A random signal has no ordinary Fourier transform. What exists is a power spectral density. From \(N\) samples you get a random function \(\hat P\). Raw Fourier methods are asymptotically unbiased and inconsistent. Averaging or a lag-window buys variance. An AR / MA / ARMA model then writes \(P(e^{j\theta})\) from a handful of lags.
Exam block, every year: one MC on the table below, then a correlogram from given \(r[k]\), then an AR or MA fit, then a sketch on \([-\pi,\pi]\).
Total energy and average power of a discrete-time signal:
$$E_s=\sum_{n=-\infty}^{\infty}|x[n]|^2,\qquad P_s=\lim_{N\to\infty}\frac1N\sum_{n=0}^{N-1}|x[n]|^2.$$| Type | Energy | Average power | Example |
|---|---|---|---|
| Energy | finite | \(P_s=0\) | one pulse, \(a^n u[n]\) with \(|a|<1\) |
| Power | infinite | finite, \(P_s>0\) | everlasting sinusoid, WSS noise |
| Neither | infinite | infinite | a ramp |
An energy signal is characterised by its energy spectral density \(\mathcal E(e^{j\theta})=|X(e^{j\theta})|^2\). A power signal is characterised by its PSD \(P(e^{j\theta})\). Noise, speech, radar clutter: if you wait forever the energy blows up; the rate of energy stays finite. Those are power signals.
Parseval (energy): \(\sum_n|x[n]|^2=\frac1{2\pi}\int_{-\pi}^{\pi}\mathcal E(e^{j\theta})\,d\theta\). Power analogue:
$$P_s=\frac1{2\pi}\int_{-\pi}^{\pi}P(e^{j\theta})\,d\theta=r_x[0].$$The area under the PSD is the average power. \(P(e^{j\theta})\,d\theta/2\pi\) is the power in a small band.
A decaying exponential \(e^{-t}u(t)\) is an energy signal (\(E=\tfrac12\), \(P=0\)). A cosine is a power signal.
A realisation of a WSS process is not square-summable, and the next realisation is different. What is well-defined is the Fourier pair of the true autocorrelation and the true PSD:
$$P_x(e^{j\theta})=\sum_{l=-\infty}^{\infty}r_x[l]e^{-j\theta l},\qquad r_x[l]=\frac1{2\pi}\int_{-\pi}^{\pi}P_x(e^{j\theta})e^{jl\theta}\,d\theta.$$The pair needs the true \(r[l]\) at infinitely many lags. We have \(N\) samples of one realisation. Replacing \(r\) by \(\hat r\) produces a random \(\hat P\) with bias and variance. That is the whole subject.
Two families. Non-parametric (Fourier): no model of how samples correlate. Parametric: assume AR / MA / ARMA driven by white noise, estimate a few coefficients, plug into a closed-form PSD (which implicitly extrapolates \(r[l]\) past the observed lags).
A power signal must be windowed to length \(N\): \(\tilde x[n]=x[n]\) for \(n=0,\dots,N-1\).
Periodogram (direct):
$$\hat P_N(e^{j\theta})=\frac1N\bigl|X_N(e^{j\theta})\bigr|^2=\frac1N\left|\sum_{n=0}^{N-1}\tilde x[n]e^{-jn\theta}\right|^2.$$Correlogram (indirect):
$$\hat P_N(e^{j\theta})=\sum_{l=-(N-1)}^{N-1}\hat r_x[l]\,e^{-j\theta l}.$$These two are identical if \(\hat r\) is the biased autocorrelation estimator. Always use the biased ACF for a correlogram.
$$\hat r_b[l]=\frac1N\sum_{n=0}^{N-1-|l|}x[n+|l|]x^*[n],\qquad \mathbb E[\hat r_b[l]]=\frac{N-|l|}{N}r[l].$$Equivalently \(\hat r_b[l]=r[l]\,w_B[l]\) with the Bartlett (triangular) lag-window \(w_B[l]=(N-|l|)/N\). The unbiased estimator \(\hat r_{ub}\) divides by \(N-|l|\) instead. Its lag-window is rectangular, and the Dirichlet kernel \(W_R(e^{j\theta})\) takes negative values, so \(\mathrm{FT}\{\hat r_{ub}\}\) can go negative — an illegal PSD. \(W_B=|W_R|^2/N\ge 0\), so the biased correlogram stays non-negative. The periodogram \(\frac1N|X|^2\) is automatically \(\ge 0\).
