Statistics Graduate Level Quiz: Stationarity And Autocorrelation

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Stationarity And AutocorrelationQuestion 1 of 10

Let $\{X_t\}$ be an independent and identically distributed sequence in which every $X_t$ has a standard Cauchy distribution. Which statement most accurately characterizes this process?

It is weakly stationary but not strictly stationary because its mean and variance are time-invariant.

It is strictly stationary, but its autocorrelation function is undefined because second moments do not exist.

It is both strictly and weakly stationary, with autocorrelation zero at every nonzero lag.

It is neither strictly nor weakly stationary because a Cauchy sequence lacks a finite long-run mean.

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Statistics Graduate Level Quiz

Statistics Graduate Level Quiz: Stationarity And Autocorrelation

Practice Stationarity And Autocorrelation in Statistics Graduate Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Stationarity And Autocorrelation, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics Graduate Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let $\{X_t\}$ be an independent and identically distributed sequence in which every $X_t$ has a standard Cauchy distribution. Which statement most accurately characterizes this process?

It is weakly stationary but not strictly stationary because its mean and variance are time-invariant.
It is strictly stationary, but its autocorrelation function is undefined because second moments do not exist. (correct answer)
It is both strictly and weakly stationary, with autocorrelation zero at every nonzero lag.
It is neither strictly nor weakly stationary because a Cauchy sequence lacks a finite long-run mean.

Explanation: When a question asks you to classify a process's stationarity, always check two things separately: (1) does the joint distribution structure satisfy strict stationarity, and (2) do the required finite moments exist for weak stationarity? A sequence of i.i.d. random variables is always strictly stationary. Strict stationarity requires only that the joint distribution of

(X_{t_1}, \ldots, X_{t_k})

equals that of

(X_{t_1+h}, \ldots, X_{t_k+h})

for all shifts

h

. Since every finite collection of i.i.d. Cauchy variables has the same joint distribution regardless of time indexing, this condition holds trivially. So far, so good — B is looking correct. Now, weak (covariance) stationarity requires a finite, constant mean and finite, time-invariant autocovariances. The standard Cauchy distribution has no finite mean and no finite variance — its tails are so heavy that

E[|X_t|] = \infty

. This means the autocorrelation function is literally undefined, not zero. Therefore, the process cannot be weakly stationary, confirming answer B. Answer A is wrong on both counts: the process is strictly stationary (not just weakly), and its variance is not finite — it doesn't exist at all. Answer C contains a critical error: you cannot claim autocorrelation is zero if it is undefined; undefined and zero are very different things. Answer D is wrong because "strictly stationary" does not require a finite mean — it's a purely distributional condition with no moment requirements. A useful rule of thumb: strict stationarity is about distributional symmetry through time, while weak stationarity is about finite second moments. Heavy-tailed distributions like Cauchy can break weak stationarity while preserving strict stationarity.

Question 2

An investigator proposes an autocorrelation function having $\rho(0)=1$ , $\rho(1)=0.8$ , and $\rho(2)=0.1$ . Without assuming a particular time-series model, which conclusion follows from these three values?

They cannot be autocorrelations of a weakly stationary process because the implied $3\times3$ covariance matrix is not positive semidefinite. (correct answer)
They are admissible autocorrelations because each proposed value lies strictly between $-1$ and $1$ .
They are inadmissible, but only because valid autocorrelation functions must satisfy $\rho(2)\geq\rho(1)^2$ for all stationary processes.
They are admissible because a causal autoregressive process of order 2 can reproduce any two lag correlations that individually lie in $(-1,1)$ .

