Основы страхования. Орлов В.П. - 18 стр.

UptoLike

Составители: 

Рубрика: 

d
(m)
= md
(m)
.
1/m
i
(m)
= (1 + i)
1
m
1.
(1 + i)
1
m
= 1 +
1
m
i +
1 m
m
2
i
2
(1 + Θi)
1
m
2
, 0 Θ 1,
i
(m)
= mi
(m)
= i
m 1
m
i
2
(1 + Θi)
1
m
2
i.
i (0, 1) i
(m)
i
i
i
δ = ln(1 + i); i = e
δ
1; i =
d
1 d
; i =
1 v
v
;
v =
1
1 + i
; v = e
δ
;
d =
i
1 + i
; d = 1 v; d = 1 e
δ
;
i
(m)
= (1 + i)
1
m
1; i =
³
1 + i
(m)
´
m
1;
d
(m)
=
i
(m)
1 + i
(m)
; d
(m)
= 1 (1 d)
1
m
;
d
(m)
= m
³
1 (1 d)
1
m
´
; i
(m)
=
³
(1 + i)
1
m
1
´
m.
                                                 18 


     Îïðåäåëåíèå 4.14. Íîìèíàëüíîé ñòàâêîé äèñêîíòèðîâàíèÿ
íàçûâàåòñÿ ÷èñëî
                                            d(m) = md(m)
                                                     ∗ .                                  (4.2.19)

     Çàìå÷àíèå 4.1. Ïîÿñíèì, îòêóäà ïîÿâëÿþòñÿ íîìèíàëüíûå ïðîöåíòíûå
ñòàâêè. Ïðè ïåðèîäå êîíâåðñèè 1/m ýôôåêòèâíàÿ ïðîöåíòíàÿ ñòàâêà ðàâíà
 (m)         1
i∗ = (1 + i) m − 1. Òàê êàê ïî ôîðìóëå Òåéëîðà
                       1          1    1−m 2          1
              (1 + i) m = 1 +       i+     i (1 + Θi) m −2 ,                0 ≤ Θ ≤ 1,
                                  m     m2
òî
                                 m−1 2          1
                      i(m) = mi(m)
                               ∗     i (1 + Θi) m −2 ≈ i.
                                   =i−
                                  m
Îòñþäà âèäíî, ÷òî ïðè ìàëûõ i ∈ (0, 1) âåëè÷èíà i(m) ïðèáëèçèòåëüíî
ðàâíà i. Ýòî óäîáíî äëÿ ëþäåé, èìåþùèõ äåëî ñ ïðàêòè÷åñêèì ñ÷åòîì.

     Çàìå÷àíèå 4.2. Âñå âûâåäåííûå âûøå îïðåäåëåíèÿ îñíîâûâàþòñÿ íà
áàíêîâñêîì ñ÷åòå (4.2.1) è îïðåäåëÿþòñÿ ïðîöåíòíîé ñòàâêîé i, ò.å. èõ âñå
ìîæíî âûðàçèòü ÷åðåç i. Òî÷íî òàê æå, âçÿâ â êà÷åñòâå îïðåäåëåííîé âåëè-
÷èíû ëþáóþ èç âûâåäåííûõ, ìû ìîæåì ÷åðåç íåå âûðàçèòü âñå îñòàëüíûå.

  çàêëþ÷åíèå ïðèâåäåì ñâîäêó îñíîâíûõ ôîðìóë.
                                                                d                 1−v
       δ = ln(1 + i);         i = eδ − 1;               i=         ;       i=         ;   (4.2.20)
                                                               1−d                 v

                                             1
                                  v=            ;             v = e−δ ;                   (4.2.21)
                                            1+i
                            i
                      d=       ;            d = 1 − v;              d = 1 − e−δ ;         (4.2.22)
                           1+i
                                        1
                                                             ³         ´m
                     i(m)
                      ∗   = (1 + i) m − 1;                i = 1 + i(m)
                                                                   ∗      − 1;            (4.2.23)

                                   (m)
                                   i∗                                         1
                       d(m)
                        ∗     =         (m)
                                            ;       d(m)
                                                     ∗   = 1 − (1 − d) m ;                (4.2.24)
                             1 + i∗
                     ³           1
                                   ´                           ³       1
                                                                           ´
            (m)                                         (m)
        d         = m 1 − (1 − d) ;
                                 m                  i         = (1 + i) − 1 m.
                                                                       m                  (4.2.25)