How is ECON4916 assessed?

The Semester 2 2026 weights are 5% quizzes, 10% Assignment 1, 25% in-semester test, 10% Assignment 2 and 50% final exam.

What is the final exam format?

It is a two-hour, in-person, supervised written exam with multiple-choice and short-answer questions.

Why is conditional expectation the squared-loss forecast?

Conditional forecast error separates into irreducible conditional variance and squared distance from the conditional mean; the latter is minimised at the conditional expectation.

Why are MA parameters harder to estimate?

Lagged innovations are unobserved and must be recovered recursively for each candidate parameter vector.

How should time-series models be validated?

Preserve time order, fit using only information available at each forecast origin, and compare aligned dates, horizons and loss functions.

ECON4916 Cheatsheet · Financial Econometrics · USyd

Q: What notes are allowed?

The course materials state: In-person, closed-book, supervised 2hr exam (two-sided A4 handwritten notes allowed).

ECON4916

Topics in Applied Financial Econometrics

University of Sydney · Economics · Level 4

Full guide + ask Sia
about ECON4916

SIDE 1/2 HANDWRITTEN A4 PLANNER · conditions → formula → direction checkcopy this side by handS2 2026 · final 50% · 2 hours

1 · Exam mapcourse data · S2 2026

FINAL 50Q5A1 10TEST 25A2 10

Assessment	Format	Wt
Final exam	Multiple-choice + short-answer	50%
In-class quizzes	Multiple-choice or fill	5%
Assignment 1	Short/long-answer	10%
In-semester test	Paper-based test	25%
Assignment 2	Short/long-answer	10%

“In-person, closed-book, supervised 2hr exam (two-sided A4 handwritten notes allowed)” ECMT6006/ECON4916 unit overview, Semester 2 2026

Final: written in the formal examination period, in person and supervised. No published question count or section split: prepare portable methods, not a guessed paper.

2 · What earns spaceall chapters

Name the task first: moments need support; forecasts need 𝓕ₜ; inference needs H₀; estimation needs an objective; selection needs a holdout rule.

Priority	Write beside the rule
1	Support / information set / null
2	Existence, stationarity, rank or orthogonality
3	Formula + sum, integral, max or min
4	Direction/unit check + scoped conclusion

Copy-side inventory in priority order

Task	Minimum handwritten entry
Probability	support; Σp=1 or ∫f=1; E[X], E[X²], Var
Dependence	Cov definition; independence ⇒ zero covariance only
Stationarity	finite second moment; constant μ; γ depends only on lag
Forecast	𝓕ₜ; E[future\|𝓕ₜ]; future shocks only set to zero
OLS	β̂−β decomposition; LLN, rank, E[xε]=0
Restriction	H₀:Rβ=r; W; q; scoped rejection sentence
AR(1)	\|φ\|<1; μ, variance, ρₖ, h-step forecast
MA(1)	shared-shock γ₀, γ₁; zero after lag 1
ARMA	candidate-specific ε̂ recursion + initial condition
Selection	criterion direction; aligned sample/horizon/loss; no leakage

Visible-working sequence

Name the stochastic object, model or restriction.
Authorise the calculation with the required condition.
Substitute without hiding intermediate moments or residuals.
Audit sign, units, horizon and direction.
Conclude only under the maintained assumptions.

For multiple-choice, use the same sequence as elimination logic: reject options with the wrong operator, missing condition, impossible sign, incorrect unit or a conclusion broader than the calculation.

3 · Probability toolkitCh 1

Discrete masses are nonnegative and sum to one across the complete support.

Object	Validity	Use
pmf p(x)	p≥0; Σp=1	sum support points
density f(x)	f≥0; ∫f=1	integrate interval
CDF F(a)	P(X≤a)	threshold probability

Momentsμ = E[X] = Σx p(x) or ∫x f(x)dx
E[g(X)] = Σg(x)p(x) or ∫g(x)f(x)dx
Var(X) = E[X²] − μ²
Cov(X,Y) = E[(X−μₓ)(Y−μᵧ)]

Units: E[X] has X-units; Var(X) has squared units; covariance records linear comovement. A density height can exceed one because probability is area; for continuous X, P(X=x)=0.

3b · Dependence trapsCh 1 FAQ

Claim	Verdict / reason
independent ⇒ Cov=0	Yes, with finite moments
Cov=0 ⇒ independent	No; nonlinear dependence may remain
variance formula always valid	No; require finite second moment
observation = random variable	No; one realisation versus full mapping

If Y=X² and X is symmetric, covariance may be zero although Y is completely determined by X. Independence is a full-distribution claim; check joint/conditional versus marginal behaviour.

