GSB518 HW3

Problem 1

(This is an overly simplified example from actuarial science.) A life insurance company sells a term insurance policy to a 21 year old man (the “insured”) that pays $100,000 to a designated beneficiary if the insured dies within the next 5 years, and pays nothing if the insured does not die before age 26. The probability that a randomly chosen 21-year-old man will die each year can be found in mortality tables. The company collects a premium of $250 at the beginning of each year of the 5 years of the term, as long as the insured is alive, as payment for the insurance. The amount \(Y\) (in dollars) that the company earns on this policy is $250 per year, less the $100,000 that it pays out in the event the insured dies. The company doesn’t pay out anything if the insured does not die in the 5 year term.

Age at death	Earnings \(Y\)	Probability
21	−99,750	0.00183
22		0.00186
23		0.00189
24		0.00191
25		0.00193
26 or older

1. Find the distribution of \(Y\) by completing the above table. For example, the value -99,750 provides an example of how \(Y\) is calculated when the insured dies in the first year of the policy.

Age at death	Earnings \(Y\)	Probability
21	−99,750	0.00183
22	-99,500	0.00186
23	-99,250	0.00189
24	-99,000	0.00191
25	-98,750	0.00193
26 or older	1,250	0.99058

2. Compute \(\text{E}(Y)\).

E(Y)=(-99750)(0.00183)+(-99500)(0.00186)+(-99250)(0.00189)+(-99000)(0.00191)+(98750)(0.00193)+(1250)(0.99058)=303.23

3. Write a sentence explaining in this context in what sense the number from the previous part is “expected”.

If the insurance company sells a lot of these policies the insurance company will expect to make a profit $303.36 per policy on average.

4. Would you be willing to sell such an insurance policy to one of your 21 year old friends? Explain why this is good business for the insurance company, but not for you.

It wouldn’t make sense to just sell one of these policies because although you are almost guaranteed to make a profit of $1250, there is a very slim chance that your friend dies before 26 and you have to pay 100000. You would not only be a friend down but 100000 dollars down. It does however make sense if you are selling thousands of these policies because your long term average would come out to around 303 dollars of profit per policy and the risk of paying out 100000 dollars would be outweighed by the amount you would be receiving from those who make it to 26.

5. Compute \(\text{Var}(Y)\)

E(Y^2)=(-99750)2(0.00183)+(-99500)^{2(0.00186)+(-99250)}2(0.00189)+(-99000)^{2(0.00191)+(98750)}2(0.00193)+(1250)^2(0.99058)=94328849 Var(Y)=E(Y^2)-(E(Y))2=94328849-(303.23)^2=94236687.6

6. Compute \(\text{SD}(Y)\).

SD(Y)=Sq route of Var(Y) = $9707.56

7. Compute the probability that \(Y\) is more than 1 standard deviation above its mean.

One SD about the mean is around 10000. The probability of one SD above the mean is 0 because it’s not possible to have that value of Y.

Problem 2

In a certain population, household income \(X\) ($ thousands) follows the pdf \[ f_X(x) = c x^{-2.5}, \quad x \ge 30 \] for an appropriate constant \(c\). Note that the measurement units are $ thousands, so 1 represents 1 thousand dollars.

1. What are the possible values of \(X\)? Sketch the pdf of \(X\).

The possible values of X are all values greater or equal to 30 (thousand dollars)

2. Find the value of \(c\).

1 = \(\$\int_{30}^{\infty}\)c(x^-2.5)dx = (c30^-1.5)/1.5

c = (1.5)(30^1.5)x-2.5 = 247.76

3. Find the probability that a household has an income over $200K.

P(X>200) = \(\int_{200}^{\infty}\) (1.5)(30^1.5)(x-2.5)dx = 0.058

Around 5.8% of households have an income over 200 thousand $s in this area.

4. Find the probability that a household has an income exactly equal to $200K.

Because X is continuous P(X=200) = 0.

5. Without integrating, find the probability that a household has an income, rounded to the nearest thousand dollars, of $200K.

fx(200)(1)=(1.5)(30^1.5)(200-2.5) = 0.00044

Around 0.044% of households have an income of 200 thousand $s to the nearest thousand in this area.

