Tutorials › Exam ALTAM - Advanced Long-Term Actuarial Mathematics › Small Decrement Approximations

Home  /  Tutorials  /  Exam ALTAM / Associated Single-Decrement Tables

Exam ALTAM · Associated Single-Decrement Tables · Tutorial 254 of 999

Small Decrement Approximations

Learn why associated and observed probabilities differ only slightly when total annual decrements are small, and quantify the difference using a Taylor approximation and error bounds.

Intermediate 10 min read

What You'll Learn

  • Relate an observed cause-specific probability to its associated single-decrement probability under constant forces.
  • Express the associated probability in terms of the total decrement probability and cause share.
  • Derive the leading approximation to the difference between associated and observed probabilities.
  • Bound the error left by the quadratic approximation using Taylor’s theorem.
  • Recognize when the approximation is unsuitable and avoid mixing constant-force and UDD conversions.

Why Small Decrements Make Associated Probabilities Close

An observed decrement probability reflects a cohort exposed to competing causes: an exit from one cause prevents a later exit from another. An associated single-decrement probability instead describes the probability of decrement from one selected cause when the competing causes are removed, using a specified assumption for the decrements within the year. These probabilities are not generally equal. When the total probability of decrement in the year is small, however, their difference is small as well.

This tutorial quantifies that statement under the constant-force assumption. The approximation is useful for understanding the size and direction of the conversion, but it does not replace the exact conversion when that is needed. In particular, the formula below is not a UDD conversion.

Definition: For one year and a set of competing decrement causes, let \(q_j\) be the observed probability of decrement from cause \(j\), and let \(Q=\sum_k q_k\) be the total observed decrement probability. For \(Q>0\), define the observed cause share \(r=q_j/Q\). The associated single-decrement probability for cause \(j\), denoted here by \(q'_j\), is the probability of decrement from cause \(j\) in the corresponding single-decrement model after all competing causes are removed.

Assume each cause has a constant force during the year. Write \(\mu_k\) for the force of cause \(k\) and \(\mu=\sum_k\mu_k\) for the total force. The probability of no decrement from any cause during the year is \(1-Q=e^{-\mu}\). The share of total force due to cause \(j\) is \(\mu_j/\mu\), and under constant forces this equals the observed cause share \(q_j/Q\). Removing the other causes leaves force \(\mu_j\), so

$$ q'_j=1-e^{-\mu_j} =1-(1-Q)^r. $$

This exact formula requires constant cause-specific forces over the year and \(0\leq Q<1\). If \(Q=0\), every observed cause probability is zero and there is no nonzero cause share to convert. If a different within-year assumption is used, the exact relationship can differ.

The Small-\(Q\) Approximation

To see why the probabilities are close, expand \(1-(1-Q)^r\) about \(Q=0\). The linear term is \(rQ=q_j\), exactly the observed probability for cause \(j\). The next term gives the leading increase in the associated probability.

$$ q'_j = rQ+\frac{r(1-r)}{2}Q^2+R, \qquad q'_j-q_j \approx \frac{r(1-r)}{2}Q^2. $$

Thus the associated and observed probabilities agree to first order in the total probability \(Q\); their difference is of second order. This is the precise sense in which they are close when annual decrements are small. The correction is nonnegative because \(0\leq r\leq1\). If the selected cause is the only possible cause, \(r=1\), the correction and the exact difference are both zero.

Why the Expansion and Error Bound Hold

Set \(f(Q)=1-(1-Q)^r\). Its first three derivatives are

$$ f'(Q)=r(1-Q)^{r-1},\qquad f''(Q)=r(1-r)(1-Q)^{r-2}, $$ $$ f'''(Q)=r(1-r)(2-r)(1-Q)^{r-3}. $$