Use the biased ACF in a correlogram, or \(\hat P\) may go negative. Truncating a true ACF to a few lags can also make the cosine polynomial dip below zero (Nov 2019 blue curve).
A finite record is a windowed infinite signal: \(\tilde x[n]=w[n]x[n]\). Multiplication in time is periodic convolution in frequency, \(\tilde X=X\,*_{2\pi}\,W\). The rectangular window of length \(N\) has the Dirichlet transform
$$W_R(e^{j\theta})=\frac{\sin(N\theta/2)}{\sin(\theta/2)}\,e^{-j(N-1)\theta/2}.$$A main lobe plus side lobes. The expected periodogram is the true PSD smeared by that kernel:
$$\mathbb E[\hat P]=\underbrace{P*W_{\mathrm{ML}}}_{\text{loss of resolution}}+\underbrace{P*W_{\mathrm{SL}}}_{\text{spectral leakage}}.$$| Window | 3 dB main-lobe bandwidth | First side-lobe |
|---|---|---|
| Rectangular | \(1.81\,\pi/(N-1)\) | \(-13\) dB |
| Triangular | \(5.01\,\pi/(N-1)\) | \(-27\) dB |
| Hanning | \(6.27\,\pi/(N-1)\) | \(-32\) dB |
Same \(N\): rectangular = best resolution, worst leakage. Hanning = worse resolution, less leakage.
To resolve two sines, \(|\theta_1-\theta_2|\) must exceed the 3 dB width. With \(\theta=2\pi f/f_s\),
$$|f_1-f_2|_{\min}=\frac{1.81\,f_s}{2(N-1)}.$$SLT5 2024–25, official: \(f_s=100\,\mathrm{Hz}\), \(N=30\) \(\Rightarrow\) \(|f_1-f_2|_{\min}=1.81\cdot 100/58\approx 3.12\,\mathrm{Hz}\approx 3\,\mathrm{Hz}\) (rectangular). Triangular, same \(N\): \(\approx 8.6\,\mathrm{Hz}\). Tapering hurts resolution.
The DFT of \(N\) samples evaluates the DTFT of the windowed signal at \(N\) frequencies \(k\cdot 2\pi/N\). Zero-padding to length \(L>N\) evaluates the same underlying DTFT on a finer grid \(k\cdot 2\pi/L\). You walk in smaller steps on the same smeared curve. You do not recover the infinite-data spectrum, and you do not split two peaks the main lobe already merged.
Zero-padding \(\neq\) resolution. “Zeropad to \(L\) samples the FTD at step \(2\pi/N\)” is false — the step becomes \(2\pi/L\). “Triangular has higher resolution than rectangular, same \(N\)” is false. “Rectangular leakage drops if you increase \(N\)” is false (side-lobe height is unchanged). True slogan: main-lobe width \(\leftrightarrow\) resolution, side-lobe height \(\leftrightarrow\) leakage.
Bias of the raw periodogram / correlogram: \(\mathbb E[\hat P]=P\,*_{2\pi}\,W\), a smoothed true PSD. As \(N\to\infty\), \(W\to\delta\), so the estimator is asymptotically unbiased. Finite \(N\): always resolution loss plus leakage.
Variance, Gaussian data through an LTI (reader 12.21–12.22):
$$\mathrm{Var}\bigl[\hat P(e^{j\theta})\bigr]=P(e^{j\theta})^2\left[1+\left(\frac1N\frac{\sin(\theta N)}{\sin\theta}\right)^2\right]\;\xrightarrow{N\to\infty}\;P(e^{j\theta})^2.$$At a generic \(\theta\), \(\mathrm{Var}\approx P^2\). It does not go to 0 as \(N\to\infty\). More data narrows the main lobe (less bias) but the estimate stays just as noisy. That is an inconsistent estimator. At \(\theta=0\) for real white Gaussian data the extra sinc inflates this to \(\mathrm{Var}=2P^2\); the exam slogan remains “variance \(\propto P^2\)”.
Periodogram variance does not fall with \(N\). A longer record is not a quieter periodogram.
Split \(N=K\cdot M\) into \(K\) non-overlapping length-\(M\) blocks. Periodogram each block, average:
$$\hat P_B(e^{j\theta})=\frac1K\sum_{i=1}^K\hat P_i(e^{j\theta}),\qquad\hat P_i=\frac1M\left|\sum_{n=0}^{M-1}x_i[n]e^{-j\theta n}\right|^2.$$Bias is that of a raw periodogram of length \(M\), not \(N\): shorter windows, wider main lobe, more bias. If the blocks are independent, \(\mathrm{Var}(\hat P_B)\approx P^2/K\). Variance drops as \(1/K\). Consistent only if you let \(K\to\infty\) (so \(M\) must not eat all of \(N\)).
