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STA130H1
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V.Radu Craiu
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Lecture

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Statistical Sciences

STA130H1

V.Radu Craiu

Fall

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STA107H1b.doc
Lecture #12 – Wednesday, October 15, 2003
P ARTITIONS
1) The sets{A1, 2 ,..n} Aare mutually exclusive
n
2) ▯ Ai= S
i=1
n
• If l ,…,nA is a partition then for any event B Ì▯S(B ˙ Ai)
i=1
Example
c
• B =(B˙ A )¨ (B˙ A )
L AW OF TOTAL P ROBABILITY
• If{B,B c} is a partition, PhAn)= P(A| BP B )+ (A| Bc)P(Bc)
• If{B1,B2,B 3} is a partition of SPaBdi> 0 for i =1,2,3,..., then
P(A)= P A| B1)P(B1)+ P(A| 2 P B 2)+ PA| B3)P(B3)
Generalization
n
• P(A)= ▯ P(A| BiP B i)
i=1
Example
A transport company needs to do safety checking for its cars. Suppose there are only two repair stores in
town. One of them is owned by a rival firm and the other is owned by an allied firm so the diagnostics will
not be objective. The friendly garage will lie with probability 20% if a car is unsafe and will tell the truth if
the car is safe. The “rival” garage with probability 35% will declare a car unsafe even if the car is safe and
will always declare the truth if the car is unsafe. Assume that 20% of the company’s cars are actually unsafe
and assume that the company will obtain a certificate from each garage for each of its cars. What is the
percentage of cases in which the two conclusions are contradictory?
• G ={Goodcars }, B ={Badcars}
• C ={Conclusionsarecontradicto}y
• P(C) = PC ˙G + P(C ˙B )becauseG and Bformsa partition of allcars
= PC |G P(G )+ PC | )P(B)
• PC |G )= 35% because the friendly garage tells the truth and the rival garage tells the lie
• PC | B)= 20% because the rival garage tells the truth and the friendly garage tells the lie
• P(C) = 35 · 80 + 20 · 20 = 32
100 100 100 100 100
B AYES R ULE
• Start with a parti{B1, 2 , 3},
• Take an event A
Page 1 of 17 STA107H1b.doc
• Instead of being interested in(A| B1), we are now interested in B 1 A )
P B |A = P B 1˙ A)
1 P A)
= P A B P1B ( 1
P A ˙ B1)+ P A ˙ B 2)+ P A ˙ B3 )
= P A B P1B ( 1
P A B P1B ( 1 + P A B P2B ( 2 + P A B P3B ( 3
Experiment In 2 Stages
• First stage – one of the B’s happens
• Second stage – we check whether A happens
P(B 1 P(B3)
P(B 2
Stage 1 B1 B1 B 1
P(A | B ) P(A | B ) P(A | B )
1 2 3
c c c
Stage 2 A A A A A A
• I know A happened. What is the probability tha1 B happened?
P B1˙ A ) P A B 1 B( 1
• P B1 |A = P A ) = P A B P B( ) + P A B P B ( ) +P A B P B ( )
1 1 2 2 3 3
• P(A )= P(A| B1)P(B1 + P (A| B2)P(B2 )+ P(A| B3)P(B 3)
• P(B ˙ A ) = P(A˙B )= P (A| B P (B )
1 1 1 1
Example
Say I have a wallet that contains either a $5 bill or a $20 bill (with equal probability), but I don’t know which
one. I add a $5 bill. Later, I reach into my wallet (without looking) and remove a bill. It’s a $5 bill. What is
the probability that the bill left in the wallet is a $5 bill?
Start 1/2 1/2
$5 $20
Add
2 · $5 $5, $20
1 0 1/2 1/2
Extract
$5 $20 $5 $20
Page 2 of 17 STA107H1b.doc
1 ×1
2 2
• P start with $5|extract )=$1 1 1 = 3
×1+ ×
2 2 2
Example
A lab blood test is 95% effective in detecting a certain disease when it is in fact, present. However the test
also yields a “false positive” result for 1% of the healthy people tested.
If 0.5% of the population has the disease, what is the probability hat a person who was tested positive actually
has the disease?
Initial Condition 0.005 0.995
sick healthy
0.95 0.05 0.01 0.95
Test
P N P N
P sick˙ positive P positive|sickP(sick)
• P sick | positi=e P positiv) = P(positiv)
• P(positiv)= 0.005·0.95+0.995·0.01
• P(sick˙positive)= 0.005·0.95
• P sick | positi=e 0.005·0.95 = 32.31%
0.005·0.95 +.995 ·.01