At \(Q=0\), \(f(0)=0\), \(f'(0)=r\), and \(f''(0)=r(1-r)\). Taylor’s theorem through the quadratic term therefore gives

$$ f(Q)=rQ+\frac{r(1-r)}{2}Q^2+R. $$

For \(0<r<1\), \(f'''(Q)\) is positive and increases over \(0\leq Q<1\), since \(r-3<0\). Taylor’s theorem with the Lagrange form of the remainder states that \(R=f'''(\xi)Q^3/6\) for some \(\xi\) between \(0\) and \(Q\). It follows that

$$ \frac{r(1-r)(2-r)}{6}Q^3 \leq R \leq \frac{r(1-r)(2-r)}{6}(1-Q)^{r-3}Q^3. $$

Here \(R\) is the error after retaining the quadratic term. It is also the amount by which the quadratic approximation understates the exact increase \(q'_j-q_j\). The bound becomes less useful as \(Q\) approaches 1, consistent with the small-decrement approximation being intended for small total probabilities.

Worked Example: Estimating the Increase for a Small Cause Probability

In a hypothetical one-year decrement model, the total observed decrement probability is \(Q=0.08\), and the observed probability for a selected cause is \(q_j=0.02\). Assume constant cause-specific forces throughout the year. Find the cause share, the exact associated probability, the quadratic approximation to the increase, and its error. No interest assumption is needed for this probability conversion.

Method. Use the exact constant-force formula \(q'_j=1-(1-Q)^r\), then compare it with the quadratic approximation. This avoids substituting a conversion formula based on a different within-year assumption.

Cause share. The share of observed decrements due to the selected cause is

$$ r=\frac{q_j}{Q}=\frac{0.02}{0.08}=0.25. $$

Exact associated probability. Substitution in the exact formula gives

$$ q'_j=1-(1-0.08)^{0.25} =1-0.92^{0.25} \approx 0.020629640. $$

Quadratic estimate of the increase. The observed probability is \(q_j=rQ=0.02\). The estimated increase above it is

$$ \frac{r(1-r)}{2}Q^2 =\frac{0.25(0.75)}{2}(0.08)^2 =0.09375(0.0064) =0.000600. $$

So the quadratic estimate of \(q'_j\) is \(0.02+0.000600=0.020600\). The exact increase is \(0.020629640-0.020000000=0.000629640\), so the estimate understates that increase by approximately \(0.000029640\), or \(0.000030\) to six decimal places. The exact associated probability exceeds the observed probability, as the positive correction predicts.

Worked Example: Bounding the Error in the Increase

A hypothetical population has total one-year decrement probability \(Q=0.10\), and one cause accounts for observed share \(r=0.40\). Assume constant cause-specific forces during the year. Find the quadratic approximation to the increase in the associated probability above the observed probability, and bound the error after that approximation.

Method. First calculate the second-order increase, then apply the Taylor remainder bounds. The remainder bounds are for the error after the quadratic term, not for the whole increase.

The observed probability for the cause is \(q_j=rQ=0.40(0.10)=0.040\). The quadratic estimate of the increase is

$$ \frac{r(1-r)}{2}Q^2 =\frac{0.40(0.60)}{2}(0.10)^2 =0.12(0.01) =0.001200. $$

The quadratic estimate of the associated probability is therefore \(0.040+0.001200=0.041200\). For the remainder, the coefficient in the lower bound is

$$ \frac{r(1-r)(2-r)}{6} =\frac{0.40(0.60)(1.60)}{6} =0.064. $$

The lower bound is \(0.064(0.10)^3=0.0000640\). Since \(r-3=-2.6\), the upper bound is

$$ 0.064(1-0.10)^{-2.6}(0.10)^3 =0.000064(0.9)^{-2.6} \approx 0.0000842. $$

Thus \(0.0000640\leq R\leq0.0000842\). The exact increase must be between \(0.0012640\) and \(0.0012842\), and the associated probability must be between \(0.0412640\) and \(0.0412842\). The bound confirms that the quadratic estimate is an understatement, with a remainder much smaller than the leading increase.

Worked Example: Comparing an Exact Conversion With the Approximation

For a hypothetical one-year model, the total observed decrement probability is \(Q=0.04\), and a selected cause has observed probability \(q_j=0.03\). Assume constant cause-specific forces during the year. Calculate the exact associated probability and compare it with the quadratic approximation. No interest rate or separate mortality assumption is involved.