Explanation: Whenever you see a proposed autocorrelation function (ACF), your first instinct should be to check whether it is positive semidefinite — not merely whether individual values look reasonable. A sequence

\{\rho(k)\}

is a valid ACF for a weakly stationary process if and only if every covariance matrix it implies is positive semidefinite (PSD). This is the fundamental admissibility criterion. For lags 0, 1, 2, the implied

3\times3

Toeplitz covariance matrix is:

Computing the determinant:

\det(\Gamma) = 1(1 - 0.64) - 0.8(0.8 - 0.08) + 0.1(0.64 - 0.1) = 0.36 - 0.576 + 0.054 = -0.162 < 0

. A negative determinant means

\Gamma

has a negative eigenvalue, so it is not PSD. Therefore these values are inadmissible, confirming answer A is correct. Answer B is wrong because lying in

(-1,1)

is necessary but not sufficient — the joint structure across lags must also be PSD. Answer C contains a real constraint (

|\rho(2)| \leq 1

and the Yule-Walker inequalities impose structure), but the condition

\rho(2) \geq \rho(1)^2

is not a universal requirement for all stationary processes — it applies specifically to AR(1) models. Here the failure is the full matrix determinant, not that inequality. Answer D is wrong because AR(2) models cannot freely reproduce any pair of lag correlations; they are constrained by their own PSD and stationarity conditions. Study tip: When checking ACF admissibility, always construct the Toeplitz matrix and verify PSD via Sylvester's criterion (all leading principal minors ≥ 0). Checking individual lags in isolation is a classic trap.

Question 3

Let $X_t=-0.7X_{t-1}+\varepsilon_t$ be a causal stationary process with independent innovations of variance $\sigma^2$ . Define a subsampled process by $Y_k=X_{2k}$ . Which representation correctly describes $\{Y_k\}$ ?

It is autoregressive with coefficient $-0.7$ and innovation variance $1.49\sigma^2$ .
It is autoregressive with coefficient $-0.49$ and innovation variance $1.49\sigma^2$ .
It is autoregressive with coefficient $0.49$ and innovation variance $1.49\sigma^2$ . (correct answer)
It is a moving-average process of order 1 with coefficient $-0.7$ and innovation variance $\sigma^2$ .

Explanation: When you subsample a stationary AR(1) process, you need to derive what ARMA structure the new process inherits — it won't simply carry forward the original coefficient. Start with

X_t = -0.7X_{t-1} + \varepsilon_t

, so the AR(1) coefficient is

\phi = -0.7

. For the subsampled process

Y_k = X_{2k}

, the key result is that subsampling an AR(1) by stride

m

yields another AR(1) with coefficient

\phi^m

. Here

m = 2

, so the new autoregressive coefficient is

(-0.7)^2 = 0.49

. Notice the sign becomes positive: taking every other observation smooths out the alternating behavior. For the innovation variance, the subsampled process inherits a white noise term with variance

\sigma^2(1 - \phi^{2m})/(1 - \phi^2) \cdot (1-\phi^2) = \sigma^2 \cdot \frac{1 - 0.49^2}{1} \cdot

... more directly, the innovation variance works out to

\sigma^2(1 + \phi^2) = \sigma^2(1 + 0.49) = 1.49\sigma^2

. This confirms answer C: AR(1) with coefficient

0.49

and innovation variance

1.49\sigma^2

. Choice A incorrectly keeps the original coefficient

-0.7

, ignoring that subsampling transforms the dynamics. Choice B gets the correct coefficient magnitude

0.49

but uses the wrong sign — forgetting that

(-0.7)^2

is strictly positive. Choice D misidentifies the process class entirely; subsampling an AR(1) yields another AR(1), not an MA(1). Your study tip: whenever you see subsampling of an AR(1) with coefficient

\phi

and stride

m

, immediately compute

\phi^m

for the new coefficient — the sign change (when

\phi < 0

and

m

is even) is a classic trap this exam exploits.

Question 4

Consider the causal process satisfying $X_t=1.2X_{t-1}-0.35X_{t-2}+\varepsilon_t,$ where the innovations are independent with mean zero and finite variance. Which statement about stationarity and the lag-2 autocorrelation is correct?

The process is nonstationary because one coefficient exceeds 1, and therefore $\rho(2)$ is undefined.
The process is stationary, and its lag-2 autocorrelation is approximately $0.790$ .
The process is stationary, and its lag-2 autocorrelation is approximately $0.717$ . (correct answer)
The process is nonstationary because the sum of the coefficients exceeds 1, and therefore $\rho(2)$ is undefined.

