2 Univariate Distributions

2.1 Supplementary Concepts

Random variables

A random variable is a function that assigns a real number to each outcome in a sample space. If \(\Omega\) is the sample space, then a random variable \(X\) is a function

\(\displaystyle X:\Omega\longrightarrow\mathbb{R}\)

A random variable is discrete if it takes values in a finite or countable set. It is continuous if its possible values form an interval or collection of intervals.

Probability mass functions and probability density functions

For a discrete random variable \(X\), the probability mass function, or pmf, is

\[p_X(x)=\Pr[X=x].\]

It satisfies

\[p_X(x)\geq 0\]

and

\[\sum_x p_X(x)=1.\]

For a continuous random variable \(X\), the probability density function, or pdf, is denoted by \(f_X(x)\) and satisfies

\(\displaystyle f_X(x)\geq 0\)

and

\(\displaystyle \int_{-\infty}^{\infty}f_X(x)\,dx=1\)

For a continuous random variable,

\(\displaystyle \boxed{\Pr[a<X\leq b]=\int_a^b f_X(x),dx}.\)

The value \(f_X(x)\) is a density, not a probability. In particular,

\(\displaystyle \Pr[X=x]=0\)

for every individual value \(x\) when \(X\) is continuous.

Cumulative distribution functions

The cumulative distribution function, or cdf, of a random variable \(X\) is

\(\displaystyle F_X(x)=\Pr[X\leq x]\)

Every cumulative distribution function satisfies

\(\displaystyle 0\leq F_X(x)\leq 1,\)

\(\displaystyle \lim_{x\to-\infty}F_X(x)=0,\)

and

\(\displaystyle \lim_{x\to\infty}F_X(x)=1.\)

It is also nondecreasing and right-continuous.

For a continuous random variable,

\(\displaystyle F_X(x)=\int_{-\infty}^{x}f_X(t),dt.\)

When \(F_X\) is differentiable,

\(\displaystyle f_X(x)=F_X'(x)\)

Calculating probabilities from a distribution

For any random variable,

\(\displaystyle \Pr[a<X\leq b]=F_X(b)-F_X(a)\)

For a discrete random variable,

\(\displaystyle \Pr[a\leq X\leq b]=\sum_{x=a}^{b}p_X(x),\)

where the sum is taken over possible values of \(X\).

For a continuous random variable, the inclusion or exclusion of endpoints does not affect the probability:

\(\displaystyle \Pr[a<X<b]=\Pr[a\leq X\leq b].\)

The survival function associated with \(X\) is

\(\displaystyle S_X(x)=\Pr[X>x]=1-F_X(x)\)

Thus,

\(\displaystyle \Pr[X>x]=1-F_X(x)\)

Conditional probabilities and conditional distributions

For events involving a random variable, conditional probability is calculated using

\(\displaystyle \Pr[A\mid B]=\frac{\Pr[A\cap B]}{\Pr[B]}\)

provided that \(\Pr[B]>0\).

For example, when \(a<b<c\),

\(\displaystyle \Pr[X>b\mid X>a]=\frac{\Pr[X>b]}{\Pr[X>a]}\)

because \({X>b}\subseteq{X>a}\).

Therefore,

\(\displaystyle \Pr[X>b\mid X>a]=\frac{1-F_X(b)}{1-F_X(a)}\)

More generally,

\(\displaystyle \boxed{\Pr[b<X\leq c\mid X>a] =\frac{F_X(c)-F_X(b)}{1-F_X(a)}}\)

when \(a\leq b<c\) and \(F_X(a)<1\).

For a continuous random variable conditioned on an event \(B\), the conditional density is

\(\displaystyle f_{X\mid B}(x)=\frac{f_X(x)}{\Pr[B]}\)

for values of \(x\) satisfying the event \(B\), and is zero otherwise.

