Principles of Actuarial Science Chapter 2

1. 1
ACTL1001
Games of Chance (2)
Outline:
1. Independence
2. Distributions in Actuarial Science (ctd)
3. Probability of Ruin
4. Simulation
Reading
(Req) Sherris, 2.3-2.7

2. 2
Independence
Trials or events are independent if the occurrence of one does not the affect the
probability of occurrence of any of the others.
The probability that a number of independent events occur is just the product of the
probabilities that each event occurs.

3. 3
Geometric
The Geometric distribution gives the probability of the number of trials to the first
success where the probability of success on any trial is p.
Geometric (p)
Pr (X = x) = (1 p)x 1
p x = 1; 2 : : :
E [X] =
1
p
V ar [X] =
1
p
1
p
1

4. 4
Example
Passing or gaining exemption from actuarial professional examinations is notoriously
difficult.
Assume that the probability of passing or gaining an exemption from a particular
actuarial professional examination is 0:6, and that the chance per attempt is the same
and are independent.
Calculate the probability that it will take at least 3 attempts to pass or gain an exemp-
tion from this examination.

5. 5
Exponential
The exponential distribution is used for the probability distribution of the time until an
event occurs where the probability does not depend on elapsed time
Exponential distribution often used to model the time of survival
Exponential ( )
f(x) =
e
x
x 0 and 0 otherwise
F(x) =
Z x
1
f(t)dt = 1 e
x
x 0
E [X] =
V ar [X] = 2

6. 6
Normal Distribution
The normal distribution is the “bell” shaped symmetric distribution that typically re-
sults from the accumulation of a large number of independent random events.
Normal( ; )
f(x) =
1
p
2
e
1
2(x
)
2
for 1 < x < 1
E [X] =
V ar [X] = 2
The case of the Normal with
= 0 and = 1
is called the standard normal density

7. 7
Tables of the Cumulative Standard Normal Distribution -
N [x] =
Z x
1
1
p
2
e
1
2(s)2
ds
are tabulated (and also widely available in software including excel).
Note that if X is a standard normal random variable then
Y = + X
is a normal random with mean and standard deviation .
Central Limit Theorem: For large values of n a
Binomial (n; p)
distribution is well approximated by a
Normal(np;
p
np (1 p))
distribution.

8. 8
Transform of Normal
In actuarial science we are interested in modelling survival times and insurance claim
amounts.
These are positive valued random variables.
The normal distribution is a symmetric distribution that can take negative values.
Survival times and losses are always positive and the distribution is not symmetric -
usually positively skewed to the higher positive values.

9. 9
Log-Normal
The exponential of a Normal( ; ) random variable has a
Log normal ( ; )
distribution and is only defined for positive values.
Z Normal - then X = eZ
is Log-normal
Log normal ( ; )
f(x) =
1
x
p
2
e
1
2(ln x
)
2
for 0 < x < 1
E [X] = e +1
2
2
V ar [X] = e2 + 2
h
e
2
1
i
E Xk
= ek +1
2k2 2

10. 10
Gambler’s Ruin
A major consideration in a financial security system is ensuring that promises to pay
benefits can be reasonably met. (Solvency)
The chance that benefits payable from an insurance company will not be met - the
probability of ruin.
Classic case is Gambler’s Ruin - illustrates the use of RECURSIVE relationships in
actuarial science.

11. 11
Example
A and B start with a and b dollars respectively
Continue to play a series of games such that the probability of A winning a game is
p.
The loser of each game pays $1 to the winner.
The game continues until one or the other has all the money $(a + b):
What is the probability that A will end up with all the money?
In this case we can say that B is ruined.

12. 12
Recurrence relation for the probabilities (can do for expected values also).
Denote the probability that A will ultimately win, assuming she currently has (n + 1)
dollars by un+1:
If A wins the next throw - with probability p - then she will have (n + 2) dollars and a
probability of winning of un+2:
If B wins - with probability q = (1 p) - then she will have n dollars and a probability
of winning of un.
Hence
un+1 = pun+2 + qun for 0 < n + 1 < a + b
and by rearranging
(un+2 un+1) =
q
p
(un+1 un)

13. 13
Recursive relations
(u2 u1) =
q
p
(u1)
(u3 u2) =
q
p
(u2 u1) =
q
p
2
(u1)
(u4 u3) =
q
p
(u3 u2) =
q
p
3
(u1)
.
.
.
(ui ui 1) =
q
p
(ui 1 ui 2) =
q
p
i 1
(u1)
Summing all of these terms we get
(ui u1) =
(
q
p
+
q
p
2
+
q
p
i 1
)
u1
ui =
(
1 +
q
p
+
q
p
2
+
q
p
i 1
)
u1

14. 14
Special Case: If p = q = 1
2 we have
ui = iu1
Note that ua+b = 1 so that
u1 =
1
a + b
We require ua which is
ua =
a
a + b
if p = q =
1
2
General Case: Assume p 6= q: We have
ui =
8
>
<
>
:
1 q
p
i
1 q
p
9
>
=
>
;
u1

15. 15
Using ua+b = 1 we get
u1 =
8
>
<
>
:
1 q
p
1 q
p
a+b
9
>
=
>
;
So that
ui =
8
>
<
>
:
1 q
p
i
1 q
p
a+b
9
>
=
>
;

16. 16
Probability of Ruin
Probability of ruin - general result (easy to remember)
ua =
8
>
<
>
:
1 q
p
a
1 q
p
a+b
9
>
=
>
;
for p 6= q

17. 17
Example: Feeling Lucky?
Example: Casino has wealth of $50 million. You have wealth of $1000 and the
chance of the casino winning each game is slightly in their favour at p = 0:51 and you
win or lose a dollar on every game:
Calculate the odds that you will go broke playing at the casino (i.e. the casino will
win all of your money).

18. 18
Simulation
Desktop computers and various software allow the evaluation of problems in risk,
insurance and finance using relatively sophisticated probability models.
These models can be used to simulate the events of interest.
Spreadsheets and statistical software packages can simulate a range of probability
distributions.
Specially designed software packages for probability modelling using simulation

19. 19
Random Number Generator
Excel includes random number generators for the Uniform, Normal and Poisson dis-
tributions (amongst others).
The uniform random number generator produces numbers that lie between 0 and 1
and with each number equally likely.
Rand()

20. 20
Example - Simulate the roll of a die and calculate the expected value on the die face
using 100 and 1000 rolls of the die.
Solution:
Generate 100 and 1000 uniform random numbers.
If the random number is between i 1
6 and i
6 then we assign i as the outcome for i = 1
to 6:
Outcome = 1+ integer part of [6 rand()] :
Calculate the expected value - add up the outcomes and divide by the number of
simulations.
The average should be around the theoretical value of
1
6
6:7
2
= 3:5