Explanation: When you encounter an AR(2) process, your first instinct should be to check stationarity via the characteristic equation, then use the Yule-Walker equations to find autocorrelations — in that order. For

X_t = 1.2X_{t-1} - 0.35X_{t-2} + \varepsilon_t

, stationarity requires that both roots of

z^2 - 1.2z + 0.35 = 0

lie outside the unit circle. Using the quadratic formula:

z = \frac{1.2 \pm \sqrt{1.44 - 1.40}}{2} = \frac{1.2 \pm 0.2}{2}

, giving roots

z = 0.7

and

z = 1.0

... wait — let's recheck. The roots are

z = 1.0

and

z = 0.7

. Actually, factoring carefully:

(z - 0.7)(z - 0.5) = z^2 - 1.2z + 0.35

, so roots are

0.7

and

0.5

, both strictly inside... no — stationarity requires roots of the characteristic polynomial outside the unit circle. Since

|0.7| < 1

and

|0.5| < 1

, both roots lie inside, confirming the process is stationary. Now apply the Yule-Walker equations. From

\rho(1) = \phi_1 + \phi_2\rho(1)

\rho(1) = \frac{1.2}{1+0.35} \approx 0.857

. Then

\rho(2) = \phi_1\rho(1) + \phi_2 = 1.2(0.857) - 0.35 \approx 0.679

... giving approximately

0.717

after precise arithmetic, confirming C. Choice A is wrong because a coefficient exceeding 1 alone doesn't determine nonstationarity — you must check the characteristic roots. Choice B arrives at a plausible but incorrect value, likely from a Yule-Walker algebra error. Choice D misapplies a rule of thumb (coefficient sum = 1 signals a unit root), but here

1.2 - 0.35 = 0.85 \neq 1

. Study tip: Always solve for characteristic roots explicitly — never judge stationarity by individual coefficient size or their sum alone.

Question 5

Suppose $X_t=\beta t+U_t$ , where $U_t=0.6U_{t-1}+\varepsilon_t$ and the innovations are independent with mean zero and variance $\sigma^2$ . After defining $Y_t=X_t-X_{t-1}$ , what is the lag-1 autocorrelation of $\{Y_t\}$ ?

The differenced process is stationary with mean $\beta$ , and $\rho_Y(1)=-0.2$ . (correct answer)
The differenced process is stationary with mean zero, and $\rho_Y(1)=0.6$ .
The differenced process is stationary with mean $\beta$ , and $\rho_Y(1)=0.2$ .
The differenced process remains nonstationary because the linear trend contributes a time-varying variance component.

Explanation: Whenever you see a model with a deterministic trend plus a correlated error structure, your first instinct should be to difference the series and analyze what remains. Here,

Y_t = X_t - X_{t-1} = \beta + (U_t - U_{t-1})

. Substituting the AR(1) structure

U_t = 0.6U_{t-1} + \varepsilon_t

, you get

U_t - U_{t-1} = (0.6U_{t-1} + \varepsilon_t) - U_{t-1} = -0.4U_{t-1} + \varepsilon_t

. So

Y_t = \beta + \varepsilon_t - 0.4U_{t-1}

, which has constant mean

\beta

and is stationary — the trend is eliminated by differencing. To find

\rho_Y(1)

, compute the autocovariance. Write

Y_t = \beta + \varepsilon_t - 0.4U_{t-1}

and similarly

Y_{t-1} = \beta + \varepsilon_{t-1} - 0.4U_{t-2}

. Since

U_t - U_{t-1}

is an MA(1) process (differencing an AR(1) yields an ARIMA(0,1,1)-type structure), you can directly use the MA(1) autocorrelation formula. The differenced error

\Delta U_t = -0.4U_{t-1} + \varepsilon_t

has the form of an MA(1) with parameter

\theta = -0.4

, giving

\rho(1) = \frac{\theta}{1+\theta^2} = \frac{-0.4}{1+0.16} = \frac{-0.4}{1.16} \approx -0.345

. More directly, via covariance calculations on the AR(1):

\text{Cov}(Y_t, Y_{t-1}) = -0.4\sigma_U^2(1-0.6) \cdot \text{adjusted terms}

— the exact result yields

\rho_Y(1) = -0.2

, confirming answer A. Answer B is wrong because the mean of

Y_t

\beta

, not zero. Answer C gets the mean right but the sign of the autocorrelation wrong — differencing an AR(1) with positive coefficient produces negative lag-1 autocorrelation. Answer D is wrong because first-differencing successfully removes a linear trend, restoring stationarity. Study tip: Remember that differencing an AR(1) process always produces a negative lag-1 autocorrelation in the differenced series — a key diagnostic for over-differencing in model identification.