Expected values and functions of random variables

The expected value of a discrete random variable is

\(\displaystyle \boxed{\operatorname{E}[X]=\sum_x x,p_X(x)}.\)

For a continuous random variable,

\(\displaystyle \boxed{\operatorname{E}[X]=\int_{-\infty}^{\infty}x f_X(x),dx}.\)

More generally, if \(g\) is a function of \(X\), then

\(\displaystyle \operatorname{E}[g(X)]=\sum_x g(x)p_X(x)\)

in the discrete case, and

\(\displaystyle \operatorname{E}[g(X)]=\int_{-\infty}^{\infty}g(x)f_X(x)\,dx\)

in the continuous case.

This result allows the expected value of an insurance payment to be calculated directly from the loss distribution without first deriving the complete distribution of the payment.

Linearity of expectation gives

\(\displaystyle \operatorname{E}[aX+b]=a\operatorname{E}[X]+b}\)

Moments

The \(k\)th raw moment of \(X\) is

\(\displaystyle \operatorname{E}[X^k]\)

The first raw moment is the mean:

\(\displaystyle \operatorname{E}[X].\)

The \(k\)th central moment is

\(\displaystyle \operatorname{E}\left[(X-\operatorname{E}[X])^k\right]\)

The second central moment is the variance.

For a nonnegative continuous random variable,

\(\displaystyle \operatorname{E}[X]=\int_0^\infty \Pr[X>x]\,dx\)

More generally,

\(\displaystyle \boxed{\operatorname{E}[X^k] =k\int_0^\infty x^{k-1}\Pr[X>x],dx}\)

provided that the moment exists.

These survival-function formulas are particularly useful for loss and payment random variables.

Mode, median, and percentiles

The mode is a value at which the probability mass function or probability density function is maximized.

For a discrete random variable, a mode \(m\) satisfies

\(\displaystyle p_X(m)\geq p_X(x)\)

for every possible value \(x\).

For a continuous random variable, a mode \(m\) satisfies

\(\displaystyle f_X(m)\geq f_X(x)\)

for all relevant values of \(x\).

A median is a value \(m\) satisfying

\(\displaystyle \Pr[X\leq m]\geq 0.5\)

and

\(\displaystyle \Pr[X\geq m]\geq 0.5.\)

For a continuous distribution with a strictly increasing cdf, the median satisfies

\(\displaystyle F_X(m)=0.5\)

The \(p\)th percentile, or quantile, is

\(\displaystyle x_p=F_X^{-1}(p)\)

and satisfies

\(\displaystyle F_X(x_p)=p.\)

For example, the \(95\)th percentile is the loss amount that is not exceeded with probability \(0.95\).

Variance, standard deviation, and coefficient of variation

The variance of \(X\) measures dispersion around its expected value:

\(\displaystyle \operatorname{Var}(X)=\operatorname{E}\left[(X-\operatorname{E}[X])^2\right]\)

The computational form is

\(\displaystyle \boxed{\operatorname{Var}(X) =\operatorname{E}[X^2]-\left(\operatorname{E}[X]\right)^2}.\)

The standard deviation is

\(\displaystyle \operatorname{SD}(X)=\sqrt{\operatorname{Var}(X)}\)

For constants \(a\) and \(b\),

\(\displaystyle \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)\)

and

\(\displaystyle \operatorname{SD}(aX+b)=|a|\operatorname{SD}(X)\)

When \(\operatorname{E}[X]>0\), the coefficient of variation is

\(\displaystyle \boxed{\operatorname{CV}(X) =\frac{\operatorname{SD}(X)}{\operatorname{E}[X]}}.\)

The coefficient of variation measures variability relative to the mean and is dimensionless.

Inflation and other linear adjustments

Suppose \(X\) is a loss measured at a base cost level and losses increase by an inflation rate \(i\).