Continuous-variable decision lines

Question wording	Required setup	Check
“equals x”	P(X=x)=0 for continuous X	point has zero width
“between a and b”	∫ from a to b of f(x)dx	shade only support overlap
“below a”	F(a)=P(X≤a)	CDF is nondecreasing
“expected g(X)”	integrate g(x)f(x), not g(E[X])	existence of the integral

For the course’s uniform illustration on [−1,3], the support width is 4 and the density is 1/4 on that interval and zero elsewhere. The height is a density; only an interval’s area is probability. Always intersect a requested interval with the stated support before integrating.

Moment interpretation boundaries

4 · Stationarity & forecastsCh 2

One realised path is not proof: weak stationarity is a population moment claim.

Weak stationarity checklistE[Xₜ]=μ constant; E[Xₜ²]<∞;
Cov(Xₜ,Xₜ₋ₖ)=γₖ depends on lag k, not t.
ρₖ = γₖ/γ₀; ρ₀=1; ρₖ=ρ₋ₖ; |ρₖ|≤1.

Concept	Licences	Does not mean
Strict	joint distributions shift-invariant	flat plot
Weak	stable first/second moments	sample convergence
Ergodicity	time averages learn moments	same as stationarity
Sample ACF	noisy pattern clue	model proof

Squared-loss forecastX̂ₜ₊₁|ₜ = E[Xₜ₊₁ | 𝓕ₜ]. MSE = irreducible conditional variance + squared distance from conditional mean.

List 𝓕ₜ. Keep recovered/current shocks known at t; replace only future zero-mean innovations by their conditional means. State: “optimal under squared loss, given 𝓕ₜ”.

4b · MA covariance methodCh 2 worked

Step	Action
1	Write Xₜ and Xₜ₋ₖ side by side
2	Circle shared innovations
3	Multiply their coefficients × σ²
4	No shared shock ⇒ covariance 0

Population versus sample

Object	Status	Safe statement
γₖ, ρₖ	population quantities	model implication at lag k
sample mean	one-path estimator	learns μ only with appropriate convergence
sample ACF	finite-sample estimator	pattern disturbed by sampling noise
residual ACF	post-fit diagnostic	remaining linear dependence check

Long-lag sample correlations can be especially noisy because fewer overlapping pairs contribute. Use the expected population pattern to generate candidates, then inspect residuals rather than declaring a model from one spike.

Why the conditional mean wins

5 · OLS & consistencyCh 3

OLS chooses β̂ to minimise total squared vertical residual lengths.

Model, fit, error decompositiony = Xβ + ε; β̂ = (X′X)⁻¹X′y
β̂−β = (X′X)⁻¹X′ε
= [(X′X)/T]⁻¹[(X′ε)/T]

Condition	Role / failure location
Time-series LLN	sample moments reach population moments
E[xₜεₜ]=0	orthogonality makes second factor vanish
E[xₜxₜ′] nonsingular	rank identifies β; inverse exists

Then β̂−β →ₚ [E(xₜxₜ′)]⁻¹E(xₜεₜ)=0. Robust standard errors do not repair an inconsistent coefficient. Consistency gives convergence, not a rate or testing law; asymptotic normality supplies the √T rate and approximate normal reference.

5b · Linear restrictionsCh 3

Null and Wald statisticH₀: Rβ = r; q = independent rows of R
W=(Rβ̂−r)′[R V̂(β̂)R′]⁻¹(Rβ̂−r)
Under H₀, W has asymptotic χ²(q) reference.

Claim	R	r
β₁=β₂	[0,1,−1]	0
β₁=0; β₂=0	[0,1,0]; [0,0,1]	[0;0]

Read rows back into words. Rejection says the restriction discrepancy is large relative to estimated uncertainty under the maintained null/model; it does not establish causality, importance, forecast value or the whole model.

Keep four claims separate

Question	Evidence needed	Do not infer
Is β̂ consistent?	LLN, rank, E[xε]=0	finite-sample unbiasedness
How uncertain?	valid covariance estimate + asymptotic law	coefficient consistency
Restriction rejected?	Rβ̂−r scaled by its covariance	alternative theory proven
Economically important?	magnitude and units/context	from p-value alone
Useful forecast?	aligned later-data loss	from in-sample significance

Restriction construction routine

Write β in its declared coefficient order.

