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2.7. FILTERING 75The Þlter has the effect of scaling the spectral density of the originalobservations by a(e −iω )a(e iω ). Depending on the properties of the Þlter,some frequencies will be magniÞed while others are downweighted.One way to classify Þlters is according to the frequencies that areallowed to pass through and those that are blocked. A high pass Þlterlets through only the high frequency components. A low pass Þlterallows through the trend or growth frequencies. A business cycle passÞlter allows through frequencies ranging from 6 to 32 quarters. Themost popular Þlter used in RBC research is the Hodrick—Prescott Þlter,which we discuss next.The Hodrick—Prescott FilterHodrick and Prescott [76] assume that the original series q t is generatedby the sum of a trend component (τ t ) and a cyclical (c t )component,q t = τ t + c t . The trend is a slow-moving low-frequency component andis in general not deterministic. The objective is to construct a Þlterto to get rid of τ t from the data. This leaves c t ,whichisthepartofthe data to be studied. The problem is that for each observation q t ,there are two unknowns (τ t and c t ). The question is how to identifythe separate components?The cyclical part is just the deviation of the original series from thelong-run trend, c t = q t − τ t . Suppose your identiÞcation scheme is tominimize the variance of the cyclical part. You would end up settingits variance to 0 which means setting τ t = q t . This doesn’t help atall—the trend is just as volatile as the original observations. It thereforemakes sense to attach a penalty for making τ t too volatile. Do this byminimizing the variance of c t subject to a given amount of prespeciÞed‘smoothness’ in τ t . Since ∆τ t is like the Þrst derivative of the trendand ∆ 2 τ t is like the second derivative of the trend, one way to get asmoothly evolving trend is to force the Þrst derivative of the trend toevolve smoothly over time by limiting the size of the second derivative.This is what Hodrick and Prescott suggest. Choose a sequence of points{τ t } to minimizeTXt=1(q t − τ t ) 2 + λTX−1t=1(∆ 2 τ t+1 ) 2 , (2.114)