Past exam of the mathematics course of the University of Cambridge 2005 iii Paper 47 2 Solution Created 2026-10-03 Updated 2026-10-14
Let . The causality root criterion for an autoregressive model,is a sufficient condition for a weakly stationary process solution, and gives the causal solution . Indeed is analytic in a disk of radius greater than one, so its Taylor coefficients decay geometrically. The series therefore has mean-square convergence and defines a linear filter of a stationary time series; multiplying by verifies the equation.
For the specified order-two model, put . Since ,When , its roots are and . Both have modulus greater than one exactly when . The case is white noise and satisfies the same inequality. ConsequentlyThis is the causal stationarity condition selected above. Without causality, a two-sided autoregressive model can also have a noncausal stationary autoregression when its polynomial has no unit-circle zeros. Thus the inequality is not a necessary condition for all conceivable weak stationarity solutions; the causal interpretation is the one required by the subsequent innovations and forecasting questions.
To find the Wold representation, expand . Matching coefficients gives , and . HenceHere are the Fibonacci numbers. The coefficient formula also follows by partial fractions in the displayed factorization. Absolute summability follows from and . The past observation span equals the past noise span, because the causal series gives one inclusion and gives the reverse inclusion. Thus these noises are the linear innovations, and there is no additional deterministic component. This is the Fibonacci-coefficient autoregression.
The best linear prediction from an infinite past is the orthogonal projection onto the closed span of . Future white noise is orthogonal to that span. Substitution in the recursion therefore givesThe respective errors are and , so their mean squared errors are and . No normal distribution assumption is needed for these linear forecasts.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 29 1 Solution 2026-10-07
For an autoregressive moving-average model, the driving-to-output transfer function is , and the inverse transfer function is . The causality and invertibility root criteria for an ARMA model require these respective rational functions to have power series about zero converging on a disk larger than the unit disk. If the two polynomials have no common factor, the conditions becomeIndeed, outside-disk roots leave a radius of convergence greater than one, so the coefficients decay geometrically and are absolutely summable. Conversely an uncancelled pole inside or on the unit disk prevents the required stable power series. If factors are common, apply the criterion after cancellation, to the noise-driven solution rather than additional homogeneous components. The backshift operator translates these power series into the desired one-sided filters.
The original representation has and , with roots and . There is no cancellation. Hence it is neither causal nor invertible relative to its specified driving noise. Stationarity is nevertheless possible through a two-sided solution: expanding the autoregressive inverse in negative powers givesThis is an anticausal time series representation with square-summable coefficients.
To establish the alternative representation on the same process, defineThe inverse of is a stable one-sided filter. For , the identities and giveThe time-series spectral density of is ; therefore the defined has constant time-series spectral density . Its mean is zero, its variance is , and all its nonzero-lag autocovariances vanish. It is thus weak white noise. Its definition directly givesThis root reflection of an ARMA representation has roots , so is causal and invertible. A constant spectrum proves whiteness, not independence of non-Gaussian coordinates; no Gaussian assumption is needed for the required white noise representation.
For best linear prediction from an infinite past, let be the closed linear span of with . The causal representation expresses every such in present and past values, while invertibility puts every with in . In particular is orthogonal to . The representation at time one then gives the orthogonal projectionConsequently the best linear predictor and its error variance areOrthogonality proves optimality among linear predictors in the closed past span, without asserting that the predictor must be the conditional mean for a non-Gaussian process.