6 · AR/MA dynamicsCh 4

Stationary AR memory decays geometrically; finite MA memory ends exactly.

AR(1): Yₜ=c+φYₜ₋₁+εₜ|φ|<1; μ=c/(1−φ)
Var(Yₜ)=σ²ε/(1−φ²); ρₖ=φᵏ
Eₜ[Yₜ₊₁]=c+φYₜ
Eₜ[Yₜ₊ₕ]=μ+φʰ(Yₜ−μ)

φ	Shock/forecast check
0<φ<1	same sign; monotone reversion
−1<φ<0	alternating; oscillating reversion
\|φ\|≥1	no stationary-moment formulae

MA(1): Xₜ=μ+εₜ+θεₜ₋₁γ₀=(1+θ²)σ²; γ₁=θσ²;
γₖ=0 for |k|>1; ρ₁=θ/(1+θ²).

Feature	AR	MA
regressor	lagged outcome	lagged innovation
shock memory	infinite decay	finite cutoff
population ACF	tails off	cuts off
restriction	stationarity	invertibility for recovery

Stationarity ≠ invertibility. A finite MA with finite-variance white noise is stationary; invertibility concerns unique stable innovation recovery.

6b · Diagnostic orderCh 4 FAQ

Use theory/ACF to propose candidates.
Estimate under explicit conditions.
Inspect residual dependence.
Validate on later data.

Shock-path comparison

Horizon h	AR(1), φ=−0.5	MA(2), θ₁=.6, θ₂=−.2
0	1	1
1	−.5	.6
2	.25	−.2
3	−.125	0
later	alternates and shrinks	exactly 0

The AR path comes from repeated substitution: one innovation appears h periods later with coefficient φʰ. The MA path is read directly from the finite printed innovation window. “Small” is not “cut off”.

Forecast and identification distinctions

Direction checks

7 · Likelihood & ARMACh 5

MLE is at the peak because the observed sample has greatest log-likelihood there.

Estimator	Uses	Main check
OLS	observable regression + squared residuals	rank, orthogonality, errors
Moments	sample analogue	identification + convergence
ML	joint/conditional density	model, maximum, objective

Likelihood logicL(ϑ)=product of model densities; ℓ(ϑ)=log L(ϑ)=sum of log densities. Log is increasing, so argmax is unchanged. Score = ∂ℓ/∂ϑ; inspect curvature after score=0.

ARMA(1,1) generated shocksYₜ=c+φYₜ₋₁+εₜ+θεₜ₋₁
ε̂ₜ(ϑ)=Yₜ−c−φYₜ₋₁−θε̂ₜ₋₁(ϑ)

For every candidate ϑ, select the stated initial condition, recover a new full residual path, then evaluate a comparable objective. Never reuse shocks from another candidate.

Convergence is not proof	Audit
local optimum / flat region	alternative starts + nearby values
boundary solution	constraints + bounds
misspecification	moments + residuals + forecasts

8 · Selection & validationCh 6

Training fit rewards flexibility; validation loss can rise once noise is fitted.

Forecast-error splitConditional MSFE combines irreducible future-shock variance with error from estimating the conditional mean. Added complexity changes the second component, not the future innovation itself.

Choice	Rule
penalised fit	same sample; reward fit, charge complexity
holdout	same later dates, horizon, target and loss
recursive	growing window; old data retained

ECON4916

Topics in Applied Financial Econometrics

University of Sydney · Economics · Level 4

Full guide + ask Sia
about ECON4916

SIDE 2/2 DRILLS · carry the numbers to a checked conclusionpractise before the exam5 worked methods · traps · fast recall

9 · Drill 1: discrete momentsCh 1 worked

Given: X takes −2, 0, 3 percentage points with probabilities 0.20, 0.50, 0.30. Find validity, mean and variance.

All masses are nonnegative and 0.20+0.50+0.30=1.00.

x	p	xp	x²p
−2	.20	−.40	.80
0	.50	0	0
3	.30	.90	2.70
sum		.50	3.50

WorkingE[X]=0.50 percentage points
E[X²]=3.50 squared percentage points
Var(X)=3.50−0.50²=3.25.

Conclusion: 0.50 is the probability-weighted repeated-draw centre; 3.25 is squared dispersion, not a possible realised return. The standard deviation restores percentage-point units.

Trap Validate support and masses first; subtract the squared mean; do not report variance in unsquared units.

