New Video: Immunization, Part 1 - Infinity is Really Big

It is very important for any business, including any insurance company, to be able to meet their financial liability obligations.

A major class of liabilities that insurance companies must plan for are the claims from their customers. For example, if a person buys car insurance, then the insurance company would be obligated to pay for, among other things, some or all of the damages to that person’s car if they get in an accident.

A complicating factor for such claims is their random nature. For any given driver, the time of an accident and the severity of that accident will be unknown. Fortunately, these quantities can be modeled using probability, and the corresponding risks can be shared when there are enough customers who are paying sufficiently high premiums.

How does an insurance company, or any company, meet their liabilities? The liabilities can be covered with assets — either cash on hand or assets that can be quickly converted to cash (“liquid assets“).

Ideally, in planning for future liabilities, a company would plan for asset cashflows that will “exactly match” the liabilities (be equal them in financial value at the appropriate times). In practice, this is usually not possible to do. We can, however, still “match” overall asset and liability cashflows over time with respect to a given interest rate (to keep things simple, we are taking the yield curve to be flat). Basically this means matching their present values and their durations, with respect to that given interest rate.

But there is something else to be aware of when doing this matching: if the interest rate changes, that will affect whether a given matching, as planned, will produce asset flows that can still cover the liability flows. Something needs to be done to guard against this possibility. The liabilities need to be “immunized” by the assets, both in terms of their amounts and in terms of their times of payment.

For a given set of liability cashflows and a given interest rate, when the asset cashflows are planned in a way that 1) matches present values, 2) matches durations, and 3) keeps the asset cashflows in a surplus position for the liability cashflows when the interest rate changes a bit, we say that the liability cashflows are Redington immunized by the asset cashflows.

Example from the Video

In the video below, titled “Actuarial Exam 2/FM Prep: Use Redington Immunization to Find a Ratio of Assets in a Portfolio”, I explain these ideas in the context of solving for two unknown asset cashflow values.

Financial Math for Actuarial Exam 2 (FM), Video #169. October 2018 SOA Sample Exam, Problem #73

The statement of the problem in this video is: “Trevor has assets at time 2 of A and at time 9 of B. He has a liability of 95,000 at time 5. Trevor has achieved Redington immunization in his portfolio using an annual effective interest rate of 4%. Calculate $\frac{A}{B}$ .”

Here is a summary of the solution presented in the video.

Solution: Start by letting $P_{A}(i)$ and $P_{L}(i)$ be the present values of the assets and the liabilities, respectively, in thousands, at an arbitrary effective periodic interest rate $i$ (the subscript “A” stands for “Assets” and is not the value of “A” in the problem statement). These are functions of $i$ and their formulas are, based on the given information, seen to be $P_{A}(i)=A(1+i)^{-2}+B(1+i)^{-9}$ and $P_{L}(i)=95(1+i)^{-5}$ .

As stated in the previous section, to start the immunization process, we want to match both present values and durations of these cashflows. This entails setting $P_{A}(.04)=P_{L}(.04)$ and $P_{A}'(.04)=P_{L}'(.04).$ Since $P_{A}'(i)=-2A(1+i)^{-3}-9B(1+i)^{-10}$ and $P_{L}'(.04)=-475(1+i)^{-6},$ this leads to the following (approximate) system of linear equations:

$\begin{cases}.92455621A+.70258674B=78.08307514 \\ -1.77799272A-6.08007752B=-375.3993997\end{cases}$

After some messy algebra, the solution of the system is seen to be approximately $(A,B)=(48.259802, 47.629956).$ The final answer to the problem is therefore approximately $\frac{A}{B}=1.0132.$

Is this a guarantee that Redington immunization has been achieved?

The answer, actually, is no. We also require that $P_{A}''(.04)>P_{L}''(.04).$

Why is this? Let $h(i)=P_{A}(i)-P_{L}(i)$ . This function represents the present value of the asset flow minus the present value of the liability flow. To be Redington immunized with respect to $i=.04,$ we would want $h(i)\geq 0$ for all $i$ that are (sufficiently) close to the value .04.

So far, what we have done guarantees that $h(.04)=P_{A}(.04)-P_{L}(.04)=0$ and $h'(.04)=P_{A}'(.04)-P_{L}'(.04)=0,$ when $A\approx 48.259802$ and $B\approx 47.629956.$ In other words, the function $h$ has both a root (a “zero”) and a critical point at $i=.04.$

If $h$ were to have a local minimum at $i=.04$ , this would guarantee $h(i)\geq 0$ for all values of $i$ near .04. If we could show that $h''(.04)>0$ , the Second Derivative Test would imply that we have achieved Redington immunization.

You should check on your own that, for the given values of $A$ and $B$ , we have $h''(.04)\approx 866.31>0.$ This portfolio is indeed Redington immunized at rate 4% = .04.

