1. STATISTICS AND PROBABILITY
Raoul LePage
Professor
STATISTICS AND PROBABILITY
click on STT315_Fall 06
Week of

2. 2-75 ( "or both" is redundant ),
suggested exercises
ans in text, don't submit
2-61, 2-63, 2-65, 2-67, 2-69, 2-73,
2-75 ( "or both" is redundant ),
2-77 ( i.e. P( B up | A up ) ), 3-1, 3-5,
3-11, 3-15, 3-17, 3-23 and s.d., 3-25
Week 2.

3. TREE DIAGRAM
“oil” = oil is present “+” = a test for oil is positive “-” = a test for oil is negative +
oil -
false negative
false positive +
no oil -

4. P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
TREE DIAGRAM CONVENTIONS
P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
P(oil +)
= (0.3) (0.9)
= 0.27
P(+ | oil) = 0.9 +
P(oil) = 0.3
oil - +
no oil -

5. P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
TOTAL OF BRANCHES = 1
P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
sum of
unconditional
probabilities
is one
0.3 oil
0.7 no oil

6. P(oil) = 0.3 P(+ | oil) = 0.9 P(- | oil) = 0.1 P(+ | no oil) = 0.4
TOTAL OF CONDITIONAL BRANCHES = 1
P(oil) = 0.3 P(+ | oil) = P(- | oil) = 0.1 P(+ | no oil) = 0.4
sum of
conditional
probabilities
is one
0.9 +
0.3
0.1 -
oil
0.7 +
no oil -

7. P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
COMPLETE TREE
P(oil) = 0.3 P(+ | oil) = 0.9 P(+ | no oil) = 0.4
0.27 oil+
unconditional
0.9 +
0.3
0.1
oil -
0.03 oil-
0.7
0.28 oil+
0.4 +
no oil
0.6 -
0.42 oil-
conditional
outcomes

8. - - + oil + no oil VENN DIAGRAM S oil + 0.03 0.27 0.28 0.42 0.27 oil+
0.27
oil+
0.9 +
0.3
0.1 -
oil-
oil
0.03
0.7
oil+
0.28
0.4 +
no oil -
0.6
0.42
oil-

9. + oil + no oil TOTAL PROBABILITY P(+) = P(oil+) + P(no oil+)
0.55 =
0.27
oil+
0.9 +
0.3
oil
0.7
oil+
0.28
0.4 +
no oil
Oil contributes 0.27 to the total P(+) = 0.55.

10. P(oil | +) = P(oil+) / P(+)
BAYES FORMULA
S oil
0.27
oil+
P(oil | +) = P(oil+) / P(+)
= 0.27 / ( ) =
oil+
0.28
Oil contributes 0.27 of the total P(+) =

11. - - + + MEDICAL TEST 0.98 0.01 disease 0.02 0.03 0.99 no disease 0.97
The test for this infrequent disease seems to be reliable having only 3% false positives and 2% false negatives. What if we test positive?

12. - - + + MEDICAL TEST 0.0098 0.01 0.98 disease 0.02 0.0002 0.0297 0.03
0.99
no disease
0.97 -
0.9603
We need to calculate P(diseased | +), the conditional probability that we have this disease GIVEN we’ve tested positive for it.

13. - - + + CALCULATING OUR CHANCES OF HAVING THE DISEASE IF + only 25% !
0.0098
0.01
disease
0.98 +
0.02 -
0.0002
0.0297
0.03 +
0.99
no disease
0.97 -
0.9603
P(+) = =
P(disease | +) = P(disease+) / P(+)
= / =
only 25% !

14. - - + + FALSE POSITIVE PARADOX 0.0098 0.98 0.01 disease 0.02 0.0002
one may overwhelm a good test by failing to screen
0.0098
0.01
disease
0.98 +
0.02 -
0.0002
0.0297
0.99
no disease
0.03 +
0.97 -
0.9603
EVEN FOR THIS ACCURATE TEST: P(diseased | +) is only around 25% because the non-diseased group is so predominant that most positives come from it.

15. - - + + FALSE POSITIVE PARADOX rare ! 0.00098 0.98 0.001 disease 0.02
one may overwhelm a good test by failing to screen
0.001
disease
0.98 +
rare !
0.02 -
0.999
no disease
0.03 +
0.97 -
WHEN THE DISEASE IS TRULY RARE: P(diseased | +) is a mere 3.2% because the huge non-diseased group has completely over-whelmed the test, which no longer has value

16. IMPLICATIONS OF THE PARADOX
FOR MEDICAL PRACTICE: Good diagnostic
tests will be of little use if the system is over-whelmed by lots of healthy people taking the test. Screen patients first.
FOR BUSINESS: Good sales people capably focus their efforts on likely buyers, leading to increased sales. They can be rendered ineffective
by feeding them too many false leads, as with massive un-targeted sales promotions.

