Post on 31-Dec-2015
2
Random Variables (Discrete)
Real-valued functions defined on a sample space are random vars. determined by outcome of experiment, we can assign probability to possible values of it (them).
Exs: Toss 3 fair coins, let Y be number of H appearing, then Y is a random var. with possible values (0,1,2,3) with probability
P{y = 0} = P(T, T, T) = 1/8
P(y = 1) = P(T, T, H), (T, H,T), (H, T, T) = 3/8
P(y = 2) = P(T, H, H), (H, T, H), (H, H, T) = 3/8
P(y = 3) = P(H, H, H) = 1/8
Since y is between 0 and 3,
3
0
3
0
)(}){(1
i i
iyPiyP
3
Distribution functions
The c.d.f. or distribution function F of random variable x is defined for all real numbers b,
F(b) = P{x b} denotes prob. that x takes on a value b
Properties
1. F is a non-decreasing function, if a b F(a) < F(b)
2.
3. . . .= 0
4. F is right continuous. For any b and any decreasing sequence bn, n 1 that converges to b,
b
1)(lim
bFb
blim
)()(lim bFbF nn
4
Exs: Distribution function of a random variable x is given
0 x 0
x/2 0 x 1
F(x) = 2/3 1 x 2
11/12 2 x 3
1 3 x
Calculate P{x 3}
1211)13(lim}13{lim nFnxP
nn
5
Calculate P{x = 1}
Calculate P{x ½}
Calculate P(2 x 4)
(draw graph)
61
21
32)
11(lim)1()1()1( n
FFxPxP
43)2
1(1}21{1}2
1{ FxPxP
121)2()4()42( FFxP
6
Discrete Random VariablesA random variable that can take on at most a countable
number of possible values.
Probability mass function P(a) = P{x = a}
If x can be x1, x2, x3, . . . thenP(xi) 0, i = 1, 2, . . .P(x) = 0, else and
If we have P(0) = ¼ P(1) = ½ P(2) = ¼ (draw graph)
The plot for a random var. of sum of 2 dice are: (draw graph)
1
1)(i
xiP
7
Cumulative Distr. Function F can be expressed in terms of mass function as F(a) = p(x) all x a
If x is discrete random variable where x1 x2 x3 . . . then distrib. function F is a step function. The value of F is constant in intervals [xi-1, xi) and with a step (jump) of size p(xi) at xi
Exs:x has probability mass functionP(1) = ¼ p(2) = ½ p(3) = 1/8 P(4) = 1/8
8
Then, the c.d.f is
0 a 1
¼ 1 a 2
F(x) = ¾ 2 a 3
7/8 3 a 4
1 4 a
Step size at any of values 1, 2, 3, 4 is equal to the probability that x assumes a particular value.
(draw graph)
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Expected Value: if x is a discrete random variable with probability mass function p(x) then
expectation or expected value
This is a weighted average of possible values x can take on each value, weighted by probability that x assumes it.
If p(0) = ½ = p(1)
E[x] = 0*1/2 + 1 ½ = ½ ordinary avg. of 2 possible values 0 and 1 x can assume
Or if p(0) = ½ p(1) = 2/3 E[x] = 0*(1/2) + 1*(2/3) = 2/3 where value of 1 is given twice as much weight as value 0 so, p(1) = 2p(0) 2/3 = 2*(1/3)
0)(:
)(][xpx
xxpxE
10
Frequency Interpretation if an infinite sequence of independent replications of an experiment is done, then for any event E, the proportion of time that E occurs is p(E). So, if x can be x1, x2, x3, . . . with p(x1), p(x2), . . . then, average expectation
Exs:
Find E[x] where x is outcome of a roll of fair die
Since p(1) = p(2) = . . . p(6) = 1/6
Then, E[x] = 1(1/6) + 2(1/6) + 3(1/6) + . . . 6(1/6) = 7/2
n
i
ii xpxxE1
)(][
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Exs
If I is a stock indicator for trading volume V then
I = 1 if V true Find E[I]
0 if VC true
Since p(1) = p(v), p(0) = 1 – p(A) we haven then E[I] = p(A)
Expected value of indicator is the probability that event occurs.
