03.08.2022 - 10:18

# A discrete temporary annuity due of $20,000 is issued to (60). The annual effective interest rate is 4 %; mortality is on the basis of the ILT. What is the probability that the present value random variable (of the annuity due) is greater than$150,000?

Question:

A discrete temporary annuity due of {eq}\displaystyle $20,000 {/eq} is issued to {eq}(60). {/eq} The annual effective interest rate is {eq}4 % {/eq} ; mortality is on the basis of the {eq}ILT {/eq}. What is the probability that the present value random variable (of the annuity due) is greater than {eq}\displaystyle$150,000 {/eq}?

• First, we need to find the present value of the annuity due. Using the formula for the present value of a discrete annuity due: PV = (1+i)*a*((1+i)^n - 1)/i where PV is the present value, i is the annual effective interest rate, a is the annuity payment, and n is the number of payments. Plugging in the values, we get: PV = (1+0.04)*20000*((1+0.04)^60 - 1)/0.04 = $904,495.71 Now, we need to find the probability that this present value random variable is greater than$150,000. This can be done using the standard normal distribution: P(Z > (150000 - PV)/SD) where Z is the standard normal variable, PV is the present value we just calculated, and SD is the standard deviation of the present value random variable. To find SD, we need to use the variance formula for a discrete annuity due: Var = ((1+i)^2 - (1+i)*(n+1) + n*i*i)*a*a/((1+i)^2) Plugging in the values, we get: Var = ((1+0.04)^2 - (1+0.04)*(60+1) + 60*0.04*0.04)*20000*20000/((1+0.04)^2) = $48,601,025.43 So, SD = sqrt(Var) =$6,967.23 Plugging everything into the formula for probability, we get: P(Z > (150000 - 904495.71)/6967.23) = P(Z > -66.12) = 1 - P(Z < -66.12) = 1 The probability that the present value random variable is greater than $150,000 is therefore 1. This means that it is extremely unlikely for the present value of this annuity to be less than$150,000.
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