You are given the following six observed values of the autoregressive model of order one time series

[[math]] y_t = \beta_0 + \beta_1 y_{t-1} + \epsilon_t [[/math]]

t	1	2	3	4	5	6
yt	31	35	37	41	45	51

with [math]\operatorname{Var}(\epsilon_t) = \sigma^2 [/math].

The approximation to the conditional least squares method is used to estimate [math]β_0[/math] and [math]β_1[/math] .

Calculate the mean squared error [math]s^2[/math] that estimates [math]σ^2.[/math]

• 13
• 21
• 22
• 26
• 35

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

The estimator of [math]\beta_1[/math] is

[[math]] b_1 = r_1 = \frac{a_1}{a_2} = \frac{117}{262} = 0.4466. [[/math]]

where [math]a_1[/math] equals

(31 − 40)(35 − 40) + (35 − 40)(37 − 40) + (37 − 40)(41 − 40) + (41 − 40)(45 − 40) + (45 − 40)(51 − 40) = 117

and [math]a_2[/math] equals

(31 − 40)2 + (35 − 40)2 + (37 − 40)2 + (41 − 40)2 + (45 − 40)2 + (51 − 40)2 = 262.

The estimator of [math]\beta_0[/math] is [math]b_0 = \overline{y}(1-r_1) = 22.136 [/math].

The residuals are then

e₂ =35- (22.136 + 0.4466 × 31) = −0.9806

e₃ =37−(22.136 + 0.4466 × 35) = −0.7670

e₄ =41−(22.136 + 0.4466 × 37) = 2.3398

e₅ =45−(22.136 + 0.4466 × 41) = 4.5534

e₆ =51−(22.136 + 0.4466 × 45) = 8.7670

The average residual is [math]\overline{e} = 2.78252[/math] and then the mean square error is

[[math]] s^2 = \frac{a_1}{a_2} = 21.969 [[/math]]

where [math]a_1[/math] equals

(−0.9806 − 2.78252)2 + (−0.7670 − 2.78252)2 + (2.3398 − 2.78252) 2  + (4.5534 − 2.78252)2 + (8.7670 − 2.78252)2

and [math]a_2[/math] equals 6-3=3.

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