta \), so \(\bar Y \xrightarrow {p} \theta \), and \(\bar Y/(1 - \bar Y) \xrightarrow {p} \theta /(1 - \theta )\) by the continuous mapping theorem.

(b) \(\log \Upsilon = \log \theta - \log (1 - \theta )\) has derivative \(\frac 1\theta + \frac {1}{1 - \theta } = \frac {1}{\theta (1 - \theta )}\). By the delta method, \[ \sqrt n(\log \hat \Upsilon - \log \Upsilon ) \xrightarrow {d} N\left (0,\ \frac {\theta (1 - \theta )}{\theta ^2(1 - \theta )^2}\right ) = N\left (0,\ \frac {1}{\theta (1 - \theta )}\right ) . \]

(c) \(\bar Y = 0.24\), \(\hat \Upsilon = 0.24/0.76 = 0.316\), \(\log \hat \Upsilon = -1.153\) with standard error \(1/\sqrt {500(0.24)(0.76)} = 0.105\). The interval for the log-odds is \(-1.153 \pm 1.96(0.105) = (-1.358, -0.948)\), and exponentiating gives \((0.257, 0.388)\) for the odds. Working on the log scale keeps the interval positive and makes it asymmetric, as it should be for a ratio.

Solution. \(S_n^2 = \frac 1n\sum X_i^2 - \bar X^2\). By the law of large numbers \(\frac 1n\sum X_i^2 \xrightarrow {p} E(X^2)\) (the \(X_i^2\) are independent with finite variance) and \(\bar X \xrightarrow {p} \mu \), so by continuous mapping \(S_n^2 \xrightarrow {p} E(X^2) - \mu ^2 = \sigma ^2\), and \(S_n = \sqrt {S_n^2} \xrightarrow {p} \sigma \) since the square root is continuous. The version with divisor \(n - 1\) is \(\frac {n}{n-1}S_n^2\), and \(\frac {n}{n-1} \to 1\), so it is consistent too. Consistency cannot distinguish between them; unbiasedness does.

Solution. (a) \(\tilde b = \beta + \frac {\frac 1n\sum X_iu_i}{\frac 1n\sum X_i^2} \xrightarrow {p} \beta + \frac {E(Xu)}{E(X^2)} = \beta \).

(b) Now \(\tilde b = \beta + \frac {\alpha \frac 1n\sum X_i + \frac 1n\sum X_iu_i}{\frac 1n\sum X_i^2} \xrightarrow {p} \beta + \frac {\alpha \mu _X}{E(X^2)}\). Omitting the intercept is inconsistent unless \(\mu _X = 0\); the error does not go away with more data. This is the large-sample counterpart of the bias noted in Problem 2.7.4.

Solution. \(P(|\bar Y - \theta | \geq 0.05) \leq \theta (1 - \theta )/(n \times 0.0025) \leq 0.25/(0.0025n) = 100/n\), using \(\theta (1 - \theta ) \leq \frac 14\). This is at most \(0.05\) if \(n \geq 2000\). The CLT gives \(1.96\sqrt {0.25/n} \leq 0.05\), so \(n \geq 385\). Chebyshev’s bound holds for every distribution with that variance and every \(n\), so it must cover the worst case and is very conservative. The CLT approximation uses the near-normal shape of \(\bar Y\), which is accurate for \(n\) in the hundreds.