13_AS_8_lec_a

# E the observed no of deaths are consistent with the

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Unformatted text preview: each age Normal approximation (expected deaths should be greater than 5 for each age or age group) Null hypothesis The actual data (i.e., the observed no. of deaths) are consistent with the ones (i.e., the expected no. of deaths) that are predicted by the assumed probabilities (the graduated o o rates q x or µx + 1 ) 2 13/72 Actuarial Statistics – Module 8: Method of Graduation Statistical tests Preliminaries Deﬁnition - Standardised Deviations Most of the statistical tests will be comparing “Actual” no. of deaths (A) vs“Expected” no. of deaths (E) in some way A: the actual number of deaths observed in a particular category E: the expected number of deaths based on the assumed probabilities (i.e., graduated rates). Deviation is A-E and then standardised deviation is A−E √ E If there is a suﬃcient number of independent lives at each age x , by central limit theorem, we have, under all our models: 14/72 the standardised deviation zx ∼ Normal (0, 1) x = x1 , x2 , · · · , xm the zx ’s at diﬀerent ages are mutually independent Actuarial Statistics – Module 8: Method of Graduation Statistical tests Preliminaries Standardised deviations under Poisson model Under the Poisson model, we have for each x : Number of Deaths: o o c c Dx ∼ Normal Ex µx + 1 , Ex µx + 1 2 2 Deviation at age x=Ax − Ex o c dx − Ex µx + 1 2 x Standardised deviation at age x= A√−Ex : E x o zx = 15/72 c dx − Ex µx + 1 2 co Ex µx + 1 2 Actuarial Statistics – Module 8: Method of Graduation Statistical tests Preliminaries Standardised deviations under binomial model In the binomial model, Actual-Expected at age x: o dx − Ex q x Standardised Deviation at age x: o zx = dx − Ex q x o o Ex q x (1 − q x ) 16/72 (1) Actuarial Statistics – Module 8: Method of Graduation Statistical tests Preliminaries Example The standardised deviation at age 30 based on following extract of a graduation: o x Ex dx qx ˆ qx 30 70,000 39 0.000557 0.000460 is o z30 = d30 − E30 q 30 o o E30 q 30 (1 − q 30 ) = 17/72 39 − 70, 000(0.000460) 70, 000 × 0.000460(1 − 0.000460) = 1.2 Actuarial Statistics – Module 8: Method of Graduation Statistical tests Chi-square (χ2 ) test 1 Introduction 2 Testing smoothness 3 Statistical tests Preliminaries Chi-square (χ2 ) test Standardised Deviations Test Signs test Cumulative Deviations Test Grouping of Signs Test Serial Correlations Test 4 Methods of graduation Preliminaries Graduation by Parametric Formula...
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