New Video: Immunization, Part 2

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In the previous post, “New Video: Immunization, Part 1“, I introduced the idea of Redington immunization of liability cashflows by asset cashflows.

It was a post that delved into the details of my previous YouTube video “Actuarial Exam 2/FM Prep: Use Redington Immunization to Find a Ratio of Assets in a Portfolio” embedded below.

Actuarial Exam 2/FM Prep: Use Redington Immunization to Find a Ratio of Assets in a Portfolio. Financial Math for Actuarial Exam 2 (FM), Video #169. October 2018 SOA Sample Exam, Problem #73

Using advanced financial terminology, we can describe the problem in this video as one where the goal is to try to “match” the asset and liability cashflows, relative to a certain interest rate (assuming a flat yield curve), by equating their present values and durations at that interest rate, and by creating a certain kind of mismatch in their convexities at that interest rate.

We can be more precise by using mathematical notation: for an arbitrary effective periodic interest rate i, let P_{A}(i) represent the present value of the asset cashflow and let P_{L}(i) represent the present value of the liability cashflow. If the given interest rate of the problem is i=i_{0}, then: 1) to match present values entails setting P_{A}(i_{0})=P_{L}(i_{0}) and 2) to match durations entails both condition (1) to be true and also setting P_{A}'(i_{0})=P_{L}'(i_{0}). In the video above, these two conditions allow us to solve for unknown asset cashflows when i_{0}=.04=4\% and then form their ratio, which is the answer to the problem that is posed.

The desired mismatch in the convexities entails: 3) to make sure conditions (1) and (2) hold, in addition to setting P_{A}''(i_{0})>P_{L}''(i_{0}). It is not necessary to use this condition to solve the problem in the video, but it is a condition that guarantees that Redington immunization has been achieved: the assets will be in a surplus position compared to the liabilities of the interest rate changes a little bit away from i=i_{0}. It can be checked in the problem from the video that this condition holds after the unknowns have been solved for.

To see why condition (3) is used, apply the Second Derivative Test to the difference function h(i)=P_{A}(i)-P_{L}(i). Since h''(i_{0})=P_{A}''(i_{0})-P_{L}''(i_{0})>0, it guarantees that, for small changes in the value of i away from i_{0}, the present value of the asset cashflow will remain in a surplus position compared to the present value of the liability cashflow. That is, h(i)>0\Longleftrightarrow P_{A}(i)>P_{L}(i) for all i sufficiently close to i_{0} (basically because the graph of h will be concave up near i=i_{0}, at least when h'' is a continuous function).

Based on the solution to the problem in the video above, the relevant graphs of these functions are shown below.

The graphs of P_{A}(i)\approx 48.26(1+i)^{-2}+47.63(1+i)^{-9} (red), P_{L}(i)=95(1+i)^{-5} (blue), and the difference h(i)=P_{A}(i)-P_{L}(i) (pink) from the video above. The black dot is at the interest rate i_{0}=.04=4\%.

Assuming interest rates do not change much and our cashflows do not change, this should also imply that we will be in a surplus position of assets over liabilities for future values as well. If interest rates do change, then we may have to alter how we structure these cashflows to remain in a surplus position.

A Similar Problem Solved Using a Different Time-Valuation Point

In the latest video, which is the “new video” from the title of this post and is embedded below, I solve a similar problem. When solving this problem, it is more convenient to use a time other than the present to evaluate the time-value of both the asset and liability cashflows.

Actuarial Exam 2/FM: Full Immunization Relative to a Nonzero Time to Find a Ratio of Amount to Time. Financial Math for Actuarial Exam 2 (FM), Video #170. October 2018 SOA Sample Exam, Problem #72.

Instead of evaluating the present values of the asset and liability cashflows, the cashflows are time-valuated at a positive moment in time T>0 (which is T=8 in the video), where i=i_{0} is the interest rate of valuation (i_{0}=.03 in the video).

The general method is similar. If V_{A}(i) is the time-value of the asset cashflow at time T and V_{L}(i) is the time-value of the liability cashflow at time T, then we want: 1) V_{A}(i_{0})=V_{L}(i_{0}), 2) V_{A}'(i_{0})=V_{L}'(i_{0}), and 3) V_{A}''(i_{0})>V_{L}''(i_{0}). Once again, only the first two conditions are needed to solve the problem in this new video by using them to find the unknowns and then computing their ratio.

