ECOS3021 Business Cycles and Asset Markets University of Sydney
2021 Semester 1
Tutorial #1
1. Describe one major difference between Growth Cycles and Classical Cycles. Draw an example of
the evolution of GDP which illustrates this difference (i.e. across two different recessions).
ANSWER:
• A growth cycle recession requires a decline in real GDP below its long-term growth trend. GDP growth may still be positive, but it has fallen relative to its trend.
• A classical cycle requires that GDP falls, so GDP growth must be negative.
2. Go to the following webpage: http://www.rba.gov.au/statistics/tables/index.html and down- load Gross Domestic Product and Income – H1. The first series in column B is real GDP for Australia and the second series in column C is year-ended real GDP growth. Insert a blank column next to column B. Calculate quarterly and four-quarter ended growth rates using the formulas shown in lecture slides.
3. How do we express these growth rates in annualized percentage growth rates? Apply this compu- tation to your growth rates from Question (2). Plot these two growth rates against each other. Which is more volatile? Why?
ANSWER:
• See Figure 3
• Quarterly log-differences should be multiplied by 400. Multiplying by 4 scales up quarterly differ- ences to the size of difference equivalent to an annual difference (i.e. the growth rate that would have obtained if GDP grew like this every quarter for a year). Multiplying by 100 gets values in percentage rates.
• Annual log-differences should be multiplied by 100 to get values in percentage rates.
• Quarterly growth rates are more volatile.
• This is because the size of the sudden drops observed in some quarters are not repeated for four quarters in a row. So the annualized growth rates are much too large relative to what actually occurs in a given year.
4. Identify all the peaks and troughs in quarterly Australian GDP data using the BBQ procedure. (Note: You may like to use the BBQ Excel Macro posted on Canvas. You may alternatively do the BBQ algorithm ‘by hand’: first calculate the quarterly growth rates yt − yt−1 and yt − yt−2, then apply the formulas from Lecture 1.)
ANSWER:
1
2à Australia Real GDP 2à
Australia Real GDP
à
à
àà
−à −2à
−à −2à
−3à 96à
−3à 96à
• Recall the BBQ algorithm from Lecture 1
9àà
• A peak occurs if:
• A trough occurs if:
98à
99à
2ààà
2àà
2à2à
9àà 98à
99à 2ààà 2àà 2à2à
[(yt −yt−2)>0,(yt −yt−1)>0], and [(yt+2 −yt)<0,(yt+1 −yt)<0]
[(yt −yt−2)<0,(yt −yt−1)<0], and [(yt+2 −yt)>0,(yt+1 −yt)>0]
• Applying to the Australian GDP data, we get the peaks and troughs plotted in Figure 4.
• Note that in some cases we find multiple troughs that occur in succession, or multiple peaks that occur in succession. Since we expect peaks and troughs to occur alternately, many versions of the BBQ algorithm ‘trim’ the extra peaks and troughs…
5. What is the difference between a deterministic and stochastic trend? ANSWER:
• A deterministic trend advances in an identical manner each period. If the trend growth rate is 2%, then the trend component increases by 2% this period, 2% next period, and by 2% in every period thereafter.
• A stochastic trend varies from period to period. The growth rate may be 2% this period, but 5% next period, and then -1% in the period after that.
• Note that ‘stochastic’ means ‘random’, and so the trend is not necessarily known in advance (although it may still be predictable, with some uncertainty, from period to period).
2
Annualized Quarterly Growth Rate
Annualized Four-Quarter Growth Rate
Australia Real GDP, Peaks and Troughs
Peak Trough
3ààà 2àà5 2à5à 2à25 2ààà àà5 à5à à25 ààà
96à 9àà 98à
99à 2ààà 2àà 2à2à
6. Business cycles may be identified by the Hodrick Prescott (HP) Filter. What two components of a macroeconomic variable does the HP filter distinguish between? The flexibility of the HP filter is determined by the choice of parameter λ. What effect does the choice of this parameter have?
ANSWER:
• The HP filter distinguishes between the growth/trend component and the cyclical component
• λ is the smoothing parameter. The higher is lambda, the more smooth is the trend series, and
thus the more volatile will be the cyclical component of the data.
7. Why might the HP filter have produced mis-leading estimates of the GDP cycle throughout 2020?
ANSWER:
• ‘The end point problem’: As data during 2020 was rapidly evolving, this would have changed both the cycle and the trend each period in which new data was released. This suggests HP filtered cycles/trends estimated in real time would not have reflected the eventual cycles/trends we are likely to infer about 2020.
8. What is a data generating process (DGP)? What is a random walk? Is the random walk with drift model of real GDP consistent with the decomposition of output into growth and cyclical components? (Use equations and words in your answer).
ANSWER:
• DGP: The supposed underlying real-world process that produces the data that we observe
• Random walk: a statistical process under which a variable takes a ‘random step’ during each observation of the data. Over time, the process will ‘wander’ further and further from its starting point.
3
log Real GDP
• The random walk model for the log of GDP is:
yt =μ+yt−1 +εt
• In this model, μ represents the deterministic growth component. Absent errors/shocks, log-GDP would always grow at the rate μ, since:
yt − yt−1 = μ
• The cyclical component of log-GDP is entirely determined by the errors/shocks εt. The cycle is
defined as:
ct =(yt −yt−1)−μ=εt
9. Suppose that the log of quarterly real GDP follows a random walk with drift μ = 0.0488, and shocks that are normally distributed with mean zero and standard deviation σ = 0.05. Suppose real GDP is currently equal to $1. What do you expect the value of real GDP to be next quarter? What do you expect the value of real GDP to be one year from now?
ANSWER:
• Take the random walk model with drift:
yt =μ+yt−1 +εt
• We can ignore shocks since they are random and mean zero: yt =μ+yt−1
• Starting at $1, we have:
yt+1 = 0.0488 + ln(1)
⇒ exp(yt+1) = exp(0.0488 + ln(1)) = exp(0.0488) = $1.05
• A year from now is 4 quarters away, so we have:
yt+4 = μ + yt+3
= μ + (μ + yt+2)
= μ + (μ + (μ + yt+1))
= μ + (μ + (μ + (μ + yt))) = 4×μ+ln(1)
⇒ exp(yt+4) = exp(4 × 0.0488 + ln(1)) = exp(0.1952) = $1.2156 4