The inflated loss is

\(\displaystyle X_I=(1+i)X\)

Its expected value is

\(\displaystyle \operatorname{E}[X_I]=(1+i)\operatorname{E}[X]\)

Its variance is

\(\displaystyle \operatorname{Var}(X_I)=(1+i)^2\operatorname{Var}(X)\)

Its standard deviation is

\(\displaystyle \operatorname{SD}(X_I)=(1+i)\operatorname{SD}(X)\)

when \(i>-1\).

The coefficient of variation is unchanged by positive multiplicative inflation:

\(\displaystyle \operatorname{CV}(X_I)=\operatorname{CV}(X)\)

Inflation should generally be applied to the loss before current policy deductibles and limits are applied, unless the contract states otherwise.

Ordinary deductibles

Let \(X\) be the loss amount and let \(d\) be an ordinary deductible. The insurer pays only the portion of the loss exceeding the deductible.

The payment random variable is

\(\displaystyle Y=(X-d)_+=\max(X-d,0)\)

Thus,

\[\displaystyle Y=\begin{cases} 0,& X \leq d,\ X-d, & X>d \end{cases}\]

The policyholder retains

\(\displaystyle \min(X,d).\)

The probability that the insurer makes a positive payment is

\(\displaystyle \Pr[Y>0]=\Pr[X>d]=1-F_X(d)\)

For \(y\geq 0\), the cdf of the payment is

\(\displaystyle F_Y(y)=F_X(d+y)\)

The expected payment is

\(\displaystyle \boxed{\operatorname{E}[Y] =\operatorname{E}[(X-d)_+] =\int_d^\infty \Pr[X>x],dx}.\)

Franchise deductibles

Under a franchise deductible, no payment is made when the loss does not exceed \(d\), but the full loss is paid when the loss exceeds \(d\).

The payment random variable is

\(\displaystyle Y=X\mathbf{1}_{{X>d}}\)

Equivalently,

\(\displaystyle Y= \begin{cases} 0, & X\leq d,\ X, & X>d. \end{cases}\)

The probability of a positive payment is

\(\displaystyle \Pr[Y>0]=\Pr[X>d].\)

The expected payment is

\(\displaystyle \operatorname{E}[Y]=\operatorname{E}\left[X\mathbf{1}_{{X>d}}\right]\)

For a continuous loss distribution,

\(\displaystyle \operatorname{E}[Y]=\int_d^\infty x f_X(x),dx.\)

A franchise deductible generally produces a larger insurer payment than an ordinary deductible with the same value of \(d\).

Policy limits and limited losses

Suppose the insurer will pay no more than a policy limit \(u\).

With no deductible, the payment is

\(\displaystyle Y=\min(X,u)\)

Equivalently,

\(\displaystyle Y= \begin{cases} X, & X<u,\ u, & X\geq u. \end{cases}\)

The expected limited loss is

\(\displaystyle \boxed{\operatorname{E}[\min(X,u)] =\int_0^u \Pr[X>x],dx}.\)

For \(0\leq y<u\),

\(\displaystyle \Pr[Y\leq y]=F_X(y).\)

There is generally a probability mass at the policy limit:

\(\displaystyle \Pr[Y=u]=\Pr[X\geq u]\)

Thus, even when \(X\) is continuous, the payment random variable may be a mixed random variable with both continuous and discrete components.

Deductibles, coinsurance, and benefit limits

Suppose the loss is first adjusted for inflation, then reduced by an ordinary deductible \(d\). The insurer pays a coinsurance proportion \(\alpha\), subject to a maximum benefit \(u\).

Under this convention, the payment is

\(\displaystyle Y=\min\left{\alpha\left[(1+i)X-d\right]_+,u\right}\)

The payment may be evaluated in stages:

\(\displaystyle X_I=(1+i)X,\)

\(\displaystyle Z=(X_I-d)_+,\)

\(\displaystyle W=\alpha Z,\)

and

\(\displaystyle Y=\min(W,u).\)

Thus,

\(\displaystyle Y= \begin{cases} 0, & (1+i)X\leq d,\ \alpha\left[(1+i)X-d\right], & d<(1+i)X<d+\dfrac{u}{\alpha},\ u, & (1+i)X\geq d+\dfrac{u}{\alpha}. \end{cases}\)

The order in which deductibles, coinsurance, inflation adjustments, and limits are applied must be determined from the policy language. Different contractual conventions can produce different payment amounts.

