2

I am running a Kwiatkowski–Phillips–Schmidt–Shin test (KPSS test) in R (urca::ur.kpss). However, I am quite unsure if it is performed correctly, because the results are the same for each data column.

> dput(datSel)
structure(list(c = c(142.8163942, 143.5711365, 145.3485827, 142.0577145, 
139.4326176, 140.1236581, 138.6560282, 136.405036, 133.9337229, 
133.8785538, 132.0608441, 130.0866307, 120.1320237, 119.6368882, 
114.3312943, 117.5084111, 114.4960017, 112.9124518, 112.8185478, 
112.3047916, 106.632639, 106.2107158, 106.8455028, 106.3879556, 
104.3451786, 102.9085952, 101.0967783, 101.7858278, 101.0749044, 
102.6441976, 102.0666152, 100, 97.14084104, 97.49972913, 96.91453836, 
96.05132443, 94.98057971, 92.78373451, 92.67526281, 91.82430571, 
91.4153859, 89.51740671, 89.01587176, 84.62259911, 91.48598494, 
89.12053042, 90.02364352, 90.92496121, 89.42963565, 91.93886583, 
88.83918306, 90.39513509, 87.54571761, 91.3386451, 87.7836994, 
91.79178376, 87.56903138, 87.77875755, 89.29938784, 90.88084014
), d = c(17703.7, 17599.8, 17328.2, 17044, 17078.3, 16872.3, 
16619.2, 16502.4, 16332.5, 16268.9, 16094.7, 15956.5, 15785.3, 
15587.1, 15460.9, 15238.4, 15230.2, 15057.7, 14888.6, 14681.1, 
14566.5, 14384.1, 14340.4, 14383.9, 14549.9, 14843, 14813, 14668.4, 
14685.3, 14569.7, 14422.3, 14233.2, 14066.4, 13908.5, 13799.8, 
13648.9, 13381.6, 13205.4, 12974.1, 12813.7, 12562.2, 12367.7, 
12181.4, 11988.4, 11816.8, 11625.1, 11370.7, 11230.1, 11103.8, 
11037.1, 10934.8, 10834.4, 10701.3, 10639.5, 10638.4, 10508.1, 
10472.3, 10357.4, 10278.3, 10031), e = c(71.0619, 70.9383, 71.162, 
71.138, 71.2286, 71.5095, 71.565, 71.3246, 71.4963, 71.3738, 
71.4276, 71.3065, 71.0246, 71.3244, 71.0619, 70.9811, 71.2149, 
70.8342, 70.5568, 70.5444, 70.3286, 70.179, 70.2555, 70.5103, 
70.8038, 70.6748, 70.9769, 70.6988, 70.2125, 70.1661, 69.6284, 
69.5613, 68.9837, 68.8606, 68.4223, 67.963, 67.6293, 67.5905, 
67.1857, 67.1248, 66.7075, 66.5857, 66.4303, 66.2826, 68.7514, 
68.8897, 69.0824, 68.9718, 68.7927, 68.6387, 68.8053, 68.7286, 
68.4141, 68.2357, 68.4785, 68.4171, 68.4782, 68.3978, 68.5344, 
68.4772), f = c(2160.080078, 2203.939941, 2500.850098, 2523.820068, 
2546.54, 2528.449951, 2223.97998, 2352.01001, 2401.21, 2089.73999, 
1975.349976, 2159.060059, 1891.68, 1947.849976, 2766.72998, 2882.179932, 
2947.24, 2541.629883, 2278.800049, 2634, 2495.56, 2637.280029, 
2098.649902, 1696.619995, 1750.83, 2767.76001, 3943.149902, 3765.909912, 
4512.98, 4527.299805, 4869.259766, 4645.5, 4463.47, 3868.27002, 
3745.719971, 4139.830078, 3667.03, 3457.449951, 3049.909912, 
2632.899902, 2431.38, 2042.869995, 1989.400024, 1866.76001, 1545.15, 
1351.890015, 1305.709961, 1163.109985, 1150.05, 1070.209961, 
1243.069946, 1289.16, 1140.36, 1084.069946, 1206.819946, 1186.540039, 
1073.3, 1161.160034, 1129.579956, 1130.069946), g = c(5.7393, 
5.7072, 5.6126, 5.6411, 5.5114, 5.4551, 5.1613, 5.4087, 5.0227, 
5.2039, 4.9501, 4.5008, 4.9143, 4.1372, 4.5604, 4.7979, 4.5454, 