6. Without integrating, determine how many times more likely it is for a household to have an income, rounded to the nearest thousand dollars, of $100K than $200K.

fx(100)/fx(200) = c(100^-2.5)/c(200-2.5) = 5.66

A household in this area is 5.66 times more likely to have an income of 100 thousand dollars rounded to the nearest thousand, than 200 thousand dollars.

Problem 3

In the meeting time problem, assume that Regina’s \(R\) and Cady’s \(Y\) arrival times (measured in fractions of the hour after noon, e.g. 0.25 represents 12:15), each follow a Uniform(0, 1) distribution, independently of each other. Let \(W = |R - Y|\) be the amount of time (hours) the first person to arrive waits for the second person to arrive. It can be shown that \(W\) has pdf \[ f_W(w) = 2(1-w), \qquad 0<w<1. \]

1. Sketch a plot of the pdf. Describe roughly what this means in context.

The density would be highest at w=0, and it decreases linearly, meaning the likelihood of waiting longer decreases as w increases.

2. Coding required. Code and run a simulation to approximate the distribution of \(W\). Does the histogram reasonably match up with the pdf? Then use the simulation results to approximate \(\text{P}(W < 0.25)\).

N_rep = 10000

x = runif(N_rep, 0, 1)
y = runif(N_rep, 0, 1)

w = abs(x - y)

hist(w,
     freq = FALSE,
     main = "")

The histogram resembles the pdf

sum(w < 0.25) / N_rep

## [1] 0.4402

P(W<0.25) = 0.4345 If they were to meet over lots of days, the first person will have to wait less than 15 minutes on 43.45% of days.

3. Use the pdf to compute \(\text{P}(W < 0.25)\) and interpret the value. Set up an integral, but sketch a picture and use geometry to compute.

P(W<0.25)=$_{0}^{0.25}$2(1-w)dw

If you add the area of the square and triangle you get 0.4375. If they were to meet over lots of days, the first person will have to wait less than 15 minutes on 43.75% of days.

Problem 4

(Continued.) In the Regina/Cady problem, let \(W=|R-Y|\) be the amount of time the first person to arrive has to wait for the second person. Recall that \(W\) is a continuous random variable with pdf \[ f_W(w) = 2(1-w), \quad 0 < w < 1. \]

1. Coding required. Use your simulation results from before to approximate \(\text{E}(W)\), \(\text{E}(W^2)\), \(\text{Var}(W)\), and \(\text{SD}(W)\).

mean(w)

## [1] 0.3351795

mean(w ^ 2)

## [1] 0.1688451

var(w)

## [1] 0.05650544

sd(w)

## [1] 0.2377087

2. Compute and interpret \(\text{E}(W)\).

E(W)= \(\int_{0}^{1}\)wfw(w)dw = \(\int_{0}^{1}\)w(2(1-w))dw = 1/3. Over many days, the long run average waiting time would be 1/3 of an hour (20 minutes).

3. Compute \(\text{E}(W^2)\).

E(W^2)= \(\int_{0}^{1}\)w^2fw(w)dw = \(\int_{0}^{1}\)w^2(2(1-w))dw = 1/6.

4. Compute \(\text{Var}(W)\)

Var(W) = E(W^2)-E(W)^2= 1/6-(1/3)^2=1/18

5. Compute \(\text{SD}(W)\).

SD(W)= 0.236 hours

6. Explain in words in detail how you could use the Uniform(0, 1) spinner and simulation to approximate \(\text{Var}(W)\).

Spin the spinner twice per simulation to simulate R and Y. Make W = |R-Y|. Repeat this a lot of times, then average the values of W. Subtract the average of W from each simulated W value and square it. Average all squared deviations to get Var(W).

7. Compute the probability that \(W\) is more than 1 SD away from its mean.

1/3-0.236= 0.0973 1/3+0.236= 0.5693 $*{0}^{0.0973}*\(2(1-w)dw+\)^{1}$2(1-w)dw= 0.37