Risk Theory Solution Manual - Modern Actuarial

Likelihood: ( L = \prod_i \frace^-\mu_i \mu_i^y_iy_i! ), log-likelihood: ( \ell = \sum_i (y_i \log \mu_i - \mu_i - \log y_i!) ). With ( \mu_i = e^\beta_0 + \beta_1 x_i1 ), derivative wrt ( \beta_0 ): ( \frac\partial \ell\partial \beta_0 = \sum_i \left( y_i \frac1\mu_i \cdot \mu_i - \mu_i \right) = \sum_i (y_i - \mu_i) = 0 ). Derivative wrt ( \beta_1 ): ( \frac\partial \ell\partial \beta_1 = \sum_i \left( y_i \frac1\mu_i \cdot \mu_i x_i1 - \mu_i x_i1 \right) = \sum_i (y_i - \mu_i) x_i1 = 0 ). Thus the GLM score equations equate observed and expected weighted sums. 4. Pedagogical Features of an Ideal Solutions Manual A truly modern solutions manual would go beyond answer keys:

Set ( 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

Lundberg equation: ( \lambda (M_Y(R) - 1) = cR ). Given ( M_Y(R) = \frac11-R ) (for exponential(1)), ( c = (1+\theta)\lambda \cdot 1 ). Plug: ( \lambda \left( \frac11-R - 1 \right) = (1+\theta)\lambda R ) → ( \fracR1-R = (1+\theta)R ). If ( R > 0 ), divide by ( R ): ( \frac11-R = 1+\theta ) → ( 1 = (1+\theta)(1-R) ) → ( R = \frac\theta1+\theta ). Remark: For exponential claims, the adjustment coefficient is simply a function of the safety loading. Chapter 7: Credibility Theory Example Exercise (Bühlmann model): For a portfolio of risks, the conditional variance ( \textVar(X_ij|\Theta) = \sigma^2(\Theta) ) and ( E[X_ij|\Theta] = \mu(\Theta) ). Given ( E[\mu(\Theta)] = \mu ), ( \textVar(\mu(\Theta)) = a ), and ( E[\sigma^2(\Theta)] = v ). Derive the Bühlmann credibility factor ( Z = \fracnn + v/a ).

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 ). Likelihood: ( L = \prod_i \frace^-\mu_i \mu_i^y_iy_i

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 ).

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. Derivative wrt ( \beta_1 ): ( \frac\partial \ell\partial

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.