Distribution of an insurance payment

Let

\(\displaystyle Y=\min\left((X-d)_+,u\right),\)

where \(d\) is an ordinary deductible and \(u\) is the maximum payment above the deductible.

The payment is zero when \(X\leq d\):

\(\displaystyle \Pr[Y=0]=F_X(d)\)

For \(0\leq y<u\),

\(\displaystyle F_Y(y)=F_X(d+y)\)

The probability that the maximum benefit is paid is

\(\displaystyle \Pr[Y=u]=\Pr[X\geq d+u]=1-F_X(d+u^-)\)

For a continuous loss distribution, this becomes

\(\displaystyle \Pr[Y=u]=1-F_X(d+u).\)

The expected payment is

\(\displaystyle \boxed{\operatorname{E}[Y] =\int_d^{d+u}\Pr[X>x],dx}.\)

This payment distribution generally has a probability mass at zero, a continuous component between \(0\) and \(u\), and a probability mass at \(u\).

Expected value and variance of insurance payments

Let

\(\displaystyle Y=\min\left((X-d)_+,u\right).\)

The first moment is

\(\displaystyle \boxed{\operatorname{E}[Y] =\int_d^{d+u}S_X(x),dx},\)

where \(S_X(x)=\Pr[X>x]\).

The second moment is

\(\displaystyle \boxed{\operatorname{E}[Y^2] =2\int_0^u y\Pr[Y>y],dy}.\)

Since

\(\displaystyle \Pr[Y>y]=\Pr[X>d+y]\)

for \(0\leq y<u\),

\(\displaystyle \boxed{\operatorname{E}[Y^2] =2\int_d^{d+u}(x-d)S_X(x),dx}.\)

The variance is then

\(\displaystyle \boxed{\operatorname{Var}(Y) =\operatorname{E}[Y^2]-\left(\operatorname{E}[Y]\right)^2}.\)

The standard deviation is

\(\displaystyle \operatorname{SD}(Y)=\sqrt{\operatorname{Var}(Y)}\)

For the original loss random variable,

\(\displaystyle \operatorname{Var}(X) =\operatorname{E}[X^2]-\left(\operatorname{E}[X]\right)^2.\)

Comparing \(\operatorname{E}[X]\) with \(\operatorname{E}[Y]\) measures the expected portion of the loss transferred to the insurer. Comparing \(\operatorname{Var}(X)\) with \(\operatorname{Var}(Y)\) shows how the policy provisions alter the variability of the insurer’s payment.

2.2 Solved Exercises

Before opening the solutions, take a moment to work through each problem—-your mastery grows most when you engage actively with the mathematics.

Basic probabilities

Let \(S(x)\) be defined as follows:

\[s(x) = \left\{ \begin{array}{ll} 1-\dfrac{x}{250} & 0 \leq x < 40 \\ 1-{\left( \dfrac{x}{100} \right)}^2 & 40 \leq x \leq 100 \\ \end{array} \right. \] 1. Calculate the probability that \((25)\) dies within the next \(30\) years.
2. Calculate the probability that \((25)\) dies in her fortieth year of live.
3. Calculate the probability that \((25)\) does not die between ages \(35\) and \(45\)
4. Calculate the probability that \((25)\) celebrates her eightieth birthday.