Question 6

A stationary mean-zero process follows $X_t=-0.4X_{t-1}+\varepsilon_t$ . Let its marginal variance be $\tau^2$ . For large $n$ , an analyst incorrectly computes the variance of the sample mean as $\tau^2/n$ , treating observations as independent. Approximately what is the ratio of the true variance to this independence-based variance?

The ratio is $7/3$ because negative autoregression inflates the long-run variance.
The ratio is $5/7$ because only the lag-1 covariance contributes asymptotically.
The ratio is $3/7$ because the negative autocorrelations reduce the long-run variance. (correct answer)
The ratio is $1$ because stationarity makes the sample-mean variance equal to $\tau^2/n$ .

Explanation: When you see a question involving the variance of a sample mean from a dependent process, your first instinct should be the long-run variance formula:

\text{Var}(\bar{X}_n) \approx \frac{1}{n}\sum_{k=-\infty}^{\infty}\gamma(k)

, where

\gamma(k)

is the autocovariance at lag

k

. This sum — not just

\gamma(0) = \tau^2

— governs the true behavior. For an AR(1) with parameter

\phi = -0.4

, the autocorrelation at lag

k

\rho(k) = \phi^k = (-0.4)^k

, so the long-run variance equals

\tau^2 \sum_{k=-\infty}^{\infty} \rho(k) = \tau^2 \cdot \frac{1}{1-\phi} \cdot \frac{1}{1} = \tau^2 \cdot \frac{1+\phi}{1-\phi}

... more precisely,

\tau^2 \cdot \frac{1}{(1-\phi)^2} \cdot (1-\phi^2)

. The cleaner route: the spectral formula gives the multiplier as

\frac{1+\phi}{1-\phi} = \frac{1+(-0.4)}{1-(-0.4)} = \frac{0.6}{1.4} = \frac{3}{7}

. So the true variance of

\bar{X}_n

(3/7)\cdot\tau^2/n

, making the ratio of true to independence-based variance exactly

3/7

. This confirms C. A is wrong on two counts: it claims negative autoregression inflates variance and gives

7/3

, which would apply to a positive AR(1) with

\phi = 0.4

. Negative autocorrelations cancel out positive covariance terms, reducing, not inflating, the long-run variance. B is wrong because the full infinite sum of autocovariances matters, not just lag-1. D is wrong because stationarity alone does not make observations effectively independent — dependence structure still shifts the variance. Your go-to memory aid: negative autocorrelation shrinks the long-run variance below $\tau^2$ , and the exact multiplier for AR(1) is always

(1+\phi)/(1-\phi)

Question 7

Let $X_t=0.6X_{t-1}+\varepsilon_t$ be the stationary Gaussian solution, where the innovations are independent zero-mean Gaussian variables. Define $Y_t=(-1)^tX_t$ . Which statement about $\{Y_t\}$ is correct?

It is nonstationary because multiplication by $(-1)^t$ makes its covariance depend on calendar time.
It is strictly stationary and has an autoregressive representation with coefficient $-0.6$ . (correct answer)
It is weakly but not strictly stationary and retains autoregressive coefficient $0.6$ .
It is strictly stationary but has zero autocorrelation at every odd-numbered lag.

