Holborn Technical Seminar
Practical Statistics for Excess Reinsurance

Your Expectations

Review of Probability Theory

Definition of probability

Probability (Density) as a mapping from events to numbers: PDF

Examples: Uniform and Binomial distributions

CDF defined and as integral of PDF

Expectation

Higher moments, Variance and SD

Central Limit Theorem

Normal Distribution

Definition of Probability

For any event,

Let S = every possible outcome of A

• For rolling a die, S = {1,2,3,4,5,6}


• S is called a Sample Space

P(S) = 1

P(Ø)=0

For any sequence of n disjoint events, A1,…, An the probability of All n events is equal to the sum of the probabilities of each event, Ai

• P( A U B) = P(A) + P(B) for disjoint A & B

An ordinary deck of playing cards is shuffled, and a card is randomly chosen. What the probability that the card selected is:

(a) a heart?

(a) Since there are 13 hearts, the probability that the card selected is a heart 13/52 = 1/4.

(b) red?

(b) Because two suits are red (hearts and diamonds), there are 13+13 = 26 red cards, so the probability that a red card is selected is 26/52 = ½.

(c) the jack of hearts?

(c) Since there are only one jack of hearts, the chances that this particular card is chosen is 1/52.

(d) not a face card?

(d) There are 3 face cards per suit: the jack, queen, and king. Since there are 4 suits (hearts, diamonds, spades, and clubs), there are 3x4=12 face cards. Therefore 52-12=40 of the cards are not face cards. We conclude that the probability of not selecting a face card is 40/52 = 10/13.

Imagine throwing three coins. If we let H denote “heads and T “tails" , then the set of possible outcomes is:

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Since there are 2^3 = 8 elements in S, if all the coins are fair, the probability that any of one of these outcomes will occur must be 1/8.

What’s the probability of getting exactly 2 “heads”? In this case, the event is the following subset of S:

E = {HHT, HTH, THH}

Which consists of 3 outcomes. Therefore, P(E) = 3/8.

If A and B are events in a probability space, then the statement “A or B" is equivalent to the event A È B, and the statement "A and B" is equivalent to the event A Ç B.

Two events, A and B, are said to be are said to be independent if the occurrence of one does not influence (Change the probability of) the occurrence of the other. Finally, two events, A and B, are mutually exclusive if the occurrence of one makes the occurrence of the other impossible; therefore, if A and B are mutually exclusive, then P(A Ç B) = 0.

1. P(A È B) = P(A) + P(B) – P(A Ç B)

2. A and B are independent if and only if P(A Ç B) = P(A) · P(B)

3. If A and B are mutually exclusive, then P(A È B) = P(A) + P(B).

A gambler throws two fair dice twice. Let A be the event that the first roll is a 7 or an 11, and let B be the event that the second toss is an 11. What’s P(A or B)?

Solution:

The sample space, S, for each toss is the set of all ordered pairs (a,b) where a and b are any of the numbers 1 through 6, so S consists of 6^2 = 36 possible outcomes, all of which are equally likely to occur. The event A is the subset of S that contains all ordered pairs (a,b) such that a+b = 7 or 11:

A= {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) (5,6), (6,5)}

which consists of 8 outcomes. This gives us P(A) = 8/36. The event B is the subset of S that contains all ordered pairs (a,b) such that a+b = 11

B={(5,6), (6,5)}

which consists of 2 outcomes. This gives us P(B) = 2/36. Since A and B are independent – contrary to popular belief, dice have no brains and therefore cannot store memories! – we know that P(A and B) = P(A) . P(B). Therefore,

P(A or B) = P(A) + P(B) – P(A and B) = 8/36 + 2/36 - 8/36 * 2/36 = 43/162

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