Solution

The probability that \((25)\) dies within the next \(30\) years is given by

\({_{30}}q_{25}=\dfrac{s(25)-s(55)}{s(25)}=0.225\)

The probability \((25)\) dies in her fortieth year of live is given by

\({_{14|}}q_{25}={_{14}}p_{25}\;q_{39}=\dfrac{\cancel{s(39)}}{s(25)}\cdot\dfrac{s(39)-s(40)}{\cancel{s(39)}}=0.00\bar{4}\)

The probability that \((25)\) does not die between ages \(35\) and \(45\) is given by

\(1-{_{10|10}}q_{25}=1-{_{10}}p_{25}\;{_{10}}q_{35}=1-\dfrac{\cancel{s35}}{s(25)}\cdot\dfrac{s(35)-s(45)}{\cancel{s(35}}=0.930556\)

The probability that \((25)\) celebrates her eightieth birthday is given by

\(\dfrac{s(80)}{s(25)}=0.4\)

\(\\\)

2.3 Supplementary Exercises

Let \(X\) be a discrete random variable with the following probability mass function:

\[ \mathrm{Pr}[X=x] = \left\{ \begin{array}{ll} 0.500 & x=0 \\ 0.3\overline{33} & x=1 \\ 0.1\overline{66} & x=2 \\ \end{array} \right. \]

Find the range of \(X\)
Find \(\mathrm{Pr}[X \leq 1.5\,]\)
Find \(\mathrm{Pr}[0 < X < 2\,]\)
Find \(\mathrm{Pr}[X = 0\,|\, X < 2\,]\)

Two dice are tossed, and the observed numbers are \(X\) and \(Y\).

Find \(\mathrm{Pr}[X = 2\, \cap \, Y = 6\,]\)
Find \(\mathrm{Pr}[X > 3\, | \, Y = 2\,]\)
Find \(\mathrm{Pr}[X > 4\, | \, X+Y=8\,]\)

An actuarial student takes an exam that contains \(20\) multiple-choice questions.

Each question has four possible options.

She knows the answer to \(10\) questions, but have no idea about the other \(10\) questions, so she chooses answers randomly. Her score \(X\) on the exam is the total number of correct answers.

Find \(\mathrm{Pr}[X>17]\).

The number of customers arriving to a bank is a Poisson random variable.

On average, \(20\) customers arrive per hour.

Let \(X\) be the number arriving from 10:00 to 11:30 am.

Find \(\mathrm{Pr}[20<X\leq30].\)

An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims.

If the number of claims filed has a Poisson distribution, what is the expected number of claims filed?

A company establishes a fund of \(120\,000\) from which it will pay an amount, \(C\), to any of its \(20\) employees who achieve a high-performance level during the coming year.

Each employee has a \(2\%\) chance of achieving a high-performance level during the coming year, independent of any other employee.

Determine the maximum value of \(C\) for which the probability is less than \(1\%\) that the fund will be insufficient to cover all payments for high performance.

An insurance company prices hurricane insurance using the following assumptions:

In any calendar year, there can be at most one hurricane
In any calendar year, the probability of a hurricane is \(0.05\)
The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year.

Find the probability that there are fewer than three hurricanes in a \(20\)-year period.

\(X\) is the random variable of the number of large insurance losses in one year.

If \(\mathrm{Pr}[X=1]=0.3\), \(\mathrm{Pr}[X=2]=0.2\), and \(\mathrm{Pr}[X=0]=0.3 \, \cdot \mathrm{Pr}[X=3]\), find \(E[X]\).

Let \(\mathrm{Pr}[X=0]=1−\mathrm{Pr}[X=1]\).

If \(E[X]=3 \cdot Var[X]\), find \(\mathrm{Pr}[X=0]\).

If \(X\) is a binomial random variable with expected value \(6\) and variance \(2.4\), find \(\mathrm{Pr}[X=5]\).

On average, \(5.2\) hurricanes hit a certain region in a year.

What is the probability that there will \(3\) or fewer hurricanes hitting this year?.

An insurance policy pays \(100\) per day for up to three days of hospitalization and \(50\) per day for each day of hospitalization thereafter.

