iii Isolating n in 14 n � parenleftBig σ m parenrightBig 2 16 gives the minimal

Iii isolating n in 14 n ? parenleftbig σ m

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(iii) Isolating n in (1.4), n λ 0 parenleftBig σ m parenrightBig 2 , (1.6) gives the minimal number of observations (exposure units) to achieve full credibility. To be able to compute the probability statements in (1.1) to (1.3), the distribution of ¯ X must be known, or an approximation be used. When n is large, the Central Limit Theorem (CLT) may be invoked to approximate it by a normal distribution with mean m and variance σ 2 /n . Let epsilon1 N (0 , 1) represent a standard normal random variable, then using the CLT in (1.3) gives p = P {| epsilon1 | ≤ y p } = P {- y p epsilon1 y p } = Φ( y p ) - Φ( - y p ) , where Φ is the cdf of the N (0 , 1) variable epsilon1 . Now since Φ( - y p ) = 1 - Φ( y p ), then p = Φ( y p ) - Φ( - y p ) = Φ( y p ) - [1 - Φ( y p )] = 2Φ( y p ) - 1 ,
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8 CHAPTER 1. AMERICAN CREDIBILITY implying that Φ( y p ) = (1 + p ) 2 . For instance, if p = 0 . 9, then y 0 . 9 is the 95% one–sided percentile of the N (0 , 1), that is y 0 . 9 = 1 . 645. If in addition, an r is chosen, then (1.4) can be used as a full credibility criteria. For instance, if r = 0 . 05 then λ 0 = (1 . 645 / 0 . 05) 2 = (32 . 9) 2 = 1 , 082 . 41 and (1.6) becomes n 1 , 082 . 41 parenleftBig σ m parenrightBig 2 , (1.7) which still requires and independent estimate of the coefficient of variation σ/m . Remark 1.2. : Note the logical twist here as the argument first assumes that n → ∞ for the asymptotic approximation to hold, but then limits the conclusion only to those n λ 0 ( σ/m ) 2 . Example 1.1. (Poisson) Consider an i.i.d. sample N 1 , N 2 , . . . , N n of Poisson( λ ) counts representing the past number of claims of a particular policyholder over the last n periods. Here m = E ( N i ) = λ and σ 2 = V ( N i ) = λ and hence (1.7) becomes n λ 0 parenleftBig σ m parenrightBig 2 = 1 , 082 . 41 λ λ 2 = 1 , 082 . 41 λ , giving the required number of observations for full credibility. Equivalently n λ 1 , 082 . 41 , gives the required expected number of claims over the last n periods for full credibility. The sample total number of claims, N 1 + N 2 + · · · + N n , can be used as an estimator of n λ , in which case the condition is N 1 + N 2 + · · · + N n 1 , 082 . 41 . triangle Exercise 1.1. Rework the example above, this time using the exact dis- tribution of the sample mean ¯ N rather than its asymptotic distribution. Hint: remember that if N i Poisson( λ ) are independent, then n 1 N i Poisson( ).
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1.1. LIMITED FLUCTUATIONS CREDIBILITY 9 Example 1.2. (Compound Poisson) Now suppose that the i.i.d. sample X 1 , X 2 , . . . , X n of represents the total claims of a particular policyholder for the past n years. The X i = Y i 1 + · · · + Y iN i are assumed compound Poisson( λ ), where the claim severity distribution has mean m Y = E ( Y ij ) and variance σ 2 Y = V ( Y ij ). Hence m X = E ( X i ) = λ m Y and σ 2 X = V ( X i ) = λ ( m 2 Y + σ 2 Y ), implying that (1.7) becomes n 1 , 082 . 41 parenleftbigg σ X m X parenrightbigg 2 = 1 , 082 . 41 λ ( m 2 Y + σ 2 Y ) ( λ m Y ) 2 = 1 , 082 . 41 λ bracketleftBigg 1 + parenleftbigg σ Y m Y parenrightbigg 2 bracketrightBigg , or equivalently n λ 1 , 082 . 41 bracketleftBigg 1 + parenleftbigg σ Y m Y parenrightbigg 2
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