Like Bartlett, but segments overlap (typically 50% or 75%) and each segment is usually windowed. Overlap \(\Rightarrow\) longer segments than Bartlett for the same \(K\) \(\Rightarrow\) less extra bias. Overlap \(\Rightarrow\) segments are dependent, even if the process is white, so the variance decreases, but not as \(1/K\).
WOSA variance is not \(1/K\). Course view at the same \(K\): WOSA variance is higher than Bartlett’s (dependence) and lower than the raw periodogram. Correct 2022 / Jan 2020 statement: Bartlett, \(K\) segments: variance \(\propto 1/K\), bias increases. “WOSA reduces variance as \(1/K\)” is false.
Raw \(\hat r_b[l]\) near \(|l|=N-1\) is one product — extremely noisy. BT tapers those lags with a symmetric lag-window \(w_L[l]\) of length \(2L-1\), \(L\ll N\):
$$\hat P_{\mathrm{BT}}(e^{j\theta})=\sum_{l=-(L-1)}^{L-1}w_L[l]\,\hat r_b[l]\,e^{-j\theta l}=\frac1{2\pi}\,\hat P*W_L.$$Need \(W_L(e^{j\theta})\ge 0\) (or \(w_L\) positive semi-definite). Rectangular / Hamming / Hann in the lag domain can make \(\hat P\) negative. Triangular is safe. If \(w_L\) is triangular and \(L=N\), BT is the raw correlogram.
Asymptotically unbiased if \(w_L[0]=1\). Variance (reader 12.36): \(\mathrm{Var}[\hat P_{\mathrm{BT}}]\approx (P^2/N)\sum w_L^2[l]\). The \(1/N\) in front \(\Rightarrow\) consistent as \(N\to\infty\) with \(L\) growing slower than \(N\). Larger \(L\): less bias, more variance.
Nov 2019 1c was this table. Every later MC tests the same cells. Official sample row (periodogram): “Estimate of PSD obtained by taking the modulus squared of FT of the observed signal. The periodogram is the Fourier transform of the biased estimator of \(r[\tau]\). The periodogram is asymptotically unbiased. The variance of the periodogram is proportional to the square of the true spectrum.”
| Method | Idea | Bias | Variance | Consistent? |
|---|---|---|---|---|
| Periodogram | \(\hat P=\frac1N\bigl|\sum_{n=0}^{N-1}x[n]e^{-j\theta n}\bigr|^2\) | FT of the biased ACF. Smoothed true PSD. Asymptotically unbiased as \(N\to\infty\). | \(\approx P^2\). Does not fall with \(N\). | No |
| Correlogram | \(\hat P=\sum_l \hat r_b[l]e^{-j\theta l}\). Same number as the periodogram if \(\hat r_b\) is used. | Same as periodogram. \(\hat r_{ub}\) can make \(\hat P<0\). | Same, \(\propto P^2\) | No |
| Bartlett | Average \(K\) periodograms of non-overlapping length-\(M\) blocks, \(N=KM\). | Worse than raw: each block is shorter, main lobe wider. | \(\approx P^2/K\) if blocks independent | Only if \(K\to\infty\) |
| Welch / WOSA | Average periodograms of overlapping, usually windowed, segments. Typical overlap 50–75%. | Extra bias smaller than Bartlett (segments longer). | Lower than raw, higher than Bartlett. Not \(1/K\). | Not “\(1/K\)-consistent” |
| Blackman–Tukey | Window the ACF, then FT. Down-weight unreliable large lags. | Asymptotically unbiased if \(w[0]=1\). Larger lag-window \(L\) \(\Rightarrow\) less bias, more variance. | \(\approx (P^2/N)\sum w^2[l]\). Scales as \(1/N\). | Yes (\(N\to\infty\)) |
Non-parametric methods treat the record as one period of a periodic signal and set \(\hat r[l]=0\) for \(|l|\ge N\). Both are usually false, and they cap the resolution. Parametric idea: guess a generating filter, estimate a few parameters, write \(P(e^{j\theta})\) from the model.