Lecture #13 – Wednesday, October 22, 2003
INDEPENDENCE
• Two events, A and B are independent P(A˙B )= P(A ×P B ).
• In general,P(A˙B )= P(A| B)×P(B .
• If A and B are independenP(A| B)= P A ) – The fact that B happened doesn’t change the
chance of the occurrence of A.
• In general{A , A ,..., } are independent if any twi Aj, A are independent.
1 2 n
Example
An urn contains five red and seven blue balls. Suppose that two balls are selected at random and with
replacement. Let A and B be the events that the first and the second balls are red, respectively. Check whether
A and B are independent or not. Redo the calculation for the case of random selection without replacement.
• With replacement:
• P A˙B =) 5·5 .
12·12
Page 3 of 17 STA107H1b.doc
• P A = 5 , P B = 5 .
12 12
• So A, B are independentA ^ B ).
• Without replacement:
• P A˙B = ) 5·4 = 20 .
12·11 12·11
5 4 5 5 7 55 5
• P A = , P B = P B | A P A + P B | A P Ac = × + × = = .
12 11 12 11 12 11·12 12
• So A and B are not independent.
Exercises
1) Show that if E and F are independent then E and F are independent. What about then E and F ?
c c
• I want to show P(E ˙F )= P E )×P(F .
P E ˙ F c)= P E ˙ F c)+ P(E ˙ F)- P(E ˙ F)
= P( )- P ( ˙ F )
= P( )- P ( ) ( )
= P( ) (1 P ( ))
= P( )×P F )
c c c
• E ^ F ▯ E ^ F ▯ E ^ F
2) Show that if E and F are mutually exclusive wi(E )> 0 and P (F )> 0 , then E, F cannot be
independent.
• P (E ˙ F)= 0 .
• If E and F are independent,(E ˙ F)= P E )×P(F) > 0 .
• So A and B are not independent if they are mutually exclusive.
3) True or False: If E and F are independent and E and G are independent, then E and F ˙G are
independent.
• False!
4) True or False: In the case of 3 events E, F, G, the eq(E ˙ F ˙G )= P E )×P(F)×P G ) doesn’t
imply that E, F, G are independent.
• True!
P ROBABILITY AS A C ONTINUOUS FUNCTION
• f :R ﬁ R .
• x ¾¾¾ﬁ x .
n nﬁ¥
• f is continuous in x lim f xn = f x )
xﬁ¥
• P :sets from S ﬁ[0,]
Page 4 of 17 STA107H1b.doc
Increasing Sequence of Events
• {E 1E 2...,En} infinite sequence is increasinEkÌ E k+1,"k ‡1 .
E
2
E
1
E 3
E 4
¥
• The “limit” is the E .
▯ k
k=1
Decreasing Sequence of Events
• {E 1E 2...,En} infinite sequence is decreasinE k E k+1,"k ‡1 .
E 3
E4
E 2
E
1
¥
• The “limit” is the E k.
▯
k 1
• {E n} is decreasing or increasing. I can define nim ▯ =E n ▯ En).
• In both cases, lim PE n)= P▯lim E n ▯.
nﬁ¥ ▯ ﬁ¥ ▯
Lecture #14 – Friday, October 24, 2003 10 26
¥
• If sets are increasing, then lin E▯= En – the smallest set containing all the sets.
nﬁ¥ n=1
¥
• If sets are increasing, then lin E▯= E n – the smallest set containing all the sets.
nﬁ¥ n=1
Example
2n2
th - 2
If the probability that the entire population will die before having offspring ie the n, what is the
probability that it will survive forever?
Page 5 of 17 STA107H1b.doc
th
• Let E n ={population wiped out by tne generatio}.
• So E Ì E Ì ...ÌE Ì E .
1 2 n n+1
P (urvivesforeve)= - P(diesout sometime in the fut)re
= - P(E1 ¨ E2 ¨ .¨. En)
¥
• = - P ▯ E ▯
▯▯ n▯
▯n=1 ▯
= - lim P(En )
nﬁ¥
2
▯ 2n ▯
• P (E n )= exp -▯ 2
6n + 3
▯ 2n 2 ▯ ▯ -2n 2 ▯ ▯ 2▯ -1
• lim PE n)= lim exp▯- 2 = exp▯lim 2 = exp▯- = e3
nﬁ¥ nﬁ¥
6n +3
nﬁ¥ 6n +3
6
Lecture #15 – Monday, October 27, 2003
R ANDOM V ARIABLES
• In many situations when an experiment is performed the interest is in some numerical function of the
outcome rather than the actual outcome itself.
• If S is the sample space of an experiment then a map X : S ﬁ R is called a random variable.
-1
• Additional Requirement: For any interval I Ì R , (I) is an event in S.
Example
• Tomorrow: rain, snow, shine.
• I bet $5 that it’s rain.
• X = my profit ={-$5,$5}
• P (X = $5)
• Y = profit from betting $100on snow
• If S is a discrete sample space, then X : S ﬁ R is a discrete random variable. The set of possible values of
X is also discrete – X : {a ,a ,...,a}
1 2 n
