Theory Examples
theory Examples
imports Life_Table
begin
section ‹Examples›
text ‹
The following lemma is a verification of the solution to the multiple choice question No.\ 3
of Exam LTAM Spring 2022 by Society of Actuaries.
›
context smooth_survival_function
begin
lemma SoA_LTAM_2022_Spring_MCQ_No3:
assumes "⋀x::real. 0 ≤ x ⟹ x ≤ 100 ⟹ ccdf (distr 𝔐 borel X) x = (1 - 0.01*x).^0.5"
shows "¦1000*$μ_25 - 6.7¦ < 0.05"
proof -
let ?svl = "ccdf (distr 𝔐 borel X)"
have [simp]: "ereal 25 < $ψ"
apply (rewrite not_le[THEN sym])
using assms by (rewrite ccdfX_0_equiv[THEN sym]) simp
have ⋆: "((λx. ln (1 - x/100)) has_real_derivative (-1/75)) (at 25)"
proof -
have "((λx. (1 - x/100)) has_real_derivative (0 - 1/100)) (at 25)"
apply (intro derivative_intros)
by (rule DERIV_cdivide) simp
hence "((λx. ln (1 - x/100)) has_real_derivative (1 / (1 - 25/100) * (-1/100))) (at 25)"
by (intro derivative_intros) auto
thus ?thesis by simp
qed
have "exp ∘ (λx::real. 0.5 * ln (1 - 0.01*x)) field_differentiable at 25"
apply (rule field_differentiable_compose, simp_all)
apply (rule derivative_intros, simp_all)
using ⋆ field_differentiable_def apply blast
using field_differentiable_within_exp by blast
hence "?svl differentiable at 25"
apply (rewrite differentiable_eq_field_differentiable_real)
by (rule field_differentiable_transform_within[where d=1])
(simp_all add: powr_def dist_real_def assms)
hence "$μ_25 = - deriv (λx. ln (?svl x)) 25" by (rule mu_deriv_ln; simp)
also have "… = 0.005 / (1 - 0.01*25)"
proof -
have "∀⇩F x in nhds 25. ln (?svl x) * 2 = ln (1 - x/100)"
proof -
{ fix x::real assume "dist x 25 < 1"
hence asm: "0 < x" "x < 100" using dist_real_def by auto
hence "ln (?svl x) * 2 = ln ((1 - 0.01*x).^0.5) * 2" using assms by (smt (verit))
also have "… = (0.5 * ln (1 - 0.01*x)) * 2" using asm by (rewrite ln_powr) auto
finally have "ln (?svl x) * 2 = ln (1 - x/100)" by simp }
thus ?thesis by (rewrite eventually_nhds_metric) (smt (verit, del_insts))
qed
hence "((λx. ln (?svl x)) has_real_derivative (-0.005 / (1 - 0.01*25))) (at 25)"
using ⋆ apply (rewrite DERIV_cong_ev[where g="λx. 0.5 * ln (1 - 0.01*x)"], simp_all)
by (rule derivative_eq_intros) auto
thus ?thesis using DERIV_imp_deriv by force
qed
finally show ?thesis by simp
qed
end
text ‹
The following lemma is a verification of the solution to the problem No.\ 2.\ (1)-1
of Life Insurance Mathematics 2016 by the Institute of Actuaries of Japan,
slightly modified; see the remark below.
