New Video: Immunization, Part 4 - Infinity is Really Big

In “New Video: Immunization, Part 3“, I showed that if a liability cashflow is not immunized by an asset cashflow, we might be able to make modifications to change this state of affairs.

In particular, the time of payment of one of the asset amounts could be changed to achieve, at least, Redington immunization. To be even more specific, I showed that if the liability cashflows are 500 at time 1 and 1000 at time 4, then an asset payment of $Y\approx \frac{3186.6\cdot 1.1^{y}}{y}\approx 1253.91$ at time $y\approx 3.57207$ along with an asset payment of $X\approx 1137.56-\frac{3186.6}{y}\approx 245.47$ at time 0 are “on the boundary” of achieving Redington immunization of the liabilities by the assets. Indeed, if these same monetary amounts are paid, with the second one paid at a time even slightly larger than 3.57207, then Redington immunization will be achieved.

In the new video embedded below, “Actuarial Exam 2/FM: Modifying an Asset Cashflow to Achieve Immunization”, I generalized this even further.

Actuarial Exam 2/FM: Modifying an Asset Cashflow to Achieve Immunization. Financial Math for Actuarial Exam 2 (FM), Video #172. Generalization of October 2018 SOA Sample Exam, Problem #127.

In this video, in addition to an arbitrary time $y$ for the second asset payment, I also took the interest rate of valuation, $i=i_{0},$ to be arbitrary.

In this situation, matching present values and durations of the asset and liability cashflows results in an asset payment of $X=f(i_{0},y)=\frac{500y-500}{y}(1+i_{0})^{-1}+\frac{1000y-4000}{y}(1+i_{0})^{-4}$ at time 0 and an asset payment of $Y=g(i_{0},y)=\frac{500}{y}(1+i_{0})^{y-1}+\frac{4000}{y}(1+i_{0})^{y-4}$ at time $y.$

The function $h(i)=P_{A}(i)-P_{L}(i)$ representing the difference in the present values of the asset and liability cashflows may have a graph that is either concave up or concave down near $i=i_{0}.$ We would like it to be concave up near $i=i_{0}$ in order to satisfy the conditions for Redington immunization, which will help make us more “immune” to small interest rate changes.

In the video, Mathematica is used to make graphs and solve equations in order to find that Redington immunization is achieved (the graph of $h$ is concave up near $i=i_{0}$ ) when $y>\frac{33+3i_{0}+3i_{0}^{2}+i_{0}^{3}}{9+3i_{0}+3i_{0}^{2}+i_{0}^{3}}.$

Graphs of $X=f(i_{0},y)$ and $Y=g(i_{0},y)$ are also made in order to understand how the size of the asset payments is dependent on both $i_{0}$ and $y.$ The most interesting results are: 1) that $X$ increases as $y$ increases while $Y$ decreases as $y$ increases and 2) for relatively small values of $i_{0}>0$ , $X$ will eventually become larger than $Y$ when $y$ is sufficiently large.

The following animation shows the graphs of $X=f(i_{0},y), Y=g(i_{0},y),$ and $Y-X=g(i_{0},y)-f(i_{0},y)$ as functions of $y.$ The animation parameter is $i_{0},$ which increases from 0 to 0.2.

The graphs of $X, Y,$ and $Y-X$ as functions of $y.$ The animation parameter is $i_{0},$ which increases from 0 to 0.2.

The graphs of $X, Y,$ and $Y-X$ as functions of $y.$ The animation parameter is $i_{0},$ which increases from 0 to 0.2.