If we confirm that the third condition holds, this is sufficient to achieve Redington immunization of the liabilities by the assets. In fact, the problem statement from the new video calls for us to achieve “full” immunization of the liabilities by the assets. This actually means we desire more: we want V_{A}(i)\geq V_{L}(i) for all i>-1 (or at least i>0). In the video, we see that this occurs because, in fact, V_{A}''(i)>V_{L}''(i) for all i>-1.

We can again think of this in terms of a difference function: let g(i)=V_{A}(i)-V_{L}(i). Then our conditions for Redington immunization become: 1) g(i_{0})=0, 2) g'(i_{0})=0, and 3) g''(i_{0})>0, and we can apply the Second Derivative Test. If, in addition, g''(i)>0 for all i>-1, then the graph of g is always concave up. Thus, g(i)\geq 0 for all i>-1 and we will actually have achieved full immunization. It does not matter how much the interest rate changes; the assets will always be in a surplus position compared to the liabilities.

Why Does This Produce the Same Answer?

Now we explore why we can time-valuate at T>0 and still get the same answer as when we time-valuate in the present t=0.

Suppose that, for t=1,2,\ldots,n, the symbols A_{t} and L_{t} represent, respectively, the asset and liability cashflows at time t. Many of these values could be zero, but they are all taken to be nonnegative. Then C_{t}=A_{t}-L_{t} represents the net cashflow at time t, which can be positive, negative, or zero.

Given an arbitrary interest rate i>-1, the present value functions are P_{A}(i)=\displaystyle\sum_{t=1}^{n}A_{t}(1+i)^{-t} and P_{L}(i)=\displaystyle\sum_{t=1}^{n}L_{t}(1+i)^{-t}. By properties of summations, we can also write the difference function as h(i)=P_{C}(i)=P_{A}(i)-P_{L}(i)=\displaystyle\sum_{t=1}^{n}C_{t}(1+i)^{-t}.

The condition h(i_{0})=0 is equivalent to \displaystyle\sum_{t=1}^{n}C_{t}(1+i_{0})^{-t}=0 and \displaystyle\sum_{t=1}^{n}A_{t}(1+i_{0})^{-t}=\displaystyle\sum_{t=1}^{n}L_{t}(1+i_{0})^{-t}.

Since h'(i)=-\displaystyle\sum_{t=1}^{n}tC_{t}(1+i)^{-t-1}=-(1+i)^{-1}\displaystyle\sum_{t=1}^{n}tC_{t}(1+i)^{-t}, the condition h'(i_{0})=0 is equivalent to \displaystyle\sum_{t=1}^{n}tC_{t}(1+i_{0})^{-t}=0 and \displaystyle\sum_{t=1}^{n}tA_{t}(1+i_{0})^{-t}=\displaystyle\sum_{t=1}^{n}tL_{t}(1+i_{0})^{-t}.

And since h''(i)=\displaystyle\sum_{t=1}^{n}t(t+1)C_{t}(1+i)^{-t-2}=(1+i)^{-2}\displaystyle\sum_{t=1}^{n}t(t+1)C_{t}(1+i)^{-t}, the condition h''(i)>0, whether i=i_{0} or not, is equivalent to \displaystyle\sum_{t=1}^{n}t(t+1)C_{t}(1+i)^{-t}>0 and \displaystyle\sum_{t=1}^{n}t(t+1)A_{t}(1+i)^{-t}>\displaystyle\sum_{t=1}^{n}t(t+1)L_{t}(1+i)^{-t}.

Furthermore, since \displaystyle\sum_{t=1}^{n}t(t+1)C_{t}(1+i)^{-t}=\displaystyle\sum_{t=1}^{n}t^{2}C_{t}(1+i)^{-t}+\displaystyle\sum_{t=1}^{n}tC_{t}(1+i)^{-t}, we can also say that when h'(i_{0})=0, the condition h''(i_{0})>0 is equivalent to \displaystyle\sum_{t=1}^{n}t^{2}A_{t}(1+i_{0})^{-t}>\displaystyle\sum_{t=1}^{n}t^{2}L_{t}(1+i_{0})^{-t}.

