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2.5. PANEL UNIT-ROOT TESTS 61Table 2.5: Selected Exact Critical Values for the IPS ¯τ NT StatisticConstantTrendT 20 40 100 20 40 100A. 5 percent5 -2.19 -2.16 -2.15 -2.82 -2.77 -2.7510 -1.99 -1.98 -1.97 -2.63 -2.60 -2.58N 15 -1.91 -1.90 -1.89 -2.55 -2.52 -2.5120 -1.86 -1.85 -1.84 -2.49 -2.48 -2.4625 -1.82 -1.81 -1.81 -2.46 -2.44 -2.43B. 10 percent5 -2.04 -2.02 -2.01 -2.67 -2.63 -2.6210 -1.89 -1.88 -1.88 -2.52 -2.50 -2.49N 15 -1.82 -1.81 -1.81 -2.46 -2.44 -2.4420 -1.78 -1.78 -1.77 -2.42 -2.41 -2.4025 -1.75 -1.75 -1.75 -2.39 -2.38 -2.38Source: Im, Pesaran and Shin [78].density function of τ. Solve for g(p), the density of p i by the method oftransformations, g(p i )=f(τ i )|J| where J = dτ i /dp i is the Jacobian ofthe transformation, and |J| is its absolute value. Since dp i /dτ i = f(τ i ),the Jacobian is 1/f(τ i )andg(p i ) = 1 for 0 ≤ p i ≤ 1. That is, p i isuniformly distributed on the interval [0, 1] (p i ∼ U[0, 1]).Next, let y i = −2ln(p i ). Again, using the method of transformations,the probability density function of y i is h(y i )=g(p i )|dp i /dy i |.But g(p i )=1and|dp i /dy i | = p i /2=(1/2)e −y i/2 , so it follows thath(y i )=(1/2)e −y i/2 which is the chi-square distribution with 2 degreesof freedom. Under cross-sectional independence of the error terms ² it ,the joint test statistic also has a chi-square distributionλ = −2NXi=1ln(p i ) ∼ χ 2 2N . (2.86)The asymptotic distribution of the IPS test statistic was establishedby sequential T →∞,N →∞asymptotics, which some econometri-