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An insurance agent sells a policy that has a \(100 deductible

and a \)5000 cap. When the policyholder files a claim, the policyholder must pay the first \(100. After the first \)100, the insurance company pays therest of the claim up to a maximum payment of $5000. Any

excess must be paid by the policyholder. Suppose that thedollar amount X of a claim has a continuous distribution

with p.d.f. \({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\frac{{\bf{1}}}{{{{\left( {{\bf{1 + x}}} \right)}^{\bf{2}}}}}\) for x>0 and 0 otherwise.

LetY be the insurance company's amount to payon the claim.

a. Write Y as a function of X, i.e., \({\bf{Y = r}}\left( {\bf{X}} \right).\)

b. Find the c.d.f. of Y.

c. Explain why Y has neither a continuous nor a discretedistribution.

Short Answer

Expert verified

(a) \(\begin{aligned}r\left( x \right) &= 0\;for\;x \le 100,\\ &= x - 100\;for\;100 \le x \le 5100,\\ &= 5000\;for\;x > 5100;\;\end{aligned}\)

(b) \(\begin{aligned}G\left( y \right) &= 0\;for\;y < 0,\\ &= 1 - \frac{1}{{\left( {y + 101} \right)}}\;for\;0 \le y \le 5000,\\ &= 1\;for\;y \ge 5000.\end{aligned}\)

(c) Y is an example of a mixed distribution that is neither discrete nor continuous.

Step by step solution

01

Given information

An insurance agent pays a deductible of $100 first. The claim reimbursement has a ceiling of $5000.

The claim is represented by a random variable X, following p.d.f.

\(\begin{aligned}f\left( x \right) &= \frac{1}{{{{\left( {1 + x} \right)}^2}}},x > 0\\ &= 0,otherwise\end{aligned}\)

02

(a) Define new variable Y and find its p.d.f

  • Y denotes the random variable that the insurance company has to reimburse.
  • X denotes the random variable that denotes the claim.
  • The claim has a minimum deductible of $100, therefore, \(X \ge 100\). Therefore, Y will be 0 whenever X<100.
  • When a claim (X)is greater than $100 and less than $5100 (cap+ deductible), then the reimbursement (Y) will be X-100 (Since reimbursement doesn't include the deductible.)
  • When a claim (X) is greater than $5100 (cap+ deductible), then the reimbursement (Y) will be $5000 (Since the reimbursement ceiling value is only $5000.)
  • Therefore, the p.d.f of Y is

\(\begin{aligned}r\left( x \right) &= 0\;for\;x \le 100,\\ &= x - 100\;for\;100 \le x \le 5100,\\ &= 5000\;for\;x > 5100.\;\end{aligned}\)

  • Therefore, the range of y lies from [0,5000]
03

(b) Find the c.d.f of Y

Let us define the cumulative distribution function of y by G(y) as follows:

\(\begin{aligned}G\left( y \right) &= 0\;for\;y < 0,\\ &= 1 - \frac{1}{{\left( {y + 101} \right)}}\;for\;0 \le y \le 5000,\\ &= 1\;for\;y \ge 5000.\end{aligned}\)

04

(c) Reasoning why Y has neither a continuous nor a discrete distribution

From the p.d.f of Y, it is visible that the range of the p.d.f takes mass values at the point \(x \le 100\) where the mass point is 0 and at \(x > 5100\) where the mass point 5000. Therefore, these are discrete points.

The p.d.f. in the interval \(100 \le x \le 5100\) lies in a range and has a continuous distribution over the interval. Hence this is a continuous interval.

Therefore, this distribution is called a mixed distribution.

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Most popular questions from this chapter

Suppose that an electronic system comprises four components, and let\({X_j}\)denote the time until component j fails to operate (j = 1, 2, 3, 4). Suppose that\({X_1},{X_2},{X_3}\)and\({X_4}\)are i.i.d. random variables, each of which has a continuous distribution with c.d.f.\(F\left( x \right)\)Suppose that the system will operate as long as both component 1 and at least one of the other three components operate. Determine the c.d.f. of the time until the system fails to operate.

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Suppose that three random variables X1, X2, and X3 have a continuous joint distribution with the following joint p.d.f.:

\({\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{c}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{ + 2}}{{\bf{x}}_{\bf{2}}}{\bf{ + 3}}{{\bf{x}}_{\bf{3}}}} \right)}&{{\bf{for0}} \le {{\bf{x}}_{\bf{i}}} \le {\bf{1}}\,\,\left( {{\bf{i = 1,2,3}}} \right)}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.\)

Determine\(\left( {\bf{a}} \right)\)the value of the constant c;

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\(\left( {\bf{c}} \right)\)\({\bf{Pr}}\left( {{{\bf{X}}_{\bf{3}}}{\bf{ < }}\frac{{\bf{1}}}{{\bf{2}}}\left| {{{\bf{X}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}} \right.} \right){\bf{.}}\)

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