Now consider time-valuation of the assets and liabilities at some other time T>0. The corresponding time-valuation functions would be V_{A}(i)=\displaystyle\sum_{t=1}^{n}A_{t}(1+i)^{T-t}=(1+i)^{T}\displaystyle\sum_{t=1}^{n}A_{t}(1+i)^{-t}=(1+i)^{T}P_{A}(i) and V_{L}(i)=\displaystyle\sum_{t=1}^{n}L_{t}(1+i)^{T-t}=(1+i)^{T}\displaystyle\sum_{t=1}^{n}L_{t}(1+i)^{-t}=(1+i)^{T}P_{L}(i). By properties of summations, we can also write the difference function as g(i)=V_{C}(i)=V_{A}(i)-V_{L}(i)=(1+i)^{T}\displaystyle\sum_{t=1}^{n}C_{t}(1+i)^{-t}=(1+i)^{T}h(i).

We could have guessed these relationships from the start, but it is nice to seem them work out by using properties of exponents and summations.

The first thing we see from the relationship g(i)=(1+i)^{T}h(i) is that g'(i_{0})=0 if and only if h'(i_{0})=0.

By the Product Rule, g'(i)=T(1+i)^{T-1}h(i)+(1+i)^{T}h'(i). Therefore, g'(i_{0})=0 when both h(i_{0})=0 and h'(i_{0})=0.

Is the reverse statement also true? That is, is h'(i_{0})=0 when both g(i_{0})=0 and g'(i_{0})=0? Yes. Just apply the Product Rule to the equivalent (and symmetric) relationship h(i)=(1+i)^{-T}g(i).

How about the second derivative? We compute, by linearity and the Product Rule:

g''(i)=T(T-1)(1+i)^{T-2}h(i)+T(1+i)^{T-1}h'(i) +T(1+i)^{T-1}h'(i)+(1+i)^{T}h''(i) =T(T-1)(1+i)^{T-2}h(i)+2T(1+i)^{T-1}h'(i)+(1+i)^{T}h''(i).

Therefore, if h(i_{0})=0, h'(i_{0})=0, and h''(i_{0})>0, then g''(i_{0})>0 as well. Once again, it works to switch around the roles of g and h in this last sentence because the same calculations could be performed on h(i)=(1+i)^{-T}g(i) (it does not matter that -T<0).

Furthermore, it can be seen that the same summation conditions as shown above for (Redington) immunization still hold. That is, 1) g(i_{0})=0 if and only if \displaystyle\sum_{t=1}^{n}A_{t}(1+i_{0})^{-t}=\displaystyle\sum_{t=1}^{n}L_{t}(1+i_{0})^{-t}, 2) g'(i_{0})=0 if and only if \displaystyle\sum_{t=1}^{n}tA_{t}(1+i_{0})^{-t}=\displaystyle\sum_{t=1}^{n}tL_{t}(1+i_{0})^{-t}, and 3) g''(i_{0})>0 if and only if \displaystyle\sum_{t=1}^{n}t^{2}A_{t}(1+i_{0})^{-t}>\displaystyle\sum_{t=1}^{n}t^{2}L_{t}(1+i_{0})^{-t}.

The graphs of both h(i) and g(i) for the new video are shown below. Note that they both are concave up at their common minimum point (i_{0},0)=(.03,0). It is also interesting to note that even though g has a graph that is always concave up, h does not. This emphasize the importance of the fact that we focused on the value i_{0}, rather than an arbitrary value of i, in the previous paragraph.

The graphs of h(i)\approx 5(1+i)^{-5}+7.04(1+i)^{-10.51}-12(1+i)^{-8} (red) and g(i)=(1+i)^{8}h(i)\approx 5(1+i)^{3}+7.04(1+i)^{-2.51}-12 (blue) from the new video above. The black dot is at the interest rate i_{0}=.03=3\%.

All that we have done here is also described on pages 409-413 of the 7th Edition of Mathematics of Investment and Credit, by Samuel A. Broverman.