Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 206 / 5

For a weakly stationary process, every observation has the same finite expectation $\mu$. Linearity of expectation, without any assumption of independence, givesThe variance is a sum of all pairwise covariances:For a real-valued weakly stationary process, the autocovariance is even. There are $n$ pairs at lag zero and $n-h$ pairs in each direction at positive lag $h$. ThereforePositive autocovariances inflate the uncertainty relative to the independent-observation formula; negative terms may reduce it. An unbiased estimator need not have variance tending to zero: a process consisting of the same random variable at every time is weakly stationary, but averaging cannot reduce its variance.

For a known mean zero, a common sample autocovariance function convention isThe weakly stationary process property gives $\mathbb{E}[Y_t{Y}_{t+h}]=\gamma (h)$, soThe lag-zero estimate is unbiased, but at nonzero lags the divisor $n$ causes a finite-sample multiplicative bias. An alternative convention uses $(n-h{)}^{-1}$ instead; for that convention the expectation is exactly $\gamma (h)$. These are two normalizations of the sample autocovariance function, so the divisor must be specified. Here no estimated-mean centering term occurs because zero is known. Replacing zero by a sample mean would introduce additional bias terms. Extend to negative lags by symmetry; there are no observed lagged pairs for $h\ge n$.

A zero-mean autoregressive process of order one satisfieswhere ${\epsilon }_t$ is white noise with variance ${\sigma }_{\epsilon }^2>0$. A causal time series uses present and past innovations only. For $|\alpha |<1$ the serieshas mean-square convergence because ${\sum }_j{\alpha }^{2j}<\infty$, and it supplies the causal weakly stationary process. Its variance is ${\sigma }_{\epsilon }^2/(1-{\alpha }^2)$. For $h\ge 0$, only the common innovations contribute to the autocovariance:Evenness then givesWhen $\alpha <0$, the signs alternate; when $\alpha =0$, the process is white noise. With nondegenerate innovations, $|\alpha |\ge 1$ does not give a causal finite-variance stationary solution. These covariance calculations need uncorrelated innovations, not Gaussianity; Gaussian innovations additionally yield a Gaussian process.

The selected model is a moving-average process of order one, with the plus-sign convention used by R:The Gaussian likelihood function models innovations as independent $N(0,{\sigma }_{\epsilon }^2)$. Its autocovariance is $\gamma (0)={\sigma }_{\epsilon }^2(1+{\theta }^2)$, $\gamma (1)={\sigma }_{\epsilon }^2\theta$, and zero at larger lags, with autocorrelation $\rho (1)=\theta /(1+{\theta }^2)\simeq 0.4631$. The estimated coefficient has modulus less than one, giving invertibility of a moving-average model; a finite moving-average process is stationary regardless of this invertibility restriction.

The original plots show fluctuations around a roughly constant level, no persuasive deterministic trend, and a conspicuous positive lag-one correlation with the other plotted correlations within the approximate sampling bands. They make a stationary nonseasonal working model reasonable for this short series. stationary=TRUE excludes differenced alternatives, and seasonal=FALSE excludes seasonal ARIMA terms. Neither option is proved by a 60-day plot: annual temperature seasonality cannot be excluded from two months, and a longer record or residual diagnostics might require a different model. The options are reasonable local simplifications, not universal statements about temperature dynamics.

Let ${\mathcal{F}}_t$ contain observations through time $t$. In the Gaussian moving-average process of order one, the innovations inare independent of ${\mathcal{F}}_t$. Hence the conditional expectation is the constant mean. For known parameters, the forecast and its mean squared prediction error areThe mean prediction error itself is zero; the requested uncertainty is naturally interpreted as expected squared error. Plugging in the fit gives forecast $15.0133$, mean squared error $6.731002$, and root mean squared error about $2.5944$.

The usual naive time-series forecast repeats $Y_t$. Its unconditional expected squared error issince $\gamma (7)=0$. ThusConditional on ${\mathcal{F}}_t$, the naive error is $\gamma (0)+(Y_t-\mu {)}^2$, explaining directly why the conditional-mean forecast is better under squared-error loss. These are fitted-model calculations that ignore parameter-estimation uncertainty. Estimating the unknown mean from the past adds its estimation mean squared error to the seven-step forecast risk; the printed summary does not provide a complete assessment of that extra uncertainty. This is distinct from claiming a signed mean error as a measure of forecasting accuracy.