New Video: Immunization, Part 3 - Infinity is Really Big

Immunization of a liability cashflow by an asset cashflow can help to cushion a company when interest rates change, but is it always possible?

My most recent video is “Actuarial Exam 2/FM: Liabilities Not Immunized by Assets in Spite of PV and Duration Matching” (Financial Math for Actuarial Exam 2 (FM), Video #171. October 2018 SOA Sample Exam, Problem #127). In this video I solve and discuss an Society of Actuaries sample exam problem where an asset cashflow can be found to match the present value and duration of a liability cashflow, but the liability cashflow is not immunized by the asset cashflow. The video is embedded below.

In this post, I want to explore why this happened from a graphical perspective and also explore whether this situation can be “fixed” by some modification of the asset cashflow.

The fact that the liability cashflow is not immunized by the asset cashflow means that, for small changes in the interest rate, the present value of the assets will be in a deficit position compared to the liabilities. This is not ideal, though it might not be too big of an issue if we expect the interest rate changes to remain small, or if we have the flexibility to increase our asset cashflow amounts.

Solution of the Problem in the Video

The liability cashflows from the video are 500 at time 1 and 1000 at time 4. The asset cashflows are unknowns: $X$ at time 0 and $Y$ at time 3. The given effective annual interest rate, assuming a flat yield curve, is $i=10\%=0.1.$

The first goal of the problem is to solve for $X$ , which also means we need to solve for $Y$ along the way. This is done by matching present values and durations.

For an arbitrary interest rate $i$ , the expression $P_{A}(i)=X+Y(1+i)^{-3}$ is the present value of the assets and the expression $P_{L}(i)=500(1+i)^{-1}+1000(1+i)^{-4}.$ We can then define a function $h$ by the formula $h(i)=P_{A}(i)-P_{L}(i),$ which represents the difference in the present values.

Matching present values at $i=i_{0}=.1$ means we want $h(.1)=0.$ The equation that follows from this condition is $X+Y\cdot 1.1^{-3}-500\cdot 1.1^{-1}-1000\cdot 1.1^{-4}=0,$ which can be written in approximate form as $X+.7513148Y\approx 1137.55891.$

Matching durations at $i=i_{0}=.1$ means we want the present values to be matched, as above, and we want $h'(.1)=0$ as well. Since $h'(i)=P_{A}'(i)-P_{L}'(i)=-3Y(1+i)^{-4}+500(1+i)^{-2}+4000(1+i)^{-5},$ the resulting equation is $3Y\cdot 1.1^{-4}=500\cdot 1.1^{-2}+4000\cdot 1.1^{-5},$ or $2.04904037Y\approx 2896.908433$ in approximate form.

This system of linear equations has solution $(X,Y)\approx (75.359151,1413.787879).$ Therefore, the cashflows of $X\approx 75.36$ at time 0 and $Y\approx 1413.79$ at time 3 will, at a 10% interest rate, match the cashflows of 500 at time 1 and 1000 at time 4, with respect to present values and durations.

But again, this particular asset cashflow does not immunize the liabilities against even small changes in the interest rate. Indeed, when the function $h(i)=P_{A}(i)-P_{L}(i)$ is plotted near $i=i_{0}=.1$ , the graph is always below the horizontal $i$ -axis, just touching it at $i=i_{0}=.1$ (see below).

The present value of the assets minus the present value of the liabilities, $h(i)=P_{A}(i)-P_{L}(i),$ is never positive for $0\leq i\leq 0.25.$ It has a root and a local maximum at $i=i_{0}=.1,$ where $h(0)=h'(0)=0.$

The present value of the assets minus the present value of the liabilities, $h(i)=P_{A}(i)-P_{L}(i),$ is never positive for $0\leq i\leq 0.25.$ It has a root and a local maximum at $i=i_{0}=.1,$ where $h(0)=h'(0)=0.$

This can be (partially) confirmed with the second derivative, which is $h''(i)=12Y(1+i)^{-5}-1000(1+i)^{-3}-20000(1+i)^{-6}\approx 16965.455(1+i)^{-5}-1000(1+i)^{-3}-20000(1+i)^{-6}.$

Since $h''(.1)\approx -1507<0,$ the Second Derivative Test implies that $h$ has a local (relative) maximum at $i=i_{0}=.1.$

This is the exact opposite of what we want for immunization! We want $h''(i_{0})>0$ instead!