Innovation representation: WSS with a rational PSD = white \(i[n]\) through a causal LTI \(L(z)\). Restrict \(L\) to rational \(\Rightarrow\) ARMA. Sign convention in this course: \(x[n]=i[n]-\sum_{k=1}^p a_k x[n-k]\), so the AR polynomial is \(A(z)=1+\sum a_k z^{-k}\).
Memorise the three PSD formulas.
AR(\(p\)) — all-pole, \(H(z)=1/A(z)\):
$$P_{\mathrm{AR}}(e^{j\theta})=\frac{\sigma_i^2}{\bigl|1+a_1 e^{-j\theta}+\cdots+a_p e^{-jp\theta}\bigr|^2}.$$MA(\(q\)) — all-zero, \(b_0=1\):
$$P_{\mathrm{MA}}(e^{j\theta})=\sigma_i^2\bigl|1+b_1 e^{-j\theta}+\cdots+b_q e^{-jq\theta}\bigr|^2.$$ARMA(\(p,q\)):
$$P_{\mathrm{ARMA}}(e^{j\theta})=\sigma_i^2\frac{\bigl|1+b_1 e^{-j\theta}+\cdots+b_q e^{-jq\theta}\bigr|^2}{\bigl|1+a_1 e^{-j\theta}+\cdots+a_p e^{-jp\theta}\bigr|^2}.$$AR(\(p\)): \(r[l]\) has infinite duration. Yule–Walker (real, \(r[-l]=r[l]\)):
$$r[l]=\begin{cases}\sigma_i^2-\sum_{k=1}^p a_k r[l-k],& l=0,\\-\sum_{k=1}^p a_k r[l-k],& |l|>0.\end{cases}$$Matrix form (Toeplitz). The right-hand side is \((\sigma_i^2,0,\dots,0)^\top\), not \((r[1],r[2],\dots)\):
$$\begin{bmatrix}r[0]&r[1]&\cdots&r[p]\\r[1]&r[0]&\cdots&r[p-1]\\\vdots&\vdots&\ddots&\vdots\\r[p]&r[p-1]&\cdots&r[0]\end{bmatrix}\begin{bmatrix}1\\a_1\\\vdots\\a_p\end{bmatrix}=\begin{bmatrix}\sigma_i^2\\0\\\vdots\\0\end{bmatrix}.$$The last \(p\) rows are linear in the \(a_k\). The first row then gives \(\sigma_i^2\).
MA(\(q\)): \(r[l]=0\) for \(|l|>q\) (finite ACF). With \(b_0=1\),
$$r[l]=\sigma_i^2\sum_{k=|l|}^q b_k b_{k-|l|},\qquad 0\le|l|\le q.$$Fitting \((b_k,\sigma_i^2)\) from lags is nonlinear (products \(b_k b_{k-\ell}\)).
ARMA(\(p,q\)): the MA contribution dies after lag \(q\). For \(|l|>q\) the ACF is purely AR, \(r[l]=-\sum_{k=1}^p a_k r[l-k]\). Estimate AR from the tail of \(r\), then peel it off to fit MA.
A 1-step Wiener FIR predictor solves the same Yule–Walker system: \(w_{\mathrm{opt}}=-\hat a\), MMSE \(J=\hat\sigma_i^2\). Exam T/F (Nov 2019 1b.E): “Wiener 1-step predictor can be used for MA parametric estimation” is false. It estimates AR coefficients.
\(x[n]=i[n]-a_1 x[n-1]\). YW at lags 0 and 1:
$$r[0]=\sigma_i^2-a_1 r[1],\qquad r[1]=-a_1 r[0].$$ $$a_1=-\frac{r[1]}{r[0]},\qquad\sigma_i^2=r[0]+a_1 r[1].$$True AR(1) process: \(r[0]=\sigma_i^2/(1-a_1^2)\), \(r[l]=r[0](-a_1)^{|l|}\).
Shape: \(a_1<0\) (pole on the positive real axis) \(\Rightarrow\) low-frequency peak. \(a_1>0\) \(\Rightarrow\) high-frequency peak. Useful identity: \(|1+\alpha e^{-j\theta}|^2=1+\alpha^2+2\alpha\cos\theta\).
From \(r[0],r[\pm 1]\) for MA(1):
$$r[0]=\sigma_i^2(1+b_1^2),\qquad r[1]=\sigma_i^2 b_1.$$Quadratic in \(b_1\): two real roots, reciprocals. They give the same \(P(e^{j\theta})\) (same magnitude response). Uniqueness: the innovation filter must be causal, stable, and minimum-phase = all zeros (and poles) inside the unit circle. Pick \(|b_1|<1\).