4.8863, 5.0496, 4.9757, 5.4705, 5.8403, 5.4328, 4.6986, 4.4481, 
4.1385, 3.8379, 4.2183, 4.5429, 5.03, 5.1821, 4.8269, 5.0469, 
5.1054, 5.3959, 5.5413, 5.8139, 5.8611, 5.8396, 5.1964, 5.6386, 
5.6615, 5.5751, 5.2251, 4.4682, 4.262, 4.3487, 4.1654, 3.9651, 
3.9105, 3.7954, 4.1595, 3.8174, 3.6349, 3.6119, 3.4004, 3.366, 
3.3953, 3.3621, 3.9338), h = c(88.548662, 90.58853576, 91.32289522, 
91.56290683, 108.4682322, 93.86541244, 100.3414441, 91.98328561, 
95.53905246, 102.6461104, 97.9505881, 108.912959, 114.4931447, 
108.0431511, 98.58118608, 107.9440773, 99.41777306, 104.868483, 
100.3338425, 98.06667712, 100.6353811, 100.6491181, 106.4241282, 
79.3180456, 80.40781739, 85.35716451, 102.9110831, 88.99947733, 
99.38928861, 87.57579615, 87.49264945, 90.29013182, 92.13878645, 
90.15141711, 83.90950016, 97.24552675, 93.38024804, 94.16745797, 
98.90106448, 94.73366108, 104.1079291, 98.20132446, 97.70974526, 
91.86162897, 101.5381154, 94.56938821, 86.91581151, 87.16428746, 
87.35114009, 85.0634706, 86.2179337, 82.34156437, 79.86840987, 
84.20717658, 85.29553997, 90.94079268, 92.84823122, 88.90113767, 
88.05502443, 92.38787475), i = c(363.81, 361.19, 362.35, 359.09, 
359.31, 355.8, 356.64, 353.83, 353.49, 348.92, 348.8, 344.85, 
343.48, 340.75, 341.1, 335.72, 331.29, 328.21, 328.95, 325.92, 
324.83, 322.83, 323.18, 321.66, 322.94, 323.14, 322.89, 318.34, 
315.85, 311.61, 311.3, 308.34, 306.1, 305.64, 305.58, 302.91, 
301.64, 300.24, 299.54, 298.58, 296.4, 293.87, 293.35, 291.61, 
289.43, 288.03, 287.69, 287.6, 285.95, 284.8, 284.63, 282.62, 
281.24, 280, 280.09, 277.65, 275.73, 273.12, 272.78, 272.25), 
    j = c(307.5, 308.6, 308.9, 309.7, 311.1, 311.6, 311.6, 313.9, 
    314.9, 314.8, 314.9, 314.5, 313.4, 313, 312.9, 309, 304.5, 
    302.76, 299.28, 293.44, 291.52, 291.71, 290.61, 294.17, 297.74, 
    300.02, 295.91, 292.9, 289.23, 287.49, 285.86, 283.84, 281.1, 
    280.37, 278.63, 275.44, 273.88, 273.24, 274.6, 275.15, 269.77, 
    267.66, 264.29, 262.27, 260.53, 260.52, 261.54, 263.27, 261.45, 
    261.81, 261.99, 261.35, 262.64, 264.74, 265.56, 265.47, 267.3, 
    265.47, 262.64, 260.72), k = c(103.3086091, 102.9085757, 
    103.6086341, 107.5089591, 107.9089924, 108.9090758, 104.3086924, 
    97.80815068, 104.8087341, 108.0090008, 103.4086174, 104.5087091, 
    105.8088174, 100.308359, 102.6085507, 100.4083674, 96.80806734, 
    99.50829236, 102.708559, 100.7083924, 103.0485874, 103.9186599, 
    104.7887324, 105.0787566, 103.3386116, 104.0186682, 102.5685474, 
    112.4193683, 105.8488207, 104.5987166, 107.3989499, 108.6490541, 
    107.2989416, 106.2388532, 101.3084424, 98.02816901, 102.1785149, 
    97.83815318, 98.70822569, 88.85740478, 92.66772231, 95.36794733, 
    91.4076173, 87.54729561, 89.66747229, 87.73731144, 87.34727894, 
    90.9275773, 78.26652221, 80.29669139, 79.90665889, 77.68647387, 
    77.59646637, 78.46653888, 77.68647387, 77.01641803, 84.45703809, 