9b · Probability rapid checksCh 1 FAQ

If asked…	Show…
P(a<X<b), continuous	support + interval area, not point height
moment	existence + units
dependence	joint/conditional logic, not covariance alone
interpret mean	repeated-draw centre, not guaranteed outcome

Nonlinear warning: zero covariance removes linear association only. Independence must be assessed through the full joint distribution or conditional versus marginal probabilities.

Rebuild the answer, do not pattern-match

Line	What earns it	Failure caught
support	{−2,0,3} and corresponding masses	probability on impossible values
validity	each mass ≥0; total exactly 1	incoherent distribution
first moment	show every xp contribution	missing sign or outcome

Definition-to-method recall

10 · Drill 2: MA(1) stationarityCh 2 worked

Given: Xₜ=2+uₜ+0.4uₜ₋₁, white noise Var(uₜ)=9. Find μ, γ₀, γ₁ and γₖ for k≥2; establish weak stationarity.

Only uₜ₋₁ is shared; multiply its coefficients and innovation variance.

WorkingE[Xₜ]=2.
γ₀=(1²+0.4²)9=10.44.
γ₁=(0.4)(1)(9)=3.6.
γₖ=0 for k≥2.
ρ₁=3.6/10.44≈0.345.

Conclusion: finite second moment, constant mean and covariance depending only on separation establish weak stationarity under the stated white-noise assumptions.

Wrong move	Repair
“looks stable”	derive mean, variance, lag covariance
add γ₁ terms	shared-shock coefficient products only
stationary ⇒ ergodic	separate dependence requirement

10b · Forecast logicCh 2

At forecast origin t	Treatment
observed/recovered shock in 𝓕ₜ	keep it
future mean-zero shock	replace by its conditional mean, zero

Stationarity proof checklist

Condition	Calculation in this MA(1)	Conclusion
finite second moment	white-noise variance 9; finite coefficient squares	γ₀=10.44 exists
constant mean	E[uₜ]=E[uₜ₋₁]=0	E[Xₜ]=2 for every t
lag-only covariance	overlap depends on separation, not date	γ₁=3.6; γₖ=0 for k≥2

The conclusion is weakly stationary under the stated white-noise assumptions. It is not proved by the visual appearance of one path, and it does not by itself show that a sample mean converges to the population mean.

ACF arithmetic and diagnostics

Normalisation checkρ₁=γ₁/γ₀=3.6/10.44≈0.345; ρ₀=1; ρₖ=0 for k≥2 in this population MA(1).

Forecast wording

11 · Drill 3: OLS consistencyCh 3 worked

Given: a time-series LLN, E[xₜεₜ]=0 and nonsingular E[xₜxₜ′]. Explain why OLS is consistent.

Rank authorises the inverse; orthogonality kills the second factor.

Substitute y=Xβ+ε into β̂=(X′X)⁻¹X′y.
Obtain β̂−β=(X′X)⁻¹X′ε.
Scale: [(X′X)/T]⁻¹[(X′ε)/T].
LLN gives E[xₜxₜ′] and E[xₜεₜ].
Nonsingularity + orthogonality give [E(xx′)]⁻¹0=0; hence β̂→ₚβ.

Symptom	Where to look
collinearity	first matrix: rank
endogeneity	second moment: E[xε]
uncertain inference	covariance / CLT conditions

Trap More data does not remove E[xₜεₜ]≠0. Standard-error changes do not fix coefficient inconsistency.

11b · Wald translationCh 3

Words	Matrix line
β₁ equals β₂	[0,1,−1]β=0
β₁ and β₂ jointly zero	[0,1,0]β=0; [0,0,1]β=0

Count independent rows q; calculate W with Cov(Rβ̂); use χ²(q) asymptotically; interpret the exact combination. A small p-value is neither P(H₀ true) nor an effect-size measure.

Why every condition is visible

Removed condition	Break in the derivation	Diagnosis
time-series LLN	sample averages need not reach fixed limits	dependence/convergence problem
nonsingularity	E[xₜxₜ′] cannot be inverted	insufficient independent regressor variation
E[xₜεₜ]=0	second factor tends to a nonzero vector	systematic regressor–error relation

Slutsky combines the converging inverse moment matrix and the vanishing orthogonality moment. Attach “under the stated LLN, rank and orthogonality conditions” to the consistency conclusion.