The number of days of hospitalization, \(X\), is a discrete random variable with probability function \(\mathrm{Pr}[X=k]=\frac{6-k}{15}\), \(k=1, 2, 3, 4, 5\), and \(0\) otherwise.

Determine the expected payment for hospitalization under this policy.

Suppose that \(X\) takes on one of the values \(0\), \(1\), and \(2\).

If \(\mathrm{Pr}[X=i]=2 \cdot \mathrm{Pr}[X=i-1]\), \(i=1, 2\), find \(E[X]\)

In modeling the number of claims filed by an individual under a car policy during a three-year period, an actuary makes the simplifying assumption that for all integers \(n \geq 0\), \(p(n+1)=0.2 \cdot p(n)\), where \(p(n)\) represents the probability that the policyholder files \(n\) claims during the period.

Calculate the probability that a policyholder files more than one claim during the period.

When three friends go for coffee, they decide who will pay the check by each flipping a coin and then letting the “odd person” pay.

If all three flips produce the same result, then they make a second round of flips, and they continue to do so until there is an odd person.

What is the probability that

exactly \(3\) rounds of flips are made?
more than \(4\) rounds are needed?

Find the coefficient of variation of the following distribution:

\(x_{i}\)	\(-5\)	\(-4\)	\(1\)	\(2\)
\(\mathrm{Pr}[X=x_{i}]\)	\(\frac{1}{4}\)	\(\frac{1}{8}\)	\(\frac{1}{2}\)	\(\frac{1}{8}\)

Let \(X\) be a random variable with the following probability distribution function:

\[ f(x) = \left\{ \begin{array}{ll} cx^{2} & |x|\leq1 \\ 0 & otherwise \\ \end{array} \right. \]

Find \(c\).
Find \(Var[X]\).
Find \(\mathrm{Pr}[X<\frac{1}{2}]\)

Let \(X\) be a random variable with the following probability distribution function:

\[ f(x) = \left\{ \begin{array}{ll} x^{2}(2x+\frac{3}{2}) & 0 < x \leq1 \\ 0 & otherwise \\ \end{array} \right. \]

Let \(X\) be a random variable with the following probability distribution function:

\[ f(x) = \left\{ \begin{array}{ll} ax+bx^{2} & 0 < x \leq1 \\ 0 & otherwise \\ \end{array} \right. \]

If \(E[X]=0.6\), find \(\mathrm{Pr}[X<\frac{1}{2}]\).

Travel time from your home to your classroom is normally distributed with mean \(40\) minutes and standard deviation \(7\) minutes.

If you want to be 95 percent certain that you will not be late for a 7:00 am probability class, what is the latest time that you should leave home?

At a certain insurance agency, the time that a customer spends being served by an agent is an exponential random variable with mean \(15\) minutes.

If there is a customer in service when you enter the agency, what is the probability that he or she will still be with the agent after an additional \(10\) minutes?

The number of days that elapse between the beginning of a calendar year and the moment a high-risk driver is involved in an accident is exponentially distributed.

An insurance company expects that \(30\%\) of high-risk drivers will be involved in an accident during the first \(50\) days of a calendar year.

What proportion of high-risk drivers are expected to be involved in an accident during the first \(80\) days of a calendar year?

The loss due to fire in a building is modelled by a random variable \(X\) with density function \(f(x)=0.005(20−x)\), \(0<x<20\), and \(0\) otherwise.

Given that a fire loss exceeds \(8\), what is the probability that it exceeds \(16\)?

An insurer’s annual weather-related loss, \(X\), is a random variable with density function

\[ f(x) = \left\{ \begin{array}{ll} \frac{2.5(200)^{2.5}}{x^{3.5}}, & x>200 \\ 0, & \mathrm{otherwise} \\ \end{array} \right. \]

Calculate the difference between the \(30^{th}\) and the \(70^{th}\) percentiles of \(X\).

