Modern Actuarial Risk Theory Solution Manual < HD 2026 >

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)

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) ). 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 ). Thus ( P = \frac1a \ln\left( \frac\lambda\lambda -

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 ). Derive the moment generating function of total claim