A time series was observed at times 0, 1, …, 100. The last four observations along with estimates based on exponential and double exponential smoothing with [math]w = 0.8[/math] are:

Time (t)	97	98	99	100
Observation ( [math]y_t[/math] )	96.9	98.1	99.0	100.2
Estimates ( [math]\hat{s_1}(t) [/math]	93.1	94.1	95.1
Estimates ( [math]\hat{s_2}(t)[/math]	88.9	89.9

All forecasts should be rounded to one decimal place and the trend should be rounded to three decimal places.

Let [math]F[/math] be the predicted value of [math]y_{102}[/math] using exponential smoothing with [math]w = 0.8.[/math]

Let [math]G[/math] be the predicted value of [math]y_{102}[/math] using double exponential smoothing with [math]w = 0.8.[/math]

Calculate the absolute difference between [math]F[/math] and [math]G[/math], [math]| F − G |[/math] .

• 0.0
• 2.1
• 4.2
• 6.3
• 8.4

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: D

The smoothed forecast at 100 is

0.2(100.2) + 0.8(95.1) = 96.1. 

This is also the forecast at 102. The smoothed values are at 99:

0.2(95.1) + 0.8(89.9) = 90.9. At 100: 0.2(96.1) + 0.8(90.9) = 91.9. 

The trend is

0.2(96.1 – 91.9)/.8 = 1.05. 

The intercept (value at 100) is

2(96.1) – 91.9 = 100.3. 

The forecast at time 102 is

100.3 + 2(1.05) = 102.4.

The difference is 6.3.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.