Consistency versus inference

12 · Drill 4: stationary AR(1)Ch 4 worked

Given: Yₜ=1.2+0.6Yₜ₋₁+εₜ, Var(εₜ)=4; Yₜ=4. Find μ, variance, ρ₂ and the one-step forecast.

Positive |φ|<1 keeps an above-mean forecast above μ while moving toward it.

Working|0.6|<1 ⇒ stationary solution.
μ=1.2/(1−0.6)=3.
Var(Y)=4/(1−0.6²)=4/0.64=6.25.
ρ₂=0.6²=0.36.
Eₜ[Yₜ₊₁]=1.2+0.6(4)=3.6.

Direction audit: 4>3 and 0<φ<1, so the forecast must lie between current value 4 and mean 3. The result 3.6 passes.

Common error	Repair
μ=c/φ	μ=c+φμ ⇒ c/(1−φ)
Var=σ²/(1−φ)	variance recursion squares φ
ρ₂=2φ	AR(1) ACF is φᵏ
replace Yₜ by μ	Yₜ is observed in 𝓕ₜ

12b · AR versus MA responseCh 4 compare

Negative AR powers alternate forever while shrinking; finite MA reaches an exact cutoff.

General AR(1) marking route

Check |φ|<1 before using stationary moments.
Take unconditional means and solve μ=c+φμ.
Demean the recursion; use uncorrelated innovations to solve γ₀=φ²γ₀+σ²ε.
Use ρₖ=φᵏ for the population ACF.
For a forecast, condition on 𝓕ₜ and set only future innovations to zero.
Check direction against μ and the sign of φ.

Feature	Stationary AR(1)	Finite MA(q)
one-shock weight at h	φʰ	printed θₕ through q; then 0
positive coefficient	same-sign decay	sign follows each θ
negative coefficient	alternating decay	negative only where printed

13 · Drill 5: Gaussian MLECh 5 worked

Given: independent y=(1,2,3), normal with unknown μ and known variance 1. Derive μ̂ and verify a maximum.

Zero score occurs at the sample mean; negative curvature confirms a maximum.

Log-likelihood up to a constantℓ(μ)=−½Σₜ(yₜ−μ)².
Score: ∂ℓ/∂μ=Σₜ(yₜ−μ).
Set to 0: 1+2+3−3μ=0 ⇒ μ̂=2.
Curvature: ∂²ℓ/∂μ²=−3<0 ⇒ maximum.

Step	Exam line
objective	write what is maximised; omit only parameter-free constants
stationary point	score=0
maximum	negative second derivative / compare objective
ARMA	candidate ϑ generates candidate ε̂ path first

Trap A zero score is not automatically a maximum. “Converged” is a stopping message, not statistical validation.

13b · ARMA recoveryCh 5

Initialisation and parameter values are part of the generated shock path.

Audit	Question
objective	same likelihood/initialisation?
numerics	multiple starts, boundaries, nearby values?
model	moments and residual dependence plausible?
forecast	later aligned observations?

Estimator choice by observables

Method	Sample target	Why results may differ
OLS	minimum sum of squared observable regression residuals	uses conditional-mean/orthogonality structure

14 · Penalised-fit drillCh 6 worked

Given: T=120. Models A/B/C have RSS 150/138/132 and k 2/5/9. Rank by s²=RSS/(T−k), minimised.

Model	RSS	k	T−k	s²
A	150	2	118	1.2712
B	138	5	115	1.2000
C	132	9	111	1.1892

Conclusion: C minimises this in-sample penalised residual variance. Its advantage over B is narrow; this does not establish best forecasts. Check residuals and time-ordered holdout loss.

At each historical origin, fit only on available past data and score the later target.

Fair-comparison gate	Required
information	no future scaling, features, tuning or ranking
sample	same effective forecast dates
target	same variable and horizon
criterion	same loss; direction stated
window	same rolling/recursive rule

15 · Fast method chooserall chapters

Extra flexibility helps fit but can worsen later performance through estimation error.

Prompt cue	Method / first line
support, probabilities	pmf/density/CDF; normalise
time-invariant moments	weak-stationarity checklist
optimal MSE forecast	E[future \| 𝓕ₜ]
OLS consistency	β̂−β decomposition
coefficient claim	H₀:Rβ=r; Wald
shock persistence	AR powers / MA window
unobserved shocks	recursive recovery for each ϑ
fit versus forecast	penalty or ordered holdout

Thirty-second answer audit

Object and notation match?
Condition before formula?
Operator/direction correct?
𝓕ₜ, H₀ or window explicit?