ESSE Mid-Term Test 2017 Tuesday 17 October :30-09:45

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1 ESSE Mid-Term Test 207 Tuesday 7 October :30-09:45 Symbols have their usual meanings. All questions are worth 0 marks, although some are more difficult than others. Answer as many questions as you can (there are 5 in total). WRITE YOUR ANSWERS (LEGIBLY!) IN THE TEST BOOKLETS PROVIDED - indicate clearly which question is being answered where. EXPLAIN your methods and calculations with words wherever appropriate. Two sides of notes (8.5x inch paper) are permitted, no other aids ) Let {X t } be the auto-regressive AR(2) process given by X t = X t X t-2 + Z t, where {Z t } is iid noise, uniformly distributed in [-,]. a) Test whether the series is stationary. b) From the probability distribution, find the expected mean and variance of {Z t } c) Find the expected mean, μ, variance, γ(0), and autocorrelation, ρ() of {X t }. γ(0) can be left as an expression involving fractions since you have no calculators, but ρ() should be simple a) Stationarity can be tested by roots of θ(z) = 0 where ARMA process is θ(b)x t = ϕ(b)w t. Here θ(z) = (- z z 2 ) = (-0.5z) 2 = 0. One repeated root, 2, 2 > 0 so stationary. (3 marks) b) E[z t ] = 0, by symmetry, or because E[z t ] = x dx=[ x 2 /2] 2 = 0 and also σ 2 = E[z t2 ] = x 2 dx=[x 3 /3] 2 = /3. (3 marks) c) Since stationary E[X t-2 ] = E[X t- ] = E[X t ] = μ; and so E[X t ] = μ = 0. E[X t - (X t X t-2 + Z t )] = 0.25E[X t ]-E[z t ] = 0, E[z t ] = 0 and so For variance; γ(0) = E[X t2 ] = E[(X t X t-2 + Z t ) 2 ] = ( + /6)γ(0) - (/2)γ() + σ 2, noting E[X t- Z t ] = E[X t- Z t ] = 0 also E[X t- X t ] = E[X t-2 ] E[X t- X t-2 ] so (5/4) γ() = γ(0) and ρ() = 4/5 To extract γ() and γ(0) we have (2/5 - /6) γ(0) = /3... γ(0) = (4 marks)

2 2) Let {X t } be the random walk defined by X t = X t- + Z t, t =,2,3,... with X 0 = 0. {Z t } is iid noise with a mean of zero and variance = σ 2. Z t is uncorrelated with X s for t>s. a) Explain why E[X t ] = 0 for t > 0. b) Show that E[X t X t ] = tσ 2, for t 0. c) Explain what could happen if we let the variance of Z t, σ 2 = /t for t a) Note that X t is NOT stationary. So mean and variance may vary with t. However E[X ] = E[X 0 ] + E[Z ] ; E[X 2 ] = E[X ] + E[Z 2 ] ; etc. and since X 0 and E[Z t ] = 0, this shows E[X ] = E[X 2 ] =... E[X t ] = 0 for all t. (3 marks) b) Since X t = X t- + Z t we have E[X t2 ] = E[X t-2 ] + E[Z t2 ] = E[X t-2 ] + σ 2 since E[X t- Z t ] = 0. Again starting from t = we have E[X 2 ] = E[X 02 ] + σ 2 = σ 2. since E[X 02 ] = 0. Then simply add σ 2 each time t increases by and E[X t X t ] = tσ 2 as required. (4 marks) c) In b the iid requirement meant that σ 2 was constant but if we relax that and z t is no longer iid we could have σ 2 = /t. In that case our iterative addition would give E[X t2 ] = + /2 + /3 + /4... the harmonic series which diverges to infinity. (3 marks)