Spectral factorisation: \(P(z)=\sigma_i^2 L(z)L(z^{-1})\) with \(L\) min-phase and \(l[0]=1\). For every root \(z_k\) of a valid PSD, \(1/z_k^*\) is also a root; keep the ones inside the disc.
If you skip parameter fitting and just take \(\hat P=\sum_{l=-q}^q \hat r[l]e^{-jl\theta}\) (rectangular-windowed correlogram of width \(2q+1\)), you get a valid MA(\(q\)) spectrum shape but it can go negative if \(\hat r\) is not a valid MA ACF. Fitting \((b_k,\sigma_i^2)\) then using \(\sigma_i^2|B(e^{j\theta})|^2\) guarantees \(\hat P\ge 0\).
Favourite exam sentence: for a true MA(\(q\)), the correlogram that uses exactly lags \(|l|\le q\) equals the MA PSD. The ACF really is zero outside. An AR fit of the same lags is a different curve.
Official key. Given \(r[0]=1\), \(r[\pm 1]=0.7\), \(r[\pm 2]=0.5\), \(r[\pm 3]=0.3\), \(r[\pm 4]=0\).
At \(\theta=0\): \(1+1.4+1+0.6=4\). The truncated ACF is not a valid infinite ARMA autocorrelation, so this cosine polynomial can go slightly negative (official sketch: blue curve, small negative sidelobes). Label \(\theta\in[-\pi,\pi]\) and the values at \(0,\pi\).
Unknowns \(a_1,a_2,\sigma_i^2\). Use lags 0, 1, 2:
$$\begin{bmatrix}1&0.7&0.5\\0.7&1&0.7\\0.5&0.7&1\end{bmatrix}\begin{bmatrix}1\\a_1\\a_2\end{bmatrix}=\begin{bmatrix}\sigma_i^2\\0\\0\end{bmatrix}.$$Rows 2–3:
$$\begin{aligned}0.7+a_1+0.7 a_2&=0 &&\Rightarrow&& a_1+0.7 a_2=-0.7,\\0.5+0.7 a_1+a_2&=0 &&\Rightarrow&& 0.7 a_1+a_2=-0.5.\end{aligned}$$From the first, \(a_1=-0.7-0.7 a_2\). Plug into the second:
$$0.7(-0.7-0.7 a_2)+a_2=-0.5\Rightarrow -0.49-0.49 a_2+a_2=-0.5\Rightarrow 0.51 a_2=-0.01.$$ $$a_2=-\frac{0.01}{0.51}=-\frac{1}{51}\approx -0.01961,\qquad a_1=-0.7-0.7(-0.01961)\approx -0.6863.$$First row: \(\sigma_i^2=1+a_1\cdot 0.7+a_2\cdot 0.5\). Unrounded this is \(\approx 0.5098\). Official key rounds to \(\sigma_i^2\approx 0.5089\). Because \(a_1\) is negative,
$$P_{\mathrm{AR2}}(e^{j\theta})=\frac{0.5089}{\bigl|1-0.6863 e^{-j\theta}-0.0196 e^{-j2\theta}\bigr|^2}.$$The AR(2) peak at \(\theta=0\) is sharper than the correlogram: the model extrapolates \(r[l]\) past lag 3, and the PSD is forced to stay non-negative.
Official sol. \(H(z)=\dfrac{1-\frac23 z^{-1}}{1-\frac12 z^{-1}}\), \(\sigma_i^2=1\). Difference equation:
$$x[k]=i[k]-\tfrac23 i[k-1]+\tfrac12 x[k-1].$$So \(b_1=-\frac23\), \(a_1=-\frac12\), \(b_0=1\). Impulse response by polynomial division:
$$H(z)=1+\Bigl(\tfrac12-\tfrac23\Bigr)z^{-1}+\cdots=1-\tfrac16 z^{-1}+\cdots\qquad\Rightarrow\qquad h[0]=1,\; h[1]=-\tfrac16.$$Modified Yule–Walker at lags 0 and 1 (\(\sigma_i^2=1\)):
$$\begin{aligned}r[0]&=b_0 h[0]+b_1 h[1]-a_1 r[1]=1+\Bigl(-\tfrac23\Bigr)\Bigl(-\tfrac16\Bigr)+\tfrac12 r[1]=\tfrac{10}{9}+\tfrac12 r[1],\\r[1]&=b_1 h[0]-a_1 r[0]=-\tfrac23+\tfrac12 r[0].\end{aligned}$$Substitute the second into the first:
$$r[0]=\tfrac{10}{9}+\tfrac12\Bigl(-\tfrac23+\tfrac12 r[0]\Bigr)=\tfrac{10}{9}-\tfrac13+\tfrac14 r[0]=\tfrac79+\tfrac14 r[0].$$ $$r[0]-\tfrac14 r[0]=\tfrac79\Rightarrow\tfrac34 r[0]=\tfrac79\Rightarrow r[0]=\tfrac{28}{27}.$$ $$r[1]=-\tfrac23+\tfrac12\cdot\tfrac{28}{27}=-\tfrac{18}{27}+\tfrac{14}{27}=-\tfrac{4}{27}.$$For \(|\tau|\ge 2\) only AR remains: \(r[\tau]=-a_1 r[\tau-1]=\frac12 r[\tau-1]\), so \(r[\tau]=\bigl(\tfrac12\bigr)^{|\tau|-1}r[1]\) for \(|\tau|\ge 1\).
