70   Life Contingencies

Dx+nLPa (Nx DNx+,)(ifx-Mx+n"1 D

= Px(Nx—Nx+„) —(Ilfx—llix+n) as before.

Dx+n

4. From the fundamental relations connecting A, a and P and the rate of interest the student should prove that n Vx =1— (Px +d) ax+

=1 _ ax+n = ax — ax-En = (Px +d) (ax a

— x+n)

ax   ax

_ (Px+n — Px) ax+n = Px+n Px

Px+n +d

Ax+n—Ax Ax+n—Ax

Explanations in words should be made for each of the above expressions for n Vx.

5. Again, consider a whole life policy taken out by (x) with an annual premium of Px

n T'x is the "terminal reserve" for the nth year

(n—i ti x+Px) is the "initial reserve" for the nth year

z (n—i Vx +n Vx+Px) is the "mid-year" or "mean reserve" for the nth year

(1—n T7x) is the "net amount at risk" for the nth year and 4x+n—i(1 —nVx) is the "cost of insurance" or "death strain" for the nth year.

We have then, in respect of the nth year

(n—1 T'x+Px) (1+i) -4x+n—1(1 —nT'x) =nVx.

At the beginning of the nth year, just before the premium is paid, there is n_1 T'x to the credit of the policy holder. After this man has paid his premium there will be n_11'x+Px to his credit.

   1—Ax   dax

   =Ax+n (1—    x

P

Px+n