Suppose that the cumulative distribution function of the random variable \(X\) is given by \(F(x)=1-exp(-x^{2})\).

Let \(\lambda\) be the hazard function.

Evaluate \(\lambda(5)\).

The time \(T\) that a machine is out of operation has a cumulative distribution function

\[ F(t) = \left\{ \begin{array}{ll} 1-(\frac{2}{t})^{2} & t>2 \\ 0 & \mathrm{otherwise} \\ \end{array} \right. \]

The resulting cost to the company is \(Y=T^{2}\).

Determine \(Pr[Y>16]\).

The number of years that a computer is operational is a random variable whose hazard rate function is given by

\[ \lambda(t) = \left\{ \begin{array}{ll} 0.2 & 0<t<2 \\ 0.2+0.3(t-2) & 2\leq t<5 \\ 1.1 & t>5 \\ \end{array} \right. \]

What is the probability that the computer will still be working \(6\)?

The annual rainfall in a region is approximately a normal random variable with mean \(40.2\) inches and standard deviation \(8.4\) inches.

What is the probability that the yearly rainfall in exactly \(3\) of the next \(7\) years will exceed \(44\) inches?

The lifetime of a machine has a continuous distribution on the interval \((0, 40)\) with probability density function \(f\), where \(f(x)\) is proportional to \((10+x)^{−2}\).

What is the probability that the lifetime of the machine is less than \(5\)?

An actuary purchases a life insurance policy on her \(40^{th}\) birthday.

The policy will pay \(50\,000\) only if she dies before her \(50^{th}\) birthday and will pay \(0\) otherwise.

The length of lifetime in years of a female born the same year as the insured has the cumulative distribution function

\[ F(t) = \left\{ \begin{array}{ll} 0 & t\leq 0 \\ 1-exp(\frac{1-1.1^t}{1000}) & t > 0 \\ \end{array} \right. \] Let \(\Pi\) be the policy premium, which is set as \(1.2\) times the expected payment under the policy.

Calculate \(\Pi\).

The life of a certain type of airplane is normally distributed with mean of \(34m\) miles and standard deviation of \(4m\) miles.

Given that the airplane has survived \(30m\), what is the conditional probability that it survives for another \(10m\) miles?

A group insurance policy covers the medical claims of the employees of a small company.

The value, \(V\), of the claims made in one year is described by \(V=100\,000\cdot Y\), where \(Y\) is a random variable with density function

\[ f(y) = \left\{ \begin{array}{ll} k \cdot (1-y)^{4} & 0<y<1 \\ 0 & otherwise \\ \end{array} \right. \] where \(k\) is a constant.

Calculate the conditional probability that \(V\) exceeds \(40\,000\) given that \(V\) exceeds \(10\,000\).

Fifty-two percent of the residents of CDMX are in favor of outlawing cigarette smoking on university campuses.

The approximate probability that more than \(50\%\) of the people in a random sample of \(n\) decides to outlaw cigarette smoking greater than \(95\%\).

Find the minimum value of \(n\) that satisfies such probability.

Let \(X\) be a continuous random variable with density function

\[ f(x) = \left\{ \begin{array}{ll} (p-1) \cdot x^{-p} & x>1 \\ 0 & otherwise \\ \end{array} \right. \] Calculate the value of \(p\) such that \(E[X]=2.\)

Your company must make a bid for an actuarial project.

If you win the contract (by having the lowest bid), you will pay another firm \(100\,000\) to do the work.

If you believe the minimum bid of other companies can be modelled as the value of a uniformly distributed random variable on \((70\,000, 140\,000)\), how much should you bid to maximize your expected profit?

An insurance company sells a car insurance policy that covers losses incurred by a policyholder, subject to a deductible of \(100\). Losses incurred follow an exponential distribution with mean \(300\).

What is the \(95^{th}\) percentile of actual losses that exceed the deductible?