AR(1) fit from \((r[0],r[1])\) only (this cannot match the MA zero):
$$a_1=-\frac{r[1]}{r[0]}=\frac{4/27}{28/27}=\frac17,\qquad\sigma_i^2=r[0]+a_1 r[1]=\frac{28}{27}+\frac17\Bigl(-\frac4{27}\Bigr)=\frac{196-4}{189}=\frac{64}{63}.$$ $$P_{\mathrm{AR1}}(e^{j\theta})=\frac{64/63}{\bigl|1+\frac17 e^{-j\theta}\bigr|^2}.$$The official sol’s last cosine simplification of this PSD is algebraically wrong. The parameters \(a_1=\frac17\), \(\sigma_i^2=\frac{64}{63}\) and the \(|1+\frac17 e^{-j\theta}|^{-2}\) form are official and correct. Sketch: \(a_1>0\) \(\Rightarrow\) more power near \(\theta=\pm\pi\).
Same ARMA(1,1) recipe on every SLT6 and on Pr21 / Nov 2022: read \(a_k,b_k\) from \(H(z)\) or the DE \(\to\) \(h[0],h[1]\) \(\to\) modified YW for \(r[0],r[1]\) \(\to\) AR tail \(\to\) fake AR(1) from the \(2\times 2\) YW block. Nov 2022 uses different numbers and has no official key; do not quote it as a numbered answer.
Jan 2019 1c, two true. (A) Bartlett: bias increases as segment length \(M\) decreases. (B) Welch with 50% overlap: variance decreases as \(1/K\). (C) Blackman–Tukey is consistent. (D) Zero-padding increases spectral resolution.
Official: A and C. B is the WOSA trap (overlap kills \(1/K\)). D is interpolation, not resolution.
SLT6 2024–25, official. \(r[0]=\frac{13}{6}\), \(r[\pm 1]=-1\), and \(r[|l|\ge 2]=0\). Fit MA(1). Which root do you keep? Show that the MA PSD equals the three-lag correlogram.
\(r[0]=\sigma_i^2(1+b_1^2)\), \(r[1]=\sigma_i^2 b_1=-1\). Eliminate \(\sigma_i^2=-1/b_1\):
$$-\frac{1+b_1^2}{b_1}=\frac{13}{6}\Rightarrow 6+6b_1^2+13 b_1=0\Rightarrow b_1=\frac{-13\pm 5}{12}.$$Two solutions: \(b_1=-\frac32\), \(\sigma_i^2=\frac23\) and \(b_1=-\frac23\), \(\sigma_i^2=\frac32\). Min-phase: zero of \(1+b_1 z^{-1}\) at \(z=-b_1\). Need \(|b_1|<1\) \(\Rightarrow\) \(b_1=-\frac23\), \(\sigma_i^2=\frac32\) (zero at \(+\frac23\)). The other is max-phase (zero at \(\frac32\)).
$$P_{\mathrm{MA}}=\sigma_i^2(1+b_1^2+2b_1\cos\theta)=\frac{13}{6}-2\cos\theta.$$Correlogram with the same three lags: \(\frac{13}{6}-e^{-j\theta}-e^{j\theta}=\frac{13}{6}-2\cos\theta\). They coincide because a true MA(1) has \(r[l]=0\) for \(|l|\ge 2\).
5CTA0 Q1 2026–27 · from the course reader, lectures and official keys.