3 3) Suppose we define a set of random numbers, x i with a probability density function (pdf) defined by f(x) = - x for x < and f(x) = 0 for x a) Sketch a well labelled diagram corresponding to f(x). b) Determine E[x i ] and E[x i2 ]. c) What is the corresponding cumulative distribution, F(x)? Provide a formula and a labelled diagram a) (2 marks) x b) E[x i ] = x( x )dx =0(by symmetry ) while f (x)dx=2 x dx=2[ x x 2 /2] 0 = 0 E[x i2 ] = x 2 ( x )dx = 2 0 x 2 ( x)dx=2[ x 3 /3 x 4 /4] 0 =/6 (4 marks) c) F(x) = 0 for x < - x t dt for x >. = (x + ) 2 /2 for -<x<0 and = /2 + x - (x) 2 /2 for 0<x< F(x) f(x) (4 marks) x 3

4 4) A time series generally contains a trend, a seasonal component and a stochastic component. The plot below shows 0 years of monthly mean temperature. Toronto Island Monthly Mean Temperature, Temperature Months Suppose that the 20 monthly values are available to you as a time series TM(i), i =,20. a) Explain, in some detail, how you would process these data to separate the 3 components indicated above? You can assume that the season is 2 months. b) Suppose it is Jan, 990 and you need some quick, simple, estimates of mean temperature for the coming month, based on the data set TM(i). What might you do? a) Should describe, with details, how to decompose a time series into a mean plus seasonal, trend and irregular components using moving averages or (preferably) start with seasonal term by getting averages for Jan, Feb, etc, then removing the seasonal component and determining a linear or other trend for the remaining, trend + random, part of the series. The trend can then be removed and the stochastic component assessed in terms of autocorrelations and possible ARIMA models. From the plot it does not appear that a multiplicative approach is appropriate so a simple additive approach is needed. 20 Could first find the mean, TTMM = TM (i). Then means for all Januaries, Februaries i= 0. Decembers through TMM(j) = i= TM (2(i )+ j) for j =,2,...2. One can then form a series with the seasonal term removed, and setting TMM(0) = TMM(2) TMSR(i) = TM(i) - TMM(i mod 2) 4

5 If we wanted to assume a linear trend we could seek a least squares fit line as the trend. This would give TMSRL(i) = A + B(i-) and a random component, TMR(i) = TMSR(i) - (A+B(i-)) The mean of TMSRL may differ slightly from zero if it is a fit to minimize least square differences so that should also be adjusted for. TMR2(i) = TMR(i) (A + 6B) One could then seek an ARIMA model for TMR2(i) to aid in understanding the random component of the time series and possibly for forecasting use. (8 marks) b) As quick, simple estimates one coud use simple persistence, or climatology, based on the seasonal component, or simple extrapolation. A better bet if you had the seasonal and trend components computed would be to use those plus persistence of the random component. (2 marks) 5

6 5) An ARMA(p,q) process can be defined by θ(b)x t = ϕ(b) Z t where Z t is WN(0,σ 2 ), B is the backward shift operator, θ and ϕ are polynomials of degree p and q respectively. a) Briefly explain the differences between AR, MA and ARMA processes. b) What is "White Noise" WN (0, σ 2 ). Explain "iid". c) Consider the ARMA(,) process with φ(b) = + 0.5B and θ(b) = -0.5B. Is this a stationary process? Explain why and what this means. d) Is the same process invertible? Explain why and what this means a) An autoregressive process AR(p) has the form θ(b)x t = Z t so that X t depends on previous values X t- etc plus white noise, Z t. A moving average MA(q) process X t = ϕ(b)z t has X t depending on a sum of white noise values, Z t, from current and previous times. An autoregressive - moving average process involves past values of both X and Z. (3marks) b) iid means independent and identically distributed so E[Z t Z t+k ] = 0 if k 0, and all have identical pdfs. WN(0,σ 2 ) is iid and further has a zero mean and variance σ 2, but no specific distribution is implied. (2 marks) c) Stationary implies that the mean and variance of multiple realizations of the random process X t are constant, not time dependent, and occurs when the roots (real or complex) of θ(z) = 0 are > in absolute value. Here root is z = 2, so {Z t } is stationary. (3 marks) d) Invertible means that one can express an ARMA process as an AR process of infinite order, and requires roots of ϕ(z) to be > in absolute value. Here root is z = -2, so {Z t } is invertible. It implies that the white noise Z t can be recovered from an infinite sequence of past X t values. (2 marks) 6