Creating a Family Trust
 | 2. Probability Density Functions: Uniform, Exponential, Normal, and Beta | | 3. Mean, Median, Variance and Standard Deviation | | Calculus and Probability Main Page | | "Real World" Page |
Creating a Family Trust Your position as financial consultant to the clients of Family Bank, Inc., often entails your having to give financial advice to clients with complex questions about savings. One of your newer clients, Malcolm Adams, recently graduated from college and 22 years old, presents you with a perplexing question. "I would like to set up my own insurance policy by opening a trust account into which I can make monthly payments starting now, so that upon my death or my ninety-fifth birthday-whichever comes sooner-the trust can be expected to be worth $500,000. How much should I invest each month?"
This is not one of those questions that you can answer by consulting a table, so you promise Malcolm an answer by the next day and begin to work on the problem. After a little thought, you realize that the question is one about expected value-the expected future value of an annuity into which monthly payments are made. Since the annuity would terminate upon his death (or his ninety-fifth birthday), you decide that you need a model for the probability distribution of the lifespan of a male in the United States. To obtain this information, you consult mortality tables and come up with the histogram in shown in the following figure (you work with the actual numbers, but they are not important for the discussion to follow).
From the data you calculate that the mean is µ = 70.778 and the standard deviation is 
= 16.5119.
Next, you decide to model these data with a suitable probability density function. You rule out the uniform and exponential density functions, since they have the wrong shape, and you first try the normal distribution. The normal distribution is
Using the above values for µ and , you obtain the following figure, which shows the graph of the normal density function superimposed on the actual data.
This does not seem like a very good fit at all! Since the actual histogram looks as though it is "pushed over" to the right, you think of the beta distribution, which has that general shape. The beta distribution is given by
+1)(+2)x(1x),
where can be obtained from the mean µ using the equation
+ 1)/( + 3).
There is one catch: the beta distribution assumes that X is between 0 and 1, whereas your distribution is between 0 and 100. This doesn't deter you: all you need to do is scale the X-values to 1/100 = 0.01 of their original value. Thus you substitute µ = (0.01)(70.776) = 0.70776 in the above equation and solve for , you obtain = 3.8437. You then plot the associated beta function (after scaling it to fit the range 0 
X 100) and again discover that, although better, the fit still leaves something to be desired.
Now you just want some function that fits the data. You turn to your statistical software and ask it to find the cubic equation that best fits the data using the least squares method. It promptly tells you that the cubic function that best fits the data is
where
3.815484
10 7
b = 5.7399145 105
c = 0.0020856085
d = 0.0190315095.
Its graph is shown below. Note that the curve, although erratic for small values of X, fits the large peak on the right more closely than the others.
Encouraged, you use the same software to obtain a quartic (degree 4) approximation, and you find:
where
9.583507109,
b = 1.650155106,
c = 8.523081105,
d = 0.0016190575
e = 0.007865381.
The following figure shows the result.
This seems like the best fit of them all-especially for the range of X you are interested in: 22
(This is a standard formula from finance. This formula assumes that interest is paid at the end of each month.) Here, P is the monthly payment-the quantity that Malcolm wants to know-i is the interest rate, and n is the number of years for which payments are made. Since Malcolm will be making investments starting at age 22, this means that n = x
As for the interest rate i, you decide to use a conservative estimate of 5%.
Now the expected value of V(X) is given by
where f(x) is the quartic approximation to the distribution function. Since Malcolm wants this to be $500,000, you set
= P
Solving for P,
You now calculate the integral numerically (using the quartic approximation to f(x)), obtaining
The next day, you can tell Malcolm that at a 5% interest rate, his family can expect the trust to be worth $500,000 upon maturity if he deposits $146.64 each month.
Exercises
1. How much smaller will the payments be if the interest rate is 6%?
2. How much larger would the payments be if Malcolm began payments at the age of 30?
3. Repeat the original calculation using the normal distribution described above. Give reasons for the discrepancy between the answers, explaining why your answer is smaller or larger than the one calculated above.
4. Repeat Exercise 3 using the cubic distribution.
5. If Malcolm wanted to terminate the trust at age 65, which model would you use for the probability density? Give reasons for your choice.
6. Suppose you were told by your superior that, since the expected male lifespan is 71, you could have saved yourself a lot of trouble by using the formula for the future value of an annuity maturing at age 71. Based on that, the payment comes out to about $198 per month. Why is it higher? Why is it the wrong amount?
7. Explain how an insurance company might use the above calculations to compute life insurance premiums.
Mail us at:
X 95. (Malcolm is 22 years old and the trust will mature at 95.)
V = 12P[ (1 + i/12)12n
1 ]/i.
22, so the future value of his annuity at age x is
V(x) = 12P[ (1 + i/12)12(x
22) 1 ]/i.
E(V) =
2295 V(x)f(x) dx.
500,000 =
2295 V(x)f(x) dx.
=
229512P[ (1 + i/12)12(x22) 1 ]f(x)/i dx
229512[ (1 + i/12)12(x22) 1 ]f(x)/i dx.
P = 500,000 / {
229512[ (1 + i/12)12(x22) 1 ]f(x)/i dx }
P ý 500,000/3,409.8019
$146.64 per month.
2. Probability Density Functions: Uniform, Exponential, Normal, and Beta
3. Mean, Median, Variance and Standard Deviation
Calculus and Probability Main Page
"Real World" Page
Stefan Waner (matszw@hofstra.edu)
Steven R. Costenoble (matsrc@hofstra.edu)