**This section of sample problems and solutions is a part of** **The Actuary’s Free Study Guide for Exam 4 / Exam C****, authored by Mr. Stolyarov. This is Section 18 of the Study Guide. See an index of all sections by following the link in this paragraph.**

“A **k-component spliced distribution** has a density function that can be expressed as follows:

fX(x) = a1f1(x), c01;

fX(x) = a2f2(x), c12;

…

fX(x) = akfk(x), ck-1k.

For j = 1, …, k, each aj > 0 and each fj(x) must be a legitimate density function with all probability on the interval (cj-1, cj). Also, a1 + … + ak = 1″ (Klugman , Panjer, and Willmot 2008, p. 69).

**Source: **

*Loss Models: From Data to Decisions,*(Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 5, pp. 69-70.

**Original Problems and Solutions from The Actuary’s Free Study Guide**

**Problem S4C18-1.** A spliced distribution consists of a uniform distribution on the interval 0 ≤ x X(x), the pdf of this spliced distribution.

**Solution S4C18-1.** We use the formula

fX(x) = a1f1(x), c01;

fX(x) = a2f2(x), c12.

The fact that F(10) = 0.5 indicates to us that the first uniform distribution’s pdf needs to be multiplied by 0.5, as it applies to only the first 50% of possible values. Thus, a1 = 0.5.

Since a1 + a2 = 1, it follows that a2 = 0.5. Now we simply need to find the pdfs of the original specified uniform distributes and multiply them by the appropriate factors a1 and a2. Since the first uniform distribution is over the interval 0 ≤ x 1(x), is 1/10. Since the first uniform distribution is over the interval 10 ≤ x 2(x), is 1/30. Thus, the following is our formula for fX(x):

**fX(x) = 1/20,** **0 ≤ x ;**

**fX(x) = 1/60,** **10 ≤ x .**

**Problem S4C18-2.** The survival function of an exponential distribution, SX(x) = e-x/θ, and the corresponding probability density function (pdf), is fX(x) = e-x/θ/θ. Derive the formula for the pdf of an exponential distribution truncated on the right at c. (That is, an exponential distribution for which the entire domain of possible values is the interval from 0 to c.)

**Solution S4C18-2.** For a pdf, it must always be the case that ab∫fX(x)dx = 1, for domain endpoints a and b. Since the regular exponential pdf is e-x/θ/θ, we posit that, for the truncated distribution, the pdf is scaled by some factor k. We need to find k. We set up the following integral and equation:

0c∫k(e-x/θ/θ)dx = 1 →

-ke-x/θ│0c = 1 →

k-ke-c/θ = 1 →

1 – e-c/θ = 1/k →

k = 1/(1 – e-c/θ).

Thus, for an exponential distribution truncated on the right at c, **fX(x) = e-x/θ/((1 – e-c/θ)*θ)**.

Formulas for pdfs of truncated distributions are not given in the Exam 4/C Tables, and so you may need to be able to derive them yourself on the exam in order to use certain truncated distributions as components of spliced distributions. This problem was intended to give you some practice with this skill.

**Problem S4C18-3.** The formula for the pdf of an exponential distribution truncated at some value c is fY(y) = e-y/θ/((1 – e-c/θ)*θ). (We derived this formula in Problem S4C18-2.) A certain spliced distribution contains two components:

1. An exponential distribution on the interval 0 ≤ x

2. A uniform distribution on the interval 15 ≤ x ≤ 95.

It is known that SX(15) = 0.7.

Find the pdf, fX(x), of this distribution.

**Solution S4C18-3.** We use the formula

fX(x) = a1f1(x), c01;

fX(x) = a2f2(x), c12.

From the given information, we can determine f1(x) = e-x/θ/((1 – e-c/θ)*θ) = e-x/5/((1 – e-15/5)*5) =

f1(x) = e-x/5/((1 – e-3)*5).

We can also determine f2(x) = 1/(95-15) = f2(x) = 1/80.

Since S(15) = 0.7, 70% of all possible values occur over the domain of the uniform distribution. Thus, a2 = 0.7, and, by implication, a1 = 1 – 0.7 = 0.3.

Hence, we have

**fX(x) = 3****e-x/5/(50(1 – e-3))****,** **0 ≤ x ;**

**fX(x) = 7/800,** **15 ≤ x ≤ 95**.

**Problem S4C18-4.** The survival function of an exponential distribution, SX(x) = e-x/θ, and the corresponding probability density function (pdf), is fX(x) = e-x/θ/θ. Derive the formula for the pdf of an exponential distribution truncated on the left at c. (That is, an exponential distribution for which the entire domain of possible values is the interval from c to ∞.)

**Solution S4C18-4.** For a pdf, it must always be the case that ab∫fX(x)dx = 1, for domain endpoints a and b. Since the regular exponential pdf is e-x/θ/θ, we posit that, for the truncated distribution, the pdf is scaled by some factor k. We need to find k. We set up the following integral and equation:

c∞∫k(e-x/θ/θ)dx = 1 →

-ke-x/θ│c∞ = 1 →

ke-c/θ = 1 →

k = ec/θ.

Thus, for an exponential distribution truncated on the left at c, fX(x) = ec/θe-x/θ/θ =

**fX(x) = e(c-x)/θ/θ**.

**Problem S4C18-5.** A certain spliced distribution contains two components:

1. An exponential distribution on the interval 0 ≤ x

This exponential distribution has mean θ1 = 50.

2. An exponential distribution on the interval 100 ≤ x.

This exponential distribution has mean θ2 = 200.

It is known that FX(100) = 0.4.

Find the pdf, fX(x), of this distribution.

**Solution S4C18-5.** We use the formula

fX(x) = a1f1(x), c01;

fX(x) = a2f2(x), c12.

We use the formulas obtained from Solution S4C18-2 and Solution S4C18-4 regarding the pdfs of exponential distributions truncated on the right and on the left, respectively:

From the given information, we can determine f1(x) = e-x/θ_1/((1 – e-c/θ_1)*θ1) =

f1(x) = e-x/50/((1 – e-100/50)*50) = f1(x) = e-x/50/((1 – e-2)*50).

From the given information, we can determine f2(x) = e(c-x)/θ_2/θ2 = f2(x) = e(100-x)/200/200.

Since FX(100) = 0.4, we know that 0.4 of the possible values of x apply to the first distribution, so a1 = 0.4, and, by implication, a2 = 1 – 0.4 = 0.6. Hence, we have

**fX(x) =** **e-x/50/(125(1 – e-2))****,** **0 ≤ x ;**

**fX(x) = 3****e(100-x)/200/1000****,** **100 ≤ x**.

**See other sections of** **The Actuary’s Free Study Guide for Exam 4 / Exam C****.**