    77.97649804, 76.72639387, 77.88649054), l = c(109.1, 109.1, 
    108.8, 108.2, 107.6, 107.2, 107.3, 106.7, 106.4, 106, 105.9, 
    104.9, 103.8, 103.5, 103, 102.3, 101.3, 100.5, 99.6, 98.6, 
    97.43314, 96.68301, 95.84954, 95.18276, 94.76602, 94.01589, 
    92.84903, 91.18208, 89.76517, 89.18174, 88.51496, 87.76484, 
    86.68132, 85.93119, 85.18107, 84.51429, 83.76416, 83.43077, 
    83.26407, 82.93068, 82.46215, 82.14979, 81.83744, 81.05654, 
    80.43183, 80.35374, 80.27565, 79.9633, 79.72903, 79.57285, 
    79.57285, 79.26049, 79.02623, 79.10432, 79.02623, 78.71387, 
    78.4796, 78.24534, 77.93298, 77.69871), m = c(108.26667, 
    107.96667, 107.46667, 106.76667, 106.66667, 106.6, 106.43333, 
    105.83333, 105, 104.8, 104.46667, 103.46667, 102.4, 102.56667, 
    102.2, 101.96667, 100.77774, 100.47032, 100.41443, 98.48607, 
    97.47997, 97.22844, 96.55771, 96.52976, 96.58566, 98.2066, 
    96.58566, 94.0704, 92.00231, 92.03026, 91.86257, 90.40932, 
    89.26348, 88.84427, 87.19538, 85.32292, 84.28887, 83.61814, 
    83.72993, 83.59019, 83.22324, 82.61167, 82.09794, 80.36107, 
    78.86882, 78.42849, 77.93923, 77.05856, 76.39806, 76.34913, 
    76.22682, 75.39507, 75.05259, 75.24829, 75.12598, 74.34316, 
    74.04961, 73.60927, 73.21786, 72.67968), n = c(108.56667, 
    108.56667, 108.23333, 107.3, 107.13333, 106.8, 106.63333, 
    105.76667, 105.46667, 105.06667, 104.8, 103.23333, 102.5, 
    102.6, 102.36667, 102.1, 100.5226, 100.32976, 100.71544, 
    98.29121, 97.35458, 97.43723, 96.80362, 96.85872, 96.36285, 
    98.75953, 97.05155, 93.6907, 91.12874, 91.29403, 91.29403, 
    89.44831, 88.07091, 87.57505, 85.86707, 83.96626, 83.4153, 
    82.64396, 82.47867, 82.17564, 82.00498, 81.76645, 81.12244, 
    79.59587, 78.02161, 77.73538, 77.18677, 76.11341, 75.39783, 
    75.42168, 75.04004, 73.94283, 73.94283, 74.08594, 73.7043, 
    72.67864, 72.2493, 71.89151, 71.43831, 70.62732), o = c(57844L, 
    57844L, 57667L, 57168L, 57080L, 56904L, 56813L, 56353L, 56193L, 
    55980L, 55838L, 55003L, 54612L, 54666L, 54541L, 54398L, 53567L, 
    53465L, 53670L, 52379L, 51878L, 51923L, 51585L, 51615L, 51351L, 
    52629L, 51718L, 49927L, 48562L, 48649L, 48640L, 47666L, 46932L, 
    46668L, 45758L, 44745L, 44428L, 44046L, 43944L, 43779L, 43690L, 
    43563L, 43219L, 42407L, 41567L, 41416L, 41123L, 40551L, 40170L, 
    40182L, 39979L, 39395L, 39394L, 39471L, 39267L, 38721L, 38514L, 
    38309L, 38061L, 37617L), p = c(59373L, 59209L, 58935L, 58551L, 
    58496L, 58458L, 58368L, 58039L, 57582L, 57472L, 57289L, 56742L, 
    56156L, 56248L, 56046L, 55919L, 55243L, 55075L, 55045L, 53988L, 
    53436L, 53298L, 52930L, 52915L, 52947L, 53834L, 52946L, 51567L, 
    50433L, 50449L, 50357L, 49557L, 48932L, 48671L, 47722L, 46772L, 
    46213L, 45865L, 45919L, 45826L, 45612L, 45276L, 44994L, 44041L, 
    43225L, 42983L, 42715L, 42232L, 41870L, 41843L, 41777L, 41321L, 
    41132L, 41240L, 41172L, 40743L, 40587L, 40352L, 40127L, 39814L