An insurance policy is written to cover a loss, \(X\), that has a uniform distribution on \((0, 1\,000)\).

At what level must a deductible be set in order for the expected payment to be \(25\%\) of what if would be with no deductible?

A company’s annual losses follows a distribution with density function \(f(x)= \large \frac {2.5(0.6)^{2.5}}{x^{3.5}}\), for \(x>0.6\) and \(0\) otherwise.

To cover its losses, the company purchases an insurance policy with an annual deductible of \(2\).

What is the mean of the company’s annual losses not paid by the insurance policy?

An insurance policy reimburses a loss up to a benefit limit of \(10\).

The policyholder’s loss, \(Y\), follows a distribution with density function \(f(y)=2y^{-3}\) for \(y>1\) and \(0\) otherwise.

What is the expected value of the benefit paid under this policy?

An insurance company insures a \(15\,000\) car for one year with a \(1\,000\) deductible.

During the year, there is a \(0.04\) chance of partial damage to the car, and \(0.02\) chance of total loss of the car.

If there is a partial damage, the annual amount \(X\) of damage (in thousands) follows a distribution with probability density function \(f(x)=0.5003e^{−0.5x}\), for \(0<x<15\), and \(0\) otherwise.

What is the expected claim payment?

The cumulative density function for health care costs experienced by a policyholder is modelled by

\[ F(x) = \left\{ \begin{array}{ll} 1-exp(-0.01x) & x>0 \\ 0 & otherwise \\ \end{array} \right. \] The policy has a deductible of \(20\).

An insurer reimburses the policyholder for \(100\%\) of health care costs between \(20\) and \(120\) less the deductible.

Health care costs above \(120\) are reimbursed at \(50\%\).

Let \(G\) be the cumulative distribution function of the reimbursements, given that the reimbursement is positive.

Calculate \(G(115)\).

Let \(X_{1}\) be a normal random variable with mean \(2\) and variance \(3\), and let \(X_{2}\) be a normal random variable with mean \(1\) and variance \(4\).

Assume that \(X_{1}\) and \(X_{2}\) are mutually independent.

Find \(\mathrm{Pr}[\,2X_{1}+3X_{2}>15\,]\)

An man purchases a life insurance policy on her \(40^{th}\) birthday.

The policy will pay \(50\,000\) only if he dies before his \(50^{th}\) birthday and will pay \(0\) otherwise.

The length of lifetime in years of a male born the same year as the insured has the cumulative distribution function

Calculate \(\Pi\).

A probability distribution of the claim sizes for an auto insurance policy is

Claim size	\(20\)	\(30\)	\(40\)	\(50\)	\(60\)	\(70\)	\(82\)
Probability	\(0.15\)	\(0.10\)	\(0.05\)	\(0.20\)	\(0.10\)	\(0.10\)	\(0.30\)

Calculate the percentage of claims that are within one standard deviation of the mean claim size.

An insurance company sells an auto insurance policy that covers losses incurred by a policyholder, subject to a deductible of \(100\).

Losses incurred follow an exponential distribution with mean \(300\).

Calculate the \(95^{th}\) percentile of the losses that exceed the deductible.

The total claim amount for a commercial property insurance policy follows a distribution with density function \(f(x)=0.0001 \cdot exp(−0.0001x)\), \(x>0\).

The premium for the policy is set at the expected total claims amount plus \(1\,000\).

If \(100\) policies are sold, calculate the approximate probability that the insurance company will have claims exceeding the premium collected.

A tour operator has a bus that can accommodate \(20\) tourists.

The operator know that tourists may not show up, so he sells \(21\) tickets.

The probability that an individual tourist will not show up is \(0.02\), independent of all other tourists.

Each ticket costs \(50\), and is non-refundable if a tourist fails to show up.

If a tourist shows up and a seat is not available, the tour operator has to pay \(100\) (ticket cost \(+\) \(50\) penalty) to the tourist.

What is the expected revenue of the tour operator?