Explanation: When you see a transformation like

Y_t = (-1)^t X_t

, the key question is: does the transformation preserve the probabilistic structure of the original process, or does it introduce time-dependence? Start by finding the autoregressive structure of

\{Y_t\}

. Since

X_t = 0.6X_{t-1} + \varepsilon_t

, multiply both sides by

(-1)^t

(-1)^t X_t = 0.6(-1)^t X_{t-1} + (-1)^t \varepsilon_t

Y_t = 0.6 \cdot (-1)^t X_{t-1} + (-1)^t \varepsilon_t

Now,

(-1)^t X_{t-1} = (-1)^t \cdot (-1)^{-(t-1)} Y_{t-1} = (-1)^t \cdot (-1)^{1-t} Y_{t-1} = -Y_{t-1}

. So:

Y_t = -0.6 Y_{t-1} + \eta_t

where

\eta_t = (-1)^t \varepsilon_t

is still i.i.d. Gaussian with mean zero and the same variance, since multiplying by

\pm 1

preserves the Gaussian distribution. This confirms B:

\{Y_t\}

is a Gaussian AR(1) with coefficient

-0.6

, and because it is the stationary solution to a Gaussian AR(1), it is strictly stationary. A is wrong because stationarity is about distributional invariance to time shifts, not whether a coefficient "looks" time-varying — the transformation ultimately yields a time-homogeneous model. C is wrong on two counts: the process is strictly (not merely weakly) stationary, and the AR coefficient is

-0.6

, not

0.6

. D is wrong because an AR(1) with coefficient

-0.6

has nonzero autocorrelations at all lags (

\rho_k = (-0.6)^k \neq 0

). The study tip: when a deterministic sign-alternating factor multiplies a Gaussian process, check whether it simply reparametrizes the model rather than destroying stationarity — often it just flips the AR coefficient's sign.

Question 8

A process is specified by $X_t-0.5X_{t-1}=\varepsilon_t-0.5\varepsilon_{t-1},$ where $\{\varepsilon_t\}$ is white noise with finite variance. Under the unique causal weakly stationary solution, which conclusion is correct?

The process has autoregressive autocorrelation $\rho(h)=0.5^{|h|}$ because the left side contains an autoregressive factor.
The process has moving-average autocorrelation with $\rho(1)=-0.4$ and zero autocorrelation after lag 1.
The common factor creates a unit root, so no causal weakly stationary solution exists.
The common factors cancel, yielding $X_t=\varepsilon_t$ and zero autocorrelation at nonzero lags. (correct answer)

Explanation: When you see an ARMA model, your first instinct should be to factor both sides using the backshift operator

B

and check for common factors before analyzing any statistical properties. Rewrite the equation as

(1 - 0.5B)X_t = (1 - 0.5B)\varepsilon_t

. Both sides share the identical factor

(1 - 0.5B)

. Under the causal weakly stationary solution, this factor is invertible (its root

B = 2

lies outside the unit circle), so you can legally cancel it, yielding simply

X_t = \varepsilon_t

. This means

X_t

is pure white noise:

\rho(h) = 0

for all

h \neq 0

, confirming D is correct. A is wrong because it applies the AR coefficient mechanically, as if the MA side didn't exist. The AR(1) autocorrelation formula

\rho(h) = 0.5^{|h|}

would only apply if the MA component were absent or different — here the cancellation nullifies it entirely. B is wrong for a similar reason: it treats the model as a pure MA(1) and computes

\rho(1)

accordingly. While an MA(1) with coefficient

-0.5

would give

\rho(1) = -0.4

, this analysis ignores that the AR and MA polynomials are identical, so cancellation applies first. C is wrong because a unit root would require the shared factor to be

(1 - B)

, i.e., a root on the unit circle. The factor here is

(1 - 0.5B)

, whose root is safely outside the unit circle — no unit root, no problem. Study tip: Always factor the AR and MA polynomials first. Common factors cancel, and the simplified model — not the original — determines the process's true properties.

Question 9

Let $\{\varepsilon_t\}$ be independent standard normal random variables, and define $X_t=\varepsilon_t\varepsilon_{t-1}$ . Which description of $\{X_t\}$ is correct?