    ), q = c(96819L, 96819L, 96090L, 94632L, 94632L, 94632L, 
    93727L, 91917L, 91917L, 91917L, 90779L, 88503L, 88416L, 88416L, 
    88270L, 87978L, 87996L, 87996L, 87566L, 86706L, 86706L, 86706L, 
    85794L, 83970L, 83970L, 83970L, 83007L, 81081L, 81081L, 81081L, 
    80423L, 79107L, 79107L, 79107L, 78321L, 76749L, 76533L, 76533L, 
    75983L, 74883L, 74883L, 74883L, 74575L, 73959L, 73959L, 73959L, 
    73167L, 71583L, 71583L, 71583L, 70858L, 69408L, 69408L, 69408L, 
    68594L, 66966L, 66831L, 66342L, 65853L, 64875L), r = c(144.5, 
    146.5, 147.3, 143.3, 140.1, 142.8, 141.2, 140.2, 137.8, 137.4, 
    136.6, 137.6, 125.5, 125.7, 120.5, 124.2, 121.5, 119.8, 121.3, 
    122, 114.1, 114.4, 114.7, 116.1, 112.8, 111.8, 110.2, 111.7, 
    112.2, 113.7, 112.7, 110.5, 107, 107.5, 108, 107.1, 106.7, 
    103.3, 104.2, 104.3, 104.1, 101.3, 100.5, 94.3, 105.6, 101, 
    102, 103.1, 101.4, 105.5, 100.5, 102.8, 100.5, 105.1, 98.8, 
    105.1, 98.2, 98.2, 100.6, 103), s = c(132.2, 133.9, 133.5, 
    126, 125, 122.6, 122.6, 123.8, 124.5, 120.2, 120.2, 123.5, 
    105.2, 116.4, 111.5, 116.4, 116.1, 114.3, 117, 117.9, 107.1, 
    104.5, 110.6, 110.5, 104.2, 105.4, 106.2, 110.3, 106.8, 111.4, 
    111.2, 108.5, 93.5, 101.5, 101.4, 101.3, 101.7, 96.8, 97.3, 
    100, 97.5, 99.4, 94.8, 93.8, 101.9, 97.4, 97.7, 98.4, 100.6, 
    100.1, 96.3, 98.1, 93.4, 99.3, 97.3, 99.6, 99.2, 97.8, 100.1, 
    102.9), t = c(149.8, 151.9, 153.2, 150.7, 146.5, 151.5, 149.2, 
    147.3, 143.6, 144.8, 143.6, 143.7, 134.1, 129.7, 124.3, 127.5, 
    123.7, 122.2, 123.1, 123.8, 117.1, 118.6, 116.4, 118.4, 116.4, 
    114.6, 111.9, 112.2, 114.5, 114.6, 113.4, 111.3, 112.8, 110.1, 
    110.8, 109.5, 108.8, 106.1, 107.1, 106.1, 107, 102.1, 103, 
    94.5, 107.2, 102.5, 103.9, 105.1, 101.7, 107.8, 102.4, 104.8, 
    103.6, 107.6, 99.5, 107.4, 97.8, 98.4, 100.8, 103), u = c(155.2, 
    157.6, 159, 156.5, 151.4, 155, 152, 149, 146.4, 147.9, 146.6, 
    146.3, 137.1, 131.1, 124.5, 127.5, 123.1, 121.9, 123, 123.5, 
    116.4, 117.7, 116.4, 118.1, 116.5, 113.7, 110.2, 111, 113.9, 
    113.9, 113.6, 110.9, 113.2, 109.9, 111.7, 109.7, 110.1, 106.3, 
    107.4, 105.9, 107.2, 101.6, 103.8, 94.1, 108.4, 102.7, 104.1, 
    105.1, 101.5, 108.8, 102.3, 105.4, 103, 107.2, 99.3, 107.6, 
    97.4, 97.6, 101.2, 103.9), v = c(112.6, 112.7, 113.6, 110.7, 
    113.4, 127.1, 130.1, 135.7, 123.7, 123.2, 123, 125.5, 113.5, 
    120.2, 123.3, 128, 128.2, 124.6, 124, 125.8, 122.2, 124.8, 
    116.6, 120.4, 115.9, 120.6, 124, 120.6, 119, 120.1, 111.6, 
    114, 110.2, 111.6, 104.5, 107.9, 100.4, 104.7, 105, 106.9, 
    105.1, 105.8, 97.3, 96.6, 99.1, 101.1, 102.5, 105.2, 103, 
    101, 102.7, 100.5, 107.4, 110.1, 101.3, 105.7, 100.3, 104.1, 
    98.4, 97.2)), .Names = c("c", "d", "e", "f", "g", "h", "i", 
"j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v"
), row.names = c(NA, -60L), class = "data.frame")
> resKpssT <- lapply(datSel,function(x){ summary(ur.kpss(x,type="tau")) })
> (resKpssT)
$c