Method. Find the cause share, use the exact constant-force formula, and then calculate the second-order correction. This comparison shows how the size of the remainder changes with the total decrement probability.

The cause share is

$$ r=\frac{0.03}{0.04}=0.75. $$

The exact associated probability is

$$ q'_j=1-(1-0.04)^{0.75} =1-0.96^{0.75} \approx 0.030152557. $$

The quadratic increase estimate is

$$ \frac{0.75(0.25)}{2}(0.04)^2 =0.09375(0.0016) =0.000150. $$

Therefore the quadratic approximation to the associated probability is \(0.03+0.000150=0.030150\). The exact increase is approximately \(0.030152557-0.030000000=0.000152557\), so the quadratic approximation understates the increase by about \(0.000002557\). The exact and approximate probabilities are close, and the sign of the error agrees with the positive Taylor remainder.

Common Mistakes and When to Use the Approximation

The approximation is most useful as a quick estimate or a reasonableness check when \(Q\) is small. For a required exact associated probability under constant forces, use \(1-(1-Q)^r\). The approximation depends on both the total probability and the cause share; a small observed probability \(q_j\) by itself does not establish that \(Q\) is small.

  • Using the wrong within-year assumption. The formula \(1-(1-Q)^r\) follows from constant cause-specific forces. A UDD-based conversion does not follow from this derivation.
  • Dropping the cause share. The approximation is \(\frac12 r(1-r)Q^2\), not a correction determined by \(Q\) alone.
  • Calling the quadratic increase the exact increase. It omits the positive remainder when \(0<r<1\) and \(Q>0\).
  • Confusing the remainder with the total increase. In the error bounds, \(R\) is the difference between the exact increase and its quadratic estimate.
  • Ignoring boundary cases. If \(r=0\), the selected cause has probability zero; if \(r=1\), there is no competing cause and \(q'_j=q_j=Q\).

For instance, in the first worked example an inappropriate shortcut is \(q_j/(1-Q/2)=0.02/0.96\approx0.0208333\). That value is not the constant-force associated probability: the constant-force calculation gives approximately \(0.0206296\). The correction is to identify the within-year assumption first, then apply its matching exact relationship before using an approximation.

Key takeaway: Under constant cause-specific forces, \(q'_j=1-(1-Q)^r\), where \(r=q_j/Q\). For small \(Q\), \(q'_j-q_j\) is approximately \(\frac12r(1-r)Q^2\), and the Taylor remainder is positive with the bounds derived above.

Check Your Understanding

For each problem, assume constant cause-specific forces over the stated one-year period. Use the exact formula where requested; round probabilities to four decimal places unless a more precise bound is requested.

  1. A hypothetical decrement model has total probability \(Q=0.06\), and one cause has observed probability \(q_j=0.03\). Find its cause share and the quadratic estimate of the associated probability. (Answer: \(r=0.50\); quadratic estimate \(0.03045\).)
  2. A cause has share \(r=0.20\) of a total probability \(Q=0.05\). Find the quadratic approximation to the increase and bound its remainder after the quadratic term. (Answer: increase \(0.000200\); \(0.0000060\leq R\leq0.0000069\).)
  3. The total decrement probability is \(Q=0.09\), and the observed probability for a selected cause is \(q_j=0.045\). Find the quadratic estimate of the associated probability. (Answer: \(0.0460125\).)
  4. A cause has observed share \(r=0.30\) when the total probability is \(Q=0.12\). Calculate the observed cause probability, the quadratic estimate of its associated probability, and bounds on the remainder. (Answer: observed probability \(0.036\); quadratic estimate \(0.037512\); approximately \(0.0001028\leq R\leq0.0001453\).)
  5. The total probability is \(Q=0.04\), and the selected cause probability is \(q_j=0.01\). Find the exact associated probability and compare it with the quadratic estimate. (Answer: exact probability approximately \(0.0101536\); quadratic estimate \(0.0101500\), an understatement of approximately \(0.0000036\).)

Previous

Next

Back to Tutorials
Clinical Biostats