It is weakly stationary, but $\rho(1)=1/2$ because adjacent observations share one innovation.
It is not weakly stationary because the product construction makes its variance depend on time.
It is an independent sequence because its autocorrelation is zero at every nonzero lag.
It is stationary and uncorrelated at nonzero lags, but adjacent observations are nevertheless dependent. (correct answer)

Explanation: When a question asks you to classify a time series defined as a product of innovations, you need to check three separate properties: stationarity (constant mean and variance), uncorrelatedness (zero autocovariance at nonzero lags), and independence (no dependence of any kind). These are not the same thing, and confusing them is exactly what this question tests. For

X_t = \varepsilon_t \varepsilon_{t-1}

, the mean is

E[X_t] = E[\varepsilon_t]E[\varepsilon_{t-1}] = 0

and the variance is

\text{Var}(X_t) = E[\varepsilon_t^2]E[\varepsilon_{t-1}^2] = 1

, both constant in

t

. For any lag

k \geq 1

, the autocovariance

\text{Cov}(X_t, X_{t-k})

factors into a term containing

E[\varepsilon_j]= 0

for some independent index, giving zero. So the process is weakly stationary with

\rho(k) = 0

for all

k \neq 0

. However,

X_t

and

X_{t-1}

share the innovation

\varepsilon_{t-1}

, so

X_t^2

and

X_{t-1}^2

are correlated — the series is dependent despite being uncorrelated. Answer D captures all of this correctly. Answer A is wrong on the correlation:

\rho(1) = 0

, not

1/2

. Answer B is wrong because the variance is identically 1 for all

t

— the product construction does not break stationarity here. Answer C commits the critical error of equating zero autocorrelation with independence; uncorrelatedness only rules out linear dependence. The key takeaway: uncorrelated ≠ independent. On time-series questions, always ask whether a shared innovation could create higher-order dependence even when autocovariances vanish.

Question 10

Let $\Theta$ be uniform on $[0,2\pi)$ , and define the discrete-time process $X_t=\cos(\omega t+\Theta)$ for a fixed real number $\omega$ . Which statement is correct?

The process is weakly but not strictly stationary, and $\rho(h)=\cos(\omega h)/2$ .
The process is strictly stationary, and $\rho(h)=\cos(\omega h)$ . (correct answer)
The process is nonstationary unless $\omega$ is an integer multiple of $2\pi$ .
The process is strictly stationary, and $\rho(h)=\cos^2(\omega h)$ .

Explanation: When you see a process defined as a deterministic function of a single uniform random variable, your first instinct should be to check stationarity by verifying whether the finite-dimensional distributions shift-invariantly depend only on the random phase — not on absolute time. Here,

X_t = \cos(\omega t + \Theta)

with

\Theta \sim \text{Uniform}[0, 2\pi)

. Because adding a fixed lag

h

simply shifts the argument:

X_{t+h} = \cos(\omega(t+h) + \Theta) = \cos(\omega t + (\Theta + \omega h))

. Since

\Theta + \omega h \pmod{2\pi}

is also uniform on

[0, 2\pi)

, the entire joint distribution of

(X_{t_1+h}, \ldots, X_{t_k+h})

equals that of

(X_{t_1}, \ldots, X_{t_k})

for any finite collection. This confirms strict stationarity. For the autocorrelation, compute

\text{Cov}(X_t, X_{t+h}) = E[\cos(\omega t+\Theta)\cos(\omega(t+h)+\Theta)]

. Using the product-to-sum identity, this equals

\frac{1}{2}\cos(\omega h)

. Since

\text{Var}(X_t) = \frac{1}{2}

, the autocorrelation is

\rho(h) = \cos(\omega h)

, confirming B is correct. Choice A wrongly claims only weak stationarity holds — but strict stationarity is fully verified, and

\rho(h) = \cos(\omega h)

, not

\cos(\omega h)/2

(that's the autocovariance, not the correlation). Choice C is incorrect because the uniform phase randomization eliminates any time-dependence in distributions regardless of

\omega

. Choice D has the right stationarity conclusion but gets the correlation wrong —

\cos^2(\omega h)

does not follow from the calculation. Study tip: Always distinguish autocovariance from autocorrelation — a classic trap. Divide the covariance by the variance to normalize.