The real-life moral of the story at this point is: do not assume that just matching present values and durations will result in immunization.

Are there any values of $i$ for which $h(i)>0$ (and $P_{A}(i)>P_{L}(i)$ )? The answer is yes, but they are altogether unrealistic values of $i.$ It turns out that $h(i)>0$ when $i>\frac{16119+11\sqrt{23011561}}{13240}\approx 5.2$ (I’ll leave it to you to double-check this). But $i=5.2$ corresponds to an interest rate of 520%!!! Below is the graph of $h(i)$ on a larger interval.

The graph of $h(i)=P_{A}(i)-P_{L}(i)$ does eventually become positive when $i$ is large enough, but this is not realistic for financial applications.

Can This Situation Be Remedied?

Is there anything we can do to “fix” this situation so that our assets are not in a deficit with respect to our liabilities?

Sure. One easy fix is to just increase our asset cashflows. In so doing, we may no longer be matching present values and durations, however.

This has a downside. By increasing our cashflows to cover these liabilities, we are using assets that could be used for other things — like other investments (physical capital, labor, or financial).

Perhaps, instead, we should try to “restructure” our asset cashflows to be occur at other times. To keep things simple, we could start by assuming that the first value $X$ is still paid at time 0. But then we could let $Y$ be paid at an arbitrary time $y.$ Our present value difference function then becomes $h(i)=X+Y(1+i)^{-y}-500(1+i)^{-1}-1000(1+i)^{-4}.$

We might then ask: are there any choices of $y$ that will result in values of $X$ and $Y$ that will lead to $h(.1)=h'(.1)=0$ and $h''(.1)>0?$

We could attempt to solve this by trial-and-error.

Or, we could use a more systematic approach, perhaps making use of technology.

In fact, I was able to solve this with Mathematica, a computer-algebra system (CAS) by Wolfram Research.

Using Mathematica‘s “Solve” command, I was able to find that, for any given payment time $y$ , the values of $X$ and $Y$ that will cause $h(.1)=h'(.1)=0$ are given by the following expressions:

$X\approx 1137.56-\frac{3186.6}{y}$ and $Y\approx \frac{3186.6\cdot 1.1^{y}}{y}.$

Based on these equations, I was then able to solve $h''(.1)=0$ on Mathematica for $y$ to get $y\approx 3.57207.$

This ultimately ends up implying that $h''(.1)>0$ for $y>3.57207.$ In other words, at an interest rate of 10%, for such values of $y$ , Redington immunization can be achieved!

For example, if $y=3.6$ years, then $X\approx 252.39$ and $Y\approx 1247.49.$ If $y=5$ years, then $X\approx 500.24$ and $Y\approx 1026.41.$

The animation below shows how the graph of $h(i)$ changes as $y$ increases from $y=3$ to $y=4.$ Note how the graph goes from being concave down to concave up at $i=i_{0}=.1$ when $y$ is just past the middle of the slider (just past $y\approx 3.57207$ ).

The graph of $h(i)=P_{A}(i)-P_{L}(i)$ near $i=i_{0}=0.1$ as the payment time $y$ for payment amount $Y$ increases from 3 to 4. When $y>3.57207$ , the graph is concave up at the red dot and Redington immunization can be achieved.

The Mathematica code for this animation is in the figure below.

I hope it’s clear that we could consider other solutions to the goal of achieving Redington immunization. For example, we could also consider changing the time of payment of the amount $X.$