[15 pts] A certain investment will pay a return of c dollars at an unknown point

Question

[15 pts] A certain investment will pay a return of c dollars at an unknown point in the future. Let X denote the time (in years) from today that the investment pays the return, and let Y denote the present value of the return 1. a. Assuming an annual inflation rate r E (0,1), express Y as a function of X b. Suppose X is uniform on the set (1... .n). Calculate E(Y) and Var(Y). c. Suppose instead X is geometric with success parameter p. Calculate E (Y) and Var(Y) d. Still supposing X is geometric, find the value of p if E(Y) -

Explanation / Answer

a) If a return c is paid after X years, the present value Y=c/(1+r)^X.

b)If X is uniform, the P(X)=1/n, E(Y)=Sum(c/(1+r)^X*1/n) from (1 to n), Var(Y)=E(Y^2)-(E(Y))^2.

E(Y^2)=Sum([c/(1+r)^X]^2*1/n) from (1,n)

c) Geometric probability P(X)=p*(1-p)^(x-1), thus E(Y)=SUM[(c/(1+r)^X*p*(1-p)^(X-1)] from (1.n) and similary

Var(Y)=E(Y^2)-(E(Y))^2 and E(Y^2)=SUM[Y^2*P(X)]=SUM([c/(1+r)^X]^2*p*(1-p)^(X-1)) from (1,n)

d) For geometric E(Y)=SUM[(c/(1+r)^X*p*(1-p)^(X-1)] or simplified it can be written as cp/(1-p)*Sum[(1-p)/(1+r)]^x] from (1,n). Using the gp sum rule and simplifying we get E(Y)=cp/(r+p)*[1-[(1-p)/(1+r)]^n], equating this equal to c/2, we get p/(r+p)*(1-[(1-p)/(1+r)]^n)=1/2, where p can be solved in terms of r.

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