›
context smooth_life_table
begin
lemma IAJ_Life_Insurance_Math_2016_2_1_1:
fixes a b :: real
assumes "-1 < a" "a < 0" "0 < b" "-b/a ≤ $ψ" and
"⋀x. 0 < x ⟹ x < -b/a ⟹ l differentiable at x" and
"⋀x. 0 ≤ x ⟹ x < -b/a ⟹ l integrable_on {x..} ∧ $e`∘_x = a*x + b"
shows "⋀x. 0 ≤ x ⟹ x < -b/a ⟹ $l_x = $l_0 * (b / (a*x + b)).^((a+1)/a)"
proof -
fix x assume asm_x: "0 ≤ x" "x < -b/a"
hence x_lt_psi[simp]: "ereal x < $ψ" using assms ereal_le_less by presburger
from asm_x have axb_pos[simp]: "a*x + b > 0"
using assms by (smt (verit, ccfv_threshold) mult.commute neg_less_divide_eq)
have mu: "⋀t. t∈{0<..<-b/a} ⟹ $μ_t = (a + 1) / (a*t + b)"
proof -
fix t assume asm_t: "t∈{0<..<-b/a}"
with assms have "((λu. $e`∘_u) has_real_derivative ($μ_t * $e`∘_t - 1)) (at t)"
using asm_t assms by (intro e_has_derivative_mu_e_l'[where a=0]; simp)
moreover have "((λu. $e`∘_u) has_real_derivative a) (at t)"
proof -
let ?d = "min t (-b/a - t)"
have "?d > 0" using assms asm_t by simp
moreover have "⋀u. dist u t < ?d ⟹ $e`∘_u = a*u + b" using assms dist_real_def by auto
ultimately have "∀⇩F u in nhds t. $e`∘_u = a*u + b" by (rewrite eventually_nhds_metric) blast
thus ?thesis
using assms apply (rewrite DERIV_cong_ev[where g="λt. a*t + b"], simp_all)
by (intro derivative_eq_intros) auto
qed
ultimately have "$μ_t * $e`∘_t - 1 = a" using DERIV_unique by blast
moreover have "$e`∘_t = a*t + b" using assms asm_t by simp
ultimately show "$μ_t = (a + 1) / (a*t + b)"
using assms by (smt (verit) mult_eq_0_iff nonzero_mult_div_cancel_right)
qed
hence "$p_{x&0} = (b / (a*x + b)).^((a+1)/a)"
proof -
have integ_mu: "integral {0..x} (λt. $μ_t) = (a + 1) / a * ln ((a*x + b) / b)"
proof -
have "integral {0..x} (λt. $μ_t) = integral {0<..x} (λt. $μ_t)"
apply (rule integral_spike_set)
apply (rule negligible_subset[where s="{0}"]; force)
by (rule negligible_subset[where s="{}"]; simp)
also have "… = integral {0<..x} (λt. ((a + 1) / a) * (a / (a*t + b)))"
using asm_x assms by (intro integral_cong) (rewrite mu; simp)
also have "… = (a + 1) / a * integral {0<..x} (λt. a / (a*t + b))"
by (metis integral_mult_right)
also have "… = (a + 1) / a * ln ((a*x + b) / b)"
proof -
have "integral {0<..x} (λt. a / (a*t + b)) = integral {0..x} (λt. a / (a*t + b))"
apply (rule integral_spike_set)
apply (rule negligible_subset[where s="{}"]; simp)
by (rule negligible_subset[where s="{0}"]; force)
also have "… = ln (a*x + b) - ln (a*0 + b)"
apply (rule integral_unique)
using assms asm_x apply (intro inverse_fun_has_integral_ln, simp_all)
using axb_pos assms apply (smt (verit) mult_less_cancel_left)
apply (intro continuous_intros)
by (intro derivative_eq_intros) auto
also have "… = ln ((a*x + b) / b)" using assms by (rewrite ln_div; simp)
finally have "integral {0<..x} (λt. a / (a*t + b)) = ln ((a*x + b) / b)" .
thus ?thesis by simp
qed
finally show ?thesis .
qed
thus ?thesis
apply (rewrite p_exp_integral_mu, simp_all add: asm_x)
unfolding powr_def using assms
by simp (smt (verit) axb_pos ln_div
nonzero_minus_divide_divide nonzero_minus_divide_right right_diff_distrib')
qed
thus "$l_x = $l_0 * (b / (a*x + b)).^((a+1)/a)"
using assms asm_x apply (rewrite in asm p_l, simp_all)
by (metis divide_mult_cancel l_0_neq_0 mult.commute)
qed
text ‹REMARK.
The original problem lacks the following hypotheses:
(i) @{text "0 < b"},
(ii) @{text "-b/a ≤ $ψ"},
(iii) @{text "∀x. 0 < x < -b/a ⟶ l differentiable at x"},
(iv) @{text "∀x. 0 ≤ x < -b/a ⟶ l integrable_on {x..}"}.
Moreover, the hypothesis @{text "∀x. 0 ≤ x < -b/a"} is originally @{text "∀x. 0 ≤ x ≤ -b/a"}.
›
end
end