Generalizing the Problem, Mathematica Usage, and Functional-Thinking

Frequent readers of this blog know that I emphasize the importance of generalization, the usefulness of the computer-algebra system Mathematica, and in the benefits of what I call “functional-thinking”. In fact, Mathematica can be a very helpful tool both in analyzing more general situations and in facilitating functional-thinking.

In the problem at hand, one way it could be generalized it to say that asset amount $A$ is paid at time $a$ and asset amount $B$ is paid at time $c.$ Furthermore, the liability amount could be called $C$ and be paid at time $c.$

Defining the same functions as in the solution above, the resulting system of equations to solve, for an arbitrary (fixed) interest rate $i$ , is

$\begin{cases}A(1+i)^{-a}+B(1+i)^{-b}=C(1+i)^{-c} \\ -aA(1+i)^{-a-1}-bB(1+i)^{-b-1}=-cC(1+i)^{-c-1}\end{cases}$

Thinking of this as a system of linear equations in the unknowns $A$ and $B,$ we can use Mathematica‘s “Solve” command to quickly get the following values, when $a\not=b:$

$(A,B)=\left(\frac{(c-b)C(1+i)^{a-c}}{a-b},\frac{(a-c)C(1+i)^{b-c}}{a-b}\right)$

This leads to the following answer for the ratio we seek, when $a\not=b$ and $a\not=c$ :

$\frac{A}{B}=\frac{(c-b)(1+i)^{a-b}}{a-c}.$

This is mathematical power: we have solved infinitely many problems all at once!

But is this useful?

The answer depends on whether we have many real-life problems like this to solve or not, and whether there are important real-life implications of the final answer.

Let’s end this post by exploring this in the context of functional thinking.

By “functional-thinking”, I mean to consider the answer above as a (multivariable) function of the quantities involved. If the function name is $f$ and if we use $R$ to represent the ratio $\frac{A}{B}$ , we could write the answer as

$R=\frac{A}{B}=f(a,b,c,i)=\frac{(c-b)(1+i)^{a-b}}{a-c}.$

This function can now be mathematically analyzed. One big tool in such an analysis is multivariable calculus (i.e., partial derivatives). We can also just look at and think about graphs. Using Mathematica, we can, for example, make a graph of $R=f(2,b,5,i)$ , as a function of $i,$ for various values of $b.$ Here is what Mathematica produces as $b$ increases from 9 to 20:

The answer $R=\frac{A}{B}$ to the general problem as a function of $i,$ when $a=2, c=5,$ and $b$ increases from 9 to 20.

The answer $R=\frac{A}{B}$ to the general problem as a function of $i,$ when $a=2, c=5,$ and $b$ increases from 9 to 20.

We observe that, for any $b$ between 9 and 20, this graph is decreasing and concave up. We also see that the graph gets “more extreme” (steeper and with generally higher curvature) for larger values of $b$ .

The fact that it is decreasing (for any fixed value of $b$ ) means that, as the interest rate $i$ goes up, the value of $\frac{A}{B}$ goes down. This means that the relative size of cashflow $A$ to cashflow $B$ to achieve Redington immunization goes down as $i$ goes up.

The practical implication: for large interest rates, you don’t need as large (relative to $B$ ) of an initial asset cashflow $A!$

The concavity is harder to interpret, though it makes graphical sense — after all, the graph is decreasing and certainly cannot become negative, so it must be concave up. A financial interpretation is this: when $i$ is large, a given small increase in $i$ will produce a smaller decrease in $R=\frac{A}{B}$ than it does when $i$ is small.

For example, when $b=9,$ it can be shown that $\frac{\partial R}{\partial i}\bigr\rvert_{i=.05}\approx -6.3,$ while $\frac{\partial R}{\partial i}\bigr\rvert_{i=.1}\approx -4.4.$ Using linear approximation, we can then say that, when $i=.05$ and $\Delta i=.001,$ then $\Delta R\approx -6.3\cdot .001=-.0063.$ On the other hand, when $i=.1$ and $\Delta i=.001,$ then $\Delta R\approx -4.4\cdot .001=-.0044.$

As $b$ increases from 9 to 20, the most obvious change is that, for very low values of $i$ , the value of $R=\frac{A}{B}$ that solves the general problem goes up from approximately the range 1 to 1.4 to approximately the range 4 to 5. This means that, when $i$ is very low, cashflow $A$ must get significantly larger, relative to cashflow $B$ , as the time of payment $b$ increases.

On the other hand, when $i$ is very large, we see from the animation that $R=\frac{A}{B}$ goes down in value as $b$ gets closer to 20.

I think is all pretty interesting stuff, especially when we work to relate it to real life. And I hope you are inspired to work hard at thinking this way and improving your skills. May God bless you in your endeavors.

For those who are interested, here is a screenshot of the Mathematica code to produce the animation.