Lec 18 Nov 12 Probability – definitions and simulation.
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Transcript of Lec 18 Nov 12 Probability – definitions and simulation.
(Discrete) Sample space
Experiment: a physical act such as tossing a coin or rolling a die.
Sample space – set of outcomes. Coin toss Sample Space S = { head, tail} Rolling a die Sample space S = {1, 2, 3, 4, 5, 6}
Tossing a coin twice. Sample space S = {(h,h), (h,t), (t,h), (t,t)}
Events and probability
Event E is any subset of sample space S. You flip 2 coins
Sample space S = {(h,h), (h,t), (t,h), (t,t)}
Event: both tosses produce same result E = {(h,h), (t,t)}
Prob(E) = |E|/ |S| In the above example, p(E) = 2/4 = 0.5Question: what is the probability of getting at
least one six in three roles of a die?
Bernoulli trial
Bernoulli trials are experiments with two outcomes. (success with prob = p and failure with prob = 1 – p.)
Example: rolling an unloaded die. Success is defined as getting a role of 1.
p(success) = 1/6
Random Variable
Random variable (RV) is a function that maps the sample space to a number. E.g. the total number of heads X you get if you
flip 100 coins
Another example: Keep tossing a coin until you get a head. The RV
n is the number of tosses. Event = { H, TH, TTH, TTTH, … } RV n(H) = 1, n(TH) = 2, n(TTH) = 3, … etc.
Common Distributions
Uniform X: U[1, N] X takes values 1, 2, …, N
E.g. picking balls of different colors from a box
Binomial distribution X takes values 0, 1, …, n
N coin tosses. What is the prob. That there are exactly k tails?
P X 1i N
P X 1n iin
i p pi
Conditional Probability
P(A|B) is the probability of event A given that B has occurred. Suppose 6 coins are tossed. Given that
there is at least one head, what is the probability that the number of heads is 3?
Definition:
p(A|B) =
)(
)(
Bp
BAp
Baye’s Rule
If X and Y are events, then p(X|Y) = p(Y|X) p(X)/p(Y)
Useful in situation where p(X), p(Y) and p(Y|X) are easier to compute than p(X|Y).
Monty Hall Problem
You're given the choice of three doors: Behind one door is a car; behind the others, goats. You want to pick the car.
You pick a door, say No. 1 The host, who knows what's behind the
doors, opens another door, say No. 3, which has a goat.
Do you want to pick door No. 2 instead?
Host mustreveal Goat B
Host mustreveal Goat A
Host revealsGoat A
orHost reveals
Goat B
Monty Hall Problem: Bayes Rule
: the car is behind door i, i = 1, 2, 3 : the host opens door j after you
pick door i
iC
ijH
1 3iP C
0
0
1 2
1 ,
ij k
i j
j kP H C
i k
i k j k
Monty Hall Problem: Bayes Rule continued
WLOG, i=1, j=3
13 1 11 13
13
P H C P CP C H
P H
13 1 11 1 1
2 3 6P H C P C
Monty Hall Problem: Bayes Rule continued
13 13 1 13 2 13 3
13 1 1 13 2 2
, , ,
1 11
6 31
2
P H P H C P H C P H C
P H C P C P H C P C
1 131 6 1
1 2 3P C H
Monty Hall Problem: Bayes Rule continued
1 131 6 1
1 2 3P C H
You should switch!
2 13 1 131 2
13 3
P C H P C H
Continuous Random Variables
What if X is continuous? Probability density function (pdf)
instead of probability mass function (pmf)
A pdf is any function that describes the probability density in terms of the input variable x.
f x
Probability Density Function
Properties of pdf
Actual probability can be obtained by taking the integral of pdf E.g. the probability of X being between
0 and 1 is
0,f x x
1f x
1
0P 0 1X f x dx
Cumulative Distribution Function
Discrete RVs
Continuous RVs
X P XF v v
X P Xi
ivF v v
X
vF v f x dx
X
dF x f x
dx
Common Distributions
Normal
E.g. the height of the entire population
2X ,N
22
1exp ,
2 2
xf x x
-5 -4 -3 -2 -1 0 1 2 3 4 50
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
x
f(x)
Moments
Mean (Expectation): Discrete RVs:
Continuous RVs: Variance:
Discrete RVs:
Continuous RVs:
XE
X P Xii iv
E v v XE xf x dx
2X XV E
2X P X
ii iv
V v v 2XV x f x dx
Properties of Moments
Mean If X and Y are independent,
Variance If X and Y are independent,
X Y X YE E E
X XE a aE
XY X YE E E
2X XV a b a V
X Y (X) (Y)V V V
Moments of Common Distributions
Uniform Mean ; variance
Binomial Mean ; variance
Normal Mean ; variance
X 1, ,U N 1 2N 2 1 12N
X ,Bin n p
np 2np
2X ,N
2
Simulating events by Matlab programs
Write a program in Matlab to distribute the 52 cards of a deck to 4 people, each getting 13 cards. All the choices must be equally likely.
One way to do this is as follows: map each card to a number 1, 2, …, 52. Generate a random permutation of the array a[1 2 … 52], then give the cards a[1:13] to first player, a[14:26] to second player etc.