17. RANDOM VARIABLE boats/month P(fewer than 3.7) = .4 P(4 to 7) = .55
probability
total
RANDOM VARIABLE
(3-17 of text)
boats/month
P(fewer than 3.7) = .4
P(4 to 7) = .55

18. P(oil) = 0.3 oil no oil OIL DRILLING EXAMPLE Cost to drill 130
Reward for oil 400
net return
“just drill”
= 270
drill oil
drill no oil
= -130
0.3
oil
0.7
no oil
A random variable is just a numerical function
over the outcomes of a probability experiment.

19. EXPECTATION Definition of E X
E X = sum of value times probability x p(x).
Key properties
E(a X + b) = a E(X) + b
E(X + Y) = E(X) + E(Y) (always, if such exist)
a. E(sum of 13 dice) = 13 E(one die) = 13(3.5).
b. E(0.82 Ford US + Ford Germany - 20M)
= E(Ford US) + E(Ford Germany) - 20M
regardless of any possible dependence.

20. total of 2 dice E ( total ) is just twice the 3.5 avg for one die
(3-15)
of text
probability product
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
/ /36
sum /36 = 7
E ( total )
is just
twice
the 3.5
avg for
one die
E(total)

21. E(number of boats this month)
probability product
total
(3-17 of text)
boats/month
we avg
4.05 boats
per month
E(number of boats this month)

22. EXPECTATION IN THE OIL EXAMPLE
Expected return from policy “just drill” is the
probability weighted average (NET) return
E(NET) = (0.3) (270) + (0.7) (-130) = = -10.
net return from policy“just drill.”
= 270
drill oil
drill no-oil
= -130
just drill
0.3
oil
0.7
no oil
E(X) = -10

23. OIL EXAMPLE WITH A "TEST FOR OIL"
A test costing 20 is available. This test has:
P(test + | oil) = 0.9
P(test + | no-oil) = 0.4.
“costs”
TEST 20
DRILL 130
OIL 400
0.27
0.9 +
0.3
0.1
oil -
0.03
0.28
0.7
0.4 +
no oil
0.6 -
0.42
Is it worth 20 to test first?

24. EXPECTED RETURN IF WE "TEST FIRST"
net return prob prod
oil+ = =
oil- = =
no oil+ = =
no oil- = =
total
drill only if
the test is +
E(NET) = .27 (250) (20) (150) (20)
= 16.5 (for the “test first” policy).
This average return is much preferred over the
E(NET) = -10 of the “just drill” policy.

25. Variance and s.d. of boats/month
(3-17)
of text
x p(x) x p(x) x2 p(x) (x-4.05)2 p(x)
total
quantity E X E X E (X - E X)2
terminology mean mean of squares variance = mean of sq dev
s.d. = root(2.7474) = root( ) =

26. VARIANCE AND STANDARD DEVIATION
Var(X) =def E (X - E X)2 =comp E (X2) - (E X)2
i.e. Var(X) is the expected square deviation of r.v. X from its own expectation.
Caution: The computing formula (right above), although perfectly accurate mathematically, is sensitive to rounding errors.
Key properties:
Var(a X + b) = a2 Var(X) (b has no effect).
sd(a X + b) = |a| sd(X).
VAR(X + Y) = Var(X) + VAR(Y) if X ind of Y.

27. EXPECTATION AND INDEPENDENCE
Random variables X, Y are INDEPENDENT if
p(x, y) = p(x) p(y) for all possible values x, y.
If random variables X, Y are INDEPENDENT
E (X Y) = (E X) (E Y) echoing the above.
Var( X + Y ) = Var( X ) + Var( Y ).

28. PRICE RELATIVES EXPECTED RETURN
Venture one returns random variable X per $1 investment. This X is termed the “price relative.” This random X may in turn be reinvested in venture two which returns random variable Y per $1 investment. The return from $1 invested at the outset is the product random variable XY.
If INDEPENDENT, E( X Y ) = (E X) (E Y).
EXPECTED RETURN

29. PARADOX OF GROWTH WE AVERAGE 14% PER PERIOD EXAMPLE: x p(x) x p(x)
$ X
E(X) = 1.14
BUT YOU WILL NOT EARN 14%. Simply put, the average is not a reliable guide to real returns in the case of exponential growth.
WE AVERAGE 14% PER PERIOD

30. EXPECTATION governs SUMS but sums are in the exponent
EXAMPLE: x p(x) Log[x] p(x)
$ X
E Log10[X] = =
With INDEPENDENT plays your RANDOM return will compound at 11.1% not 14%.
(more about this later in the course)

31. THREE RANDOM EVOLUTIONS
COMPARING 1.14^n WITH
THREE RANDOM EVOLUTIONS
you can see that 14% exceeds reality