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3717 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$d

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1771 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$e

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.158 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$f

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.2767 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$g

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1737 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$h

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.0815 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$i

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.2921 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$j

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1445 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$k

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3354 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$l

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3125 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$m

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1857 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$n

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1818 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$o

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1822 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$p

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1847 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$q

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.0801 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$r

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3628 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$s

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3033 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$t

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3514 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$u

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.3544 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


$v

####################### 
# KPSS Unit Root Test # 
####################### 

Test is of type: tau with 3 lags. 

Value of test-statistic is: 0.1649 

Critical value for a significance level of: 
                10pct  5pct 2.5pct  1pct
critical values 0.119 0.146  0.176 0.216


> cv.kpss.tau <- sapply(resKpssT, function(x) x@cval)
> (cv.kpss.tau)
         c     d     e     f     g     h     i     j     k     l     m     n     o     p     q     r     s     t
[1,] 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119 0.119
[2,] 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146 0.146
[3,] 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176 0.176
[4,] 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216 0.216
         u     v
[1,] 0.119 0.119
[2,] 0.146 0.146
[3,] 0.176 0.176
[4,] 0.216 0.216

You can see that all critical values are the same and they preach the critical values. Hence, all data should be non-stationary.

However, I do not think that this might be correct, because when looking for example at time series, q.

enter image description here

Any suggestion what I am doing wrong?

UPDATE

I created a table of my series for the KPSS test values.

enter image description here

Is this correct?

I also ran my results through a Dick Fuller test, which basically shows me complementary results:

enter image description here

My excel formula is in pseudocode:

=IF("calculated p-value" <= "critical value"; H1 ; H0 )

Here you can find an excel sheet, which I am using for the calculations:

Google Spreadsheet

The two pictures show complementary results to me. Hence, I am guessing I am doing sth wrong.

Any recommendation what I am doing wrong?

Ferdi
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Carol.Kar
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    From wikipedia- In econometrics, Kwiatkowski–Phillips–Schmidt–Shin (KPSS) tests are used for testing a null hypothesis that an observable time series is stationary around a deterministic trend. – meh Jul 11 '15 at 15:00
  • @aginensky Thx for your reply! Just changed my question above. However, from my output all data should be non-stationary. Any recommendation about my output? Is it correct? – Carol.Kar Jul 11 '15 at 15:09
  • Well, one thing you could do to convince yourself is use ols to find the trend and then detrend the series. You should get the same statistics. More importantly, graphing it would show you if it looked reasonable. – meh Jul 12 '15 at 00:53

1 Answers1

2

The critical values are the tabulated values of the theoretical distribution of the test statistic under the null hypothesis. In this case you are using the version of the test where the null hypothesis is stationarity around a deterministic trend. The critical values for the 0.10, 0.05, 0.025 and 0.01 nominal levels are given in Table I of the original paper.