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Expectation of a function of a Random Variable: Having a discrete random variable and its probability mass
function, we need expected value of some function x, g(x).One method:*calculate the prob. mass function of g(x) since it is a discrete
random variable*calculate E[g(x)] by using expected value definition
Exs X is a random variable with any value in –1, 0, 1 with x -1 x 0 x 1P(-1) = .2 P(0) = .5 P(1) = .3
Calculate E[x2] :
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Let y = x2, it follows that prob. mass function of y is
P(y = 1) = P(x = -1) + P{x = 1} = .5
P{y = 0} = P{x = 0} = .5
hence, E[x2] = E[y] = 1(.5) + 0(.5) = .5
note that .5 = E[x2] (E[x])2 .01
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Method 2:
if x is discrete random variable that takes on values xi, i ≥ 1 with P(xi) then for any real-valued function g,
Going back to previous example:
E[x2] = (-1)2 (.2) + 02 (.5) + 12 (.3)
= 1(.2 + .3) + 0 (.5)
= .5
i
ii xpxgxgE )()()]([
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Corollary
if a and b are constants then E[aX + b] = a E[X] + b
Because E[aX + b] =
=
The expected value of a random variable x, E[x] is also referred to as the mean or 1st moment of x. The quantity E[xn], n 1 is the nth moment of x and, E[xn] =
0)(:
)()(xpx
xpbax
0)(: 0)(:
][)()( xpx xpx
bxaExpbxxpa
)(0)(:
xpxxpx
n
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Variance
Given x with distribution function F and E[x]. E[x] is weighted average of all possible values for x. We need the variation or spread of these values. If x has a mean μ, then variance of x, Var(x) = E[(x – μ)2]
Or
That is also Var (x) = E[x2] – (E[x])2
22222
22
22
2
][2][
)()(2)(
)()2(
)()(
xExE
xpxxpxpx
xpxx
xpx
xx x
x
x
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Exs
Calculate Var(x) if x is outcome of a fair die rolled
from previous example, E[x] = 7/2 and
E[x2] = 12(1/6) + 22 (1/6) + 32(1/6) + . . . 62 (1/6) = 91/6
Hence, Var(x) = 91/6 – (7/2)2 = 35/12
Identity: for any constants a and b Var(ax + b) = a2 Var(x)
*note that analogous to the mean being center of gravity of a distrib. of mass, Variance represents moment of inertia
*The SQRT of Var(x) is the STD. deviation of x denoted by SD(x) =
Discrete random variables are classified according to their prob. mass function.
)(xVar
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Exs:
52-card deck is well-shuffled and then cards are turned face up one by one til Ace appears. Find expect number of cards that are face up
Let x be number cards face up til ace appears
A = { no ace among first i – 1 cards turned up }
B = { ith card is ace }
P(x = i) = P(AB) = P(B/A) P(A) =
P(x) =
)(
)(.
)1(52
452
1
481
i
i
i
49
152
1
481 116.10
)52)((
4)(
i i
i
i
i
19
Exs:
What are Expected number, Var., and S.D. of the number of spades in a poker hand? (P.H. = set of 5 cards randomly picked from 52)
A = pick any 12 spades
B = rest of cards
Thus, Var(x) = 2.43 - (1.25)2 = 0.864 and σx =
430.2)(
))((][
25.1)(
))((][
5
0525
395
1322
5
0525
395
13
i
ii
i
ii
ixE
ixE
93.864.
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Exs
A professor made 30 exams, 8 tough, 12 medium, 10 easy; exams are mixed up and 4 are selected at random. How many will be difficult?
x number of difficult ones
We need E(x), so
x can be 0, 1, 2, 3, 4 and its probability function is
Values of all p(i.s) i 0 1 2 3 4
p(i) 0.27 .45 .24 .04 .003
E(x) = 0(.27) + 1(.45) + 2(.24) + 3(.04) + 4(.003) = 1.06
4,3,2,1,0,)(
))(()()(
304
224
8
iixpip ii
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Continuous Random Variables
Set of possible values is uncountable (lifetime of a transistor, mars rovers, etc)
X is continuous random variable if there exists a nonnegative function f, defined for all real having the property that any set B of real numbers:
Probability density function of x
(states that probability that x will be in B may be obtained by integrating the p.d.f. over set B) it must also satisfy
(all probability stmts about x can be answered in terms of f)
),( x
B
dxxfBxP )(}{
dxxfxP )()},({1
22
If B = [a, b] then
if we let a = b then
(means that probability of a continuous random variable will assume any fixed value is zero).
b
adxxfaxP 0)(}{
adxxfaFaxPaxP )()(}{}{
b
adxxfbxaP )(}{
23
Exs:
Suppose x is continuous random variable whose p.d.f. is
else
What is value of C?
Since f is a p.d.f., we have
and thus,
Find P(x 1)?
)24(0
2
)( xxCxf 20 x
1)( dxxf
2
0
2 1)24( dxxxC
]3
22[
32 xxC 8
3120
Cxx
1
2
1
28
32
1)24()()1( dxxxdxxfxP
24
Exs
Lifetime in days of a MEMS wireless transceiver is a random variable having p.d.f. given by
f(s) 0 x 100
100/x2 x > 100
What’s probability that 2 of 5 such devices needs replacing within 150 days of continuous operation?
25
Exs
Assume that events Ei, i = 1, 2, 3, 4, 5
ith transceiver will need replacement within the 150 days are independent
thus, from independence of events Ei, the probability is
P(E) P(EC)
150
0
150
100
2
31100)()( dxxdxxfEiP
24380)3
2()3
1)(( 3
3
0
252
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Exs Loss in a stock option, in thousands of dollars, is a continuous random var. x with density function:
f(x) = k(2x – 3x2) -1 < x < 0
0 else
Calculate k and find prob. the loss is at most $500
Since f is a p.d.f. ,but
1)( dxxf
32
0
1
0
1
22
[
)32()32()(
xxk
dxxxkdxxxdxxf
k201
28
Relationship between c.d.f. F and p.d.f. f is
Differentiating both sides,
Density is the derivative of cumulative distr. function. Also, we can say that, from
adxxfaxPaF )()},({)(
)()( afaFda
d
2
2
)()(}22
{
)(}{
a
a
b
a
afdxxfaxaP
dxxfbxaP