Question

我在运行时间序列模型时遇到问题，预测结果（C.I。95％）为负值。它没有任何意义。我不确定哪个部分犯了错误。

week total_amount_by_week
52     10000
52     12000
52     12300
52     9800
52     23400
51
51...
.
.
.
00      10000
00      12930

数据集如下所示：因为plot.ts向后看，所以我对数据集做了相反的反应。

order_ts1 <- ts(order_sum$Total_Amount_by_week,start = c(00),end =c(52))
order_ts1[] <-rev(order_ts1)
plot.ts(order_ts1,col ='blue')

然后我计算了差异，ACF和PACF。

order_tsdiff3 <-diff(order_ts1,differences=3)
plot.ts(order_tsdiff3) # d=3

# calcualate ACF
acf(order_tsdiff3,lag.max=53)
acf(order_tsdiff3,lag.max=53,plot=FALSE)

# calculate PACF
pacf(order_tsdiff3,lag.max=53)
pacf(order_tsdiff3,lag.max=53,plot=FALSE)

fit_ma <- arima(order_ts1, order = c(1, 3, 1))
fit_ma

order_arimaforecast1 <- forecast.Arima(fit_ma,h=3,level=c(99.5))
order_arimaforecast1

所以情节看起来像这样：

所以我得到了一些估计值，但我认为这是错误的。为什么它会出现负值？

Call:
arima(x = order_sumts1, order = c(1, 3, 1))

Coefficients:
          ar1      ma1
      -0.6673  -1.0000
s.e.   0.1135   0.0539

sigma^2 estimated as 84368661:  log likelihood = -529.98,  aic = 1065.96

  Point Forecast   Lo 99.5  Hi 99.5
53      -1420.589 -27459.41 24618.23
54      -7983.391 -51772.69 35805.91
55     -21921.514 -93114.57 49271.54

所有估计金额应为正值。

Answer 1

尝试使用转换来帮助确保始终获得正值

## some sample data
a<-ts(c, frequency = 5)
## use ar() model with BoxCox() transform on the data & setting lambda = 0 (equivalent to taking the log)
fit <- ar(BoxCox(a,lambda=0))
## compute forecast and set lambda to 0 so that a back-transform is done (since we transformed the original data set we now want to back transform it to the original scale)
forecast(fit,h=3, lambda=0)

   Point Forecast     Lo 80    Hi 80     Lo 95    Hi 95
3.0       2.015135 1.1019547 3.685060 0.8005621 5.072398
3.2       1.385040 0.6984705 2.746480 0.4861378 3.946073
3.4       1.692792 0.8355244 3.429637 0.5749517 4.983975

如果你在没有变换的情况下完成这些步骤，即只预测“a”，你也会得到负值，但上述情况下的变换有助于此。

a<-ts(c, frequency = 5)
## lambda = 1 means no transformation
fit <- ar(BoxCox(a,lambda=1))
forecast(fit,h=3, lambda=1)

    Point Forecast     Lo 80    Hi 80      Lo 95    Hi 95
3.0            1.8 0.4764192 3.123581 -0.2242421 3.824242
3.2            1.8 0.4764192 3.123581 -0.2242421 3.824242
3.4            1.8 0.4764192 3.123581 -0.2242421 3.824242

您可以根据具体情况调整此代码：

arima(x = BoxCox(order_sumts1, lambda=0), order = c(1, 3, 1))

希望它有所帮助！

时间序列估计是R中的负值

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