The critical values are returned by ur.kpps for convenience so that you can easily compare the value of the test statistic obtained with your data with the theoretical critical values.

For example, given a 0.05 level: for the series c, the null hypothesis is rejected because $0.3717 > 0.146$; for the series d, the null is rejected as well because $0.1771 > 0.146$. Upon these results, we conclude that, at the 5% significance level, these series are not stationary around a deterministic trend.

The null hypothesis is not rejected for the series h because $0.0815 < 0.146$. Similarly for the series q we have $0.0801 < 0.146$. Hence, in these series the hypothesis of stationarity around a deterministic trend cannot be rejected at the 5% significance level. The augmented Dickey and Fuller test could be obtained for these series to check whether the null of a unit root is rejected.

javlacalle
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  • Thx for your great answer! What I do not understand is, how the DF test might contribute to the KPSS test. For example, what if the DF test shows non-stationarity, which is complementary to the KPSS test results? – Carol.Kar Jul 12 '15 at 08:22
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    @Kare The null hypothesis of the ADF test is the opposite of the KPSS test. Thus, a way to proceed is to test first for the null of stationarity: if the null hypothesis is rejected then conclude that the series is not stationary, otherwise test for the null of a unit root by means of the ADF test. In the latter case, if the null of a unit root is rejected then conclude that the series is stationary, otherwise the data are not informative enough to reach a conclusion since none of the hypotheses could be rejected. – javlacalle Jul 12 '15 at 09:17
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    The same procedure could be applied but starting with the ADF test, see [this post](http://stats.stackexchange.com/questions/107551/can-a-trend-stationary-series-be-modeled-with-arima/107696#107696). – javlacalle Jul 12 '15 at 09:18
  • Thx for your reply! Just for clarification series `c` if $0.3717 > 0.146$ is then `non-stationary` and should be corrected by a `log`. Is my statement true? I am a little bit confused at the moment.... – Carol.Kar Jul 18 '15 at 14:41
  • Please see also my updated picture. – Carol.Kar Jul 18 '15 at 15:14
  • The decision to taking logs does not depend on whether there is a unit root or not. For the series where the conclusion is that the trend is stochastic (unit root) you can render them stationary by taking first differences to the data. – javlacalle Jul 19 '15 at 10:22
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    For the series `c`you have `0.3717>0.146` and hence the null of stationarity around a deterministic linear trend is rejected. This is in agreement with the result of the ADF test, for which the null of a unit root is not rejected (`0.94` will be larger that the 5% critical value for any sample size). – javlacalle Jul 19 '15 at 10:22
  • Thx for your answer! I guess I still not get it. My logic current logic is like that, for KPSS `IF(0,3717 > 0,146; H1; H0)` and for DF `IF(0,94 > 0,05; H1; H0)`. So the result should be for both tests `H1`, which is conclusive. What is my flaw in my thinking? Please also have a look at the attached google spreadsheet. – Carol.Kar Jul 19 '15 at 13:28
  • Btw from your logic it should be for the DF test: `IF(0,94 > 0,05; H0; H1)`. Why here the other way round? – Carol.Kar Jul 19 '15 at 13:30
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    You should compare the ADF test statistic with the critical value given a significance level chosen beforehand, e.g. `0.05`. [This table](https://en.wikipedia.org/wiki/Augmented_Dickey%E2%80%93Fuller_test#Examples) reports critical values for different sample sizes. For example, for a series with 60 observations the 5% critical value is between $-2.93$ and $-2.89$, let's take $-2.93$ for illustration. Then, since $0,94 < -2.93$ the null hypothesis of a unit root is not rejected. – javlacalle Jul 19 '15 at 17:18
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    Notice that the KPSS test is a right tailed test (the critical region is in the right tail of the distribution, i.e., values of the test statistic larger than the tabulated critical value involve rejection of the null hypothesis) while the ADF test is a left tailed test. – javlacalle Jul 19 '15 at 17:19