• Some S’s may not be discrete.
Example
Suppose that three cards are drawn from an ordinary deck of 52 cards one-by-one at random and with
replacement. Let X be the number of spades drawn. Fin(X = ),i = 0,1,2,3.
• Possible values for random variable X {0,1,2}. X : S ﬁ {0,1,2}.
Page 6 of 17 STA107H1b.doc
39·39·39 13·39·39 13·13·39
• P X = 0) = , P(X =1 )= ·3, P (X = 2)= ·3,
52·52·52 52·52·52 52·52·52
13·13·13
P X = 0 =
52·52·52
R ANDOM S ELECTION OF P OINTS IN NTERVALS
• Fix a < b and a, b such that a £ a < b £ b . The probability that a point is randomly selected in the
intervala,b ) is b-a .
b -a
a a b b
• Sample space is not discrete but continuous (interval).
• Let C be a fixed point in the inta,b). If X is a point randomly selected in the (a,b)vthen the
probability that X is selected to be exactly C is)= 0 .
• If(a,b) was discrete{a ,a ,...,a}, thenP(X = a )= 1
1 2 n i n
• Define E = C▯- 1 ,C + 1▯.
n ▯ n n▯
1 1
C - n+1 C + n+1
C -1 C C + 1
n n
▯ 1 1 ▯ ▯ 1 1 ▯
• EnÉ E n+1 so En+1 =▯C - n+1 ,C + n+1 ▯. Sequence of events n = ▯ - n ,C +n ▯ which
is decreasing.
2
▯¥ ▯
• P(X = C )= P(X ˛ E1and...and X˛ En)= P ▯▯ X ˛ E n▯= lim P(X ˛E n)= lim n = 0
▯n=1 ▯ nﬁ¥ nﬁ¥ b-a
Example
A train passes through a town at random time between 10:00 am and 10:12 pm. If I drive through town
between 9:55 am and 10:05 am, what is the probability that I see the train?
• (a,b)= 10:00am,10:12am )
9:55 10:00 10:05 10:12
5
• P Isee the train = P train goes through between10:00and10:05 = – does not depend on the scale of
12
measurement.
Page 7 of 17 STA107H1b.doc
Example
A patient with flu may have a fever between 39°C and 42°C. Let X be the temperature of a randomly selected
flu patient. What is the probability that the temperature is less than 40°C?
• S ={allpossible temperatur}s={39°C,40°C }
• a,b )= 39,42), a,b )= 39,40 )
• P(X ˛ (39,40))= 40-39 = 1
42-39 3
• P(X = 39.5)= 0
• P X ˛ 39.49,39.51 = 0.02
3
• For continuous random variable, we look atP(interva) instead of P(point).
Definition
• For a discrete random variableX : S ﬁ {a1,a2,...,n}, we can define a functioP: {a1,a2,...,n}ﬁ 0,1]
such that P ai) = P(X = ai).
• P is called probability function of the random variable X.
Lecture #16 – Wednesday, October 29, 2003
D ISTRIBUTION F UNCTIONS
• The distribution function F :R ﬁ 0,1] of a random variable X is defined aF(t)= P(X £ t).
• For example, F 1)= P X £1 ).
Properties of F
t £ t ▯ F t )£ F t )
1) F is non-decreasing – i.e.1 2 1 2 . So F is increasing or constant.
2) lim F t =1
tﬁ¥
3) lim F t = 0
tﬁ-¥
4) F is continuous to the right.
Proof of (1)
1 £ t2. I want to showF t1)£ F t2 )
• F(t1)= P(X £ t1), F t2) = P(X £ t2).
• E = {X £ 1}, F ={X £ t1}
F
• E Ì F ▯ P E )£ P(F ) E
-▯ ▯
P(E )= P(X £ t1)= P(t1)▯
• P(F )= P(X £ t )= P t )▯ ▯ F (1) £ F(t2)
2 2 ▯
• Attention: t1< t2 does not imply F (1) < F(t2); t1< t2 implies F (1 )£ F(t2)
Page 8 of 17 STA107H1b.doc
Proof of (2)
• X : S ﬁ R – real random variable.
• Always, there will be t large enough to have P(X £ t) is almost 1.
• Let E = {X £ n},n˛ Z .
n
• lim F t = lim F n = lim P n ( ).
tﬁ¥ nﬁ¥ nﬁ¥
• E n E n+1 ▯ the sequence E in increasing. I can use the continuity property of the probability function.
▯ ¥ ▯ ▯ ¥ ▯
• lim P (En )= P ▯im E n▯ = P▯ ▯ En ▯= P ▯▯ X £ n =1
nﬁ¥ ▯nﬁ¥ ▯ ▯ n 1 ▯ ▯n 1 ▯
Proof of (3)
• The proof of lim F t = 0 is the mirror of the proof oflim F t =1 .
tﬁ-¥ tﬁ¥
Proof of (4)
• Convergence from the left means for all x n x 0 and xnﬁ x 0 . In fact, for all practical purposes, one can
consider only increasing sequences. x n x n+1n˛ Z and xn£ x 0 and nﬁ¥ x n 0 .
• Notation: x ›nx 0 – converges to x0from the left. lim f (xn )= f(x 0 ).
n›x0
• If f(x 0 )= f(x 0), then f is continuous to the left.
• Continuity to the right. x
0
• xn‡ x n+1▯ "n˛ Z
xn‡ x 0 ▯

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