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Deutsch EnglishSet ( E[1 - e^-a(W-X)] = 1 - e^-a(W-P) ). Simplify: ( E[e^-a(W-X)] = e^-a(W-P) ) → ( e^-aW E[e^aX] = e^-aW e^aP ) → ( E[e^aX] = e^aP ). For ( X \sim \textExp(\lambda) ), ( M_X(a) = \frac\lambda\lambda - a ) for ( a < \lambda ). Thus ( P = \frac1a \ln\left( \frac\lambda\lambda - a \right) ). Interpretation: Premium increases with risk aversion ( a ) and volatility of ( X ). Chapter 4: Collective Risk Model Example Exercise: Claim number ( N \sim \textPoisson(\lambda) ), claim sizes ( Y_i \sim \textExp(\mu) ). Derive the moment generating function of total claim ( S = \sum_i=1^N Y_i ). Then compute ( \textVar(S) ).
panjer_poisson <- function(lambda, fY, max_claims) pn <- dpois(0:max_claims, lambda) fs <- numeric(max_claims+1) fs[1] <- pn[1] # P(S=0) for (n in 1:max_claims) for (k in 1:n) fs[n+1] <- fs[n+1] + (lambda * k / n) * fY[k] * fs[n - k + 1] fs[n+1] <- fs[n+1] * pn[1] # adjust for Poisson return(fs) modern actuarial risk theory solution manual
MGF: ( M_S(t) = \exp\left( \lambda (M_Y(t) - 1) \right) ). For ( Y \sim \textExp(\mu) ), ( M_Y(t) = \frac\mu\mu - t ) for ( t < \mu ). Hence ( M_S(t) = \exp\left( \lambda \left( \frac\mu\mu - t - 1 \right) \right) = \exp\left( \frac\lambda t\mu - t \right) ). Variance: ( E[S] = \lambda E[Y] = \lambda/\mu ), ( \textVar(S) = \lambda E[Y^2] = \lambda \cdot \frac2\mu^2 ). Check: Using ( \textVar(S) = E[N]\textVar(Y) + \textVar(N)(E[Y])^2 = \lambda \cdot \frac1\mu^2 + \lambda \cdot \frac1\mu^2 = \frac2\lambda\mu^2 ). Correct. Chapter 5: Ruin Theory Example Exercise: For the classical compound Poisson risk process ( U(t) = u + ct - S(t) ) with ( c = (1+\theta)\lambda E[Y] ), premium loading ( \theta ), claim sizes ( Y \sim \textExp(1) ). Show that the adjustment coefficient ( R ) satisfies ( 1 + (1+\theta)R = \frac11-R ). Solve for ( R ). Set ( E[1 - e^-a(W-X)] = 1 - e^-a(W-P) )
This is a request for a on a topic that, strictly speaking, does not exist as a standard published work. There is no widely recognized, single textbook titled Modern Actuarial Risk Theory with an accompanying official solutions manual. However, the closest and most likely reference is the textbook Modern Actuarial Risk Theory by Rob Kaas, Marc Goovaerts, Jan Dhaene, and Michel Denuit (often referred to as "Kaas et al."), published by Springer. Thus ( P = \frac1a \ln\left( \frac\lambda\lambda -
The best linear unbiased predictor of ( X_i,n+1 ) is ( Z\barX i + (1-Z)\mu ). The credibility factor ( Z ) minimizes ( E[(X i,n+1 - (Z\barX_i + (1-Z)\mu))^2] ). Using the law of total variance: ( \textVar(\barX_i) = E[\textVar(\barX_i|\Theta)] + \textVar(E[\barX_i|\Theta]) = E[\sigma^2(\Theta)/n] + \textVar(\mu(\Theta)) = v/n + a ). Covariance: ( \textCov(\barX i, X i,n+1) = E[\textCov(\barX i, X i,n+1|\Theta)] + \textCov(E[\barX i|\Theta], E[X i,n+1|\Theta]) = 0 + \textVar(\mu(\Theta)) = a ). Then ( Z = \frac\textCov(\barX i, X i,n+1)\textVar(\barX_i) = \fracav/n + a = \fracnn + v/a ). Interpretation: As ( n \to \infty ), ( Z \to 1 ) (full reliance on own data); as ( a \to 0 ) (no heterogeneity), ( Z \to 0 ). Chapter 10: Generalized Linear Models in Actuarial Science Example Exercise: For a Poisson GLM with log link: ( \log(\mu_i) = \beta_0 + \beta_1 x_i1 ). Derive the score equations for ( \beta ) and show that they correspond to ( \sum_i (y_i - \mu_i) = 0 ) and ( \sum_i (y_i - \mu_i) x_i1 = 0 ).
This paper provides a for a solutions manual that does not exist yet—but should. If you need a specific chapter fully solved or a different textbook addressed, let me know.