从分布到置信区间的寓言

Question

我设法使用寓言进行预测，然后得到结果\n

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我能否获得有关如何将此分布更改为 80% 95% 置信区间的指导？谢谢你！

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您可以使用此处的示例代码来获取发行版

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result <\xe2\x80\x93USAccDeaths %>% as_tsibble %>% \n  model(arima = ARIMA(log(value) ~ pdq(0,1,1) + PDQ(0,1,1)))%>%\n  forecast(h=12)\n

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Answer 1

该hilo()函数允许您从预测分布中提取置信区间。它可以用于分布向量或寓言本身。

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library(tidyverse)\nlibrary(fable)\nresult <- as_tsibble(USAccDeaths) %>%\n  model(arima = ARIMA(log(value) ~ pdq(0,1,1) + PDQ(0,1,1)))%>%\n  forecast(h=12)\n\nresult %>% \n  mutate(`80%` = hilo(value, 80))\n#> # A fable: 12 x 5 [1M]\n#> # Key:     .model [1]\n#>    .model    index             value  .mean                    `80%`\n#>    <chr>     <mth>            <dist>  <dbl>                   <hilo>\n#>  1 arima  1979 Jan   t(N(9, 0.0014))  8290. [ 7899.082,  8689.169]80\n#>  2 arima  1979 Feb t(N(8.9, 0.0018))  7453. [ 7055.860,  7859.100]80\n#>  3 arima  1979 Mar   t(N(9, 0.0022))  8276. [ 7789.719,  8774.054]80\n#>  4 arima  1979 Apr t(N(9.1, 0.0025))  8584. [ 8036.304,  9144.752]80\n#>  5 arima  1979 May t(N(9.2, 0.0029))  9499. [ 8849.860, 10166.302]80\n#>  6 arima  1979 Jun t(N(9.2, 0.0033))  9900. [ 9180.375, 10639.833]80\n#>  7 arima  1979 Jul t(N(9.3, 0.0037)) 10988. [10145.473, 11857.038]80\n#>  8 arima  1979 Aug t(N(9.2, 0.0041)) 10132. [ 9315.840, 10974.140]80\n#>  9 arima  1979 Sep t(N(9.1, 0.0045))  9138. [ 8368.585,  9933.124]80\n#> 10 arima  1979 Oct t(N(9.1, 0.0049))  9391. [ 8567.874, 10243.615]80\n#> 11 arima  1979 Nov t(N(9.1, 0.0052))  8863. [ 8056.754,  9699.824]80\n#> 12 arima  1979 Dec t(N(9.1, 0.0056))  9356. [ 8474.732, 10271.739]80\n\nresult %>% \n  hilo(level = c(80, 95))\n#> # A tsibble: 12 x 6 [1M]\n#> # Key:       .model [1]\n#>    .model    index             value  .mean                    `80%`\n#>    <chr>     <mth>            <dist>  <dbl>                   <hilo>\n#>  1 arima  1979 Jan   t(N(9, 0.0014))  8290. [ 7899.082,  8689.169]80\n#>  2 arima  1979 Feb t(N(8.9, 0.0018))  7453. [ 7055.860,  7859.100]80\n#>  3 arima  1979 Mar   t(N(9, 0.0022))  8276. [ 7789.719,  8774.054]80\n#>  4 arima  1979 Apr t(N(9.1, 0.0025))  8584. [ 8036.304,  9144.752]80\n#>  5 arima  1979 May t(N(9.2, 0.0029))  9499. [ 8849.860, 10166.302]80\n#>  6 arima  1979 Jun t(N(9.2, 0.0033))  9900. [ 9180.375, 10639.833]80\n#>  7 arima  1979 Jul t(N(9.3, 0.0037)) 10988. [10145.473, 11857.038]80\n#>  8 arima  1979 Aug t(N(9.2, 0.0041)) 10132. [ 9315.840, 10974.140]80\n#>  9 arima  1979 Sep t(N(9.1, 0.0045))  9138. [ 8368.585,  9933.124]80\n#> 10 arima  1979 Oct t(N(9.1, 0.0049))  9391. [ 8567.874, 10243.615]80\n#> 11 arima  1979 Nov t(N(9.1, 0.0052))  8863. [ 8056.754,  9699.824]80\n#> 12 arima  1979 Dec t(N(9.1, 0.0056))  9356. [ 8474.732, 10271.739]80\n#> # \xe2\x80\xa6 with 1 more variable: `95%` <hilo>\n

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要从对象中提取数值<hilo>，您可以使用该unpack_hilo()函数，或使用<hilo>$lower,<hilo>$upper和获取每个部分<hilo>$level。

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result %>% \n  hilo(level = c(80, 95)) %>% \n  unpack_hilo("80%")\n#> # A tsibble: 12 x 7 [1M]\n#> # Key:       .model [1]\n#>    .model    index             value  .mean `80%_lower` `80%_upper`\n#>    <chr>     <mth>            <dist>  <dbl>       <dbl>       <dbl>\n#>  1 arima  1979 Jan   t(N(9, 0.0014))  8290.       7899.       8689.\n#>  2 arima  1979 Feb t(N(8.9, 0.0018))  7453.       7056.       7859.\n#>  3 arima  1979 Mar   t(N(9, 0.0022))  8276.       7790.       8774.\n#>  4 arima  1979 Apr t(N(9.1, 0.0025))  8584.       8036.       9145.\n#>  5 arima  1979 May t(N(9.2, 0.0029))  9499.       8850.      10166.\n#>  6 arima  1979 Jun t(N(9.2, 0.0033))  9900.       9180.      10640.\n#>  7 arima  1979 Jul t(N(9.3, 0.0037)) 10988.      10145.      11857.\n#>  8 arima  1979 Aug t(N(9.2, 0.0041)) 10132.       9316.      10974.\n#>  9 arima  1979 Sep t(N(9.1, 0.0045))  9138.       8369.       9933.\n#> 10 arima  1979 Oct t(N(9.1, 0.0049))  9391.       8568.      10244.\n#> 11 arima  1979 Nov t(N(9.1, 0.0052))  8863.       8057.       9700.\n#> 12 arima  1979 Dec t(N(9.1, 0.0056))  9356.       8475.      10272.\n#> # \xe2\x80\xa6 with 1 more variable: `95%` <hilo>\n

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^{由reprex 包(v0.3.0)于 2020-04-08 创建}

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