Transcript Lecture 17 Expected value and variance

Expected values and
variances
Formula
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For a discrete random variable X and pmf p(X):
Expected value:
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Variance:
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Alternate formula for variance:
 Var(x)=E(X^2)-[E(X)]^2
Standard Deviation: =sqrt(variance)
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Answers to some questions
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Assuming you are playing a poker game,
each player gets 13 cards. On average,
how many aces should you have at your
hand?
Average number of Aces
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Let X= number of aces a player has at
hand. Then X is a discrete random
variable since it can only take values of 0,
1, 2, 3 and 4. We want to find its pmf and
then find its expected value.
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pmf
X
0
1
2
3
4
P(X)
0.3038
0.4388
0.2135
0.0412
0.0026
Average number of Aces
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The average number of aces a player has
at hand is the expected value of this
discrete random variable X,
 0*0.3038+1*0.4388+2*0.2134+3*0.0412+4*0.
0026=0.9998, almost 1.
 That means if you play poker many, many
times, and if you add up the total number of
Aces you have got and find its average, it
should be around 1.
Calculating variance and standard
deviation for number of Aces
P(X) X Xp(X) X-mu
0.3038
0.4388
0
0 -0.696
1 0.4388 -0.561
0.2135
2
(X-mu)^2 (X-mu)^2P(x)
0.484416
0.314721
0.147165581
0.138099575
0.427 -0.7863 0.61826769
0.132000152
0.0412
3 0.1236 -0.9586 0.91891396
0.037859255
0.0026
4 0.0104 -0.9972 0.99440784
0.00258546
Average number of Aces
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The variance and standard deviation of the
number of Aces are therefore:
Var(X)= 0.457710023, Std. Dev.= 0.676542699
The interpretation of variance and standard
deviation will be more meaningful if we have an
underlying distribution. But here, we do not.
The best we can say is that the number of Aces
changes quite a bit, since you can only have at
most 4 Aces.
Average number of Aces
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An important thing here is that, the variance
and standard deviation we calculate here are
for “number of Aces”.
It is a very common mistake that people may
say they are for “the average number of
Aces”. That is WRONG!!!.
Both expected value and variance(std. dev.)
are used to summarize the number of Aces
you have at hand.
The summary for the “average number of
Aces” is totally a different issue.
Average number of suits
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Assuming you are playing a poker game,
each player gets 13 cards. On average,
how many different suits should you have
at your hand?
Average number of suits
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Let Y=number of suits a player has at
hand. It is also a discrete random variable
whose pmf is:
Y
1
2
3
4
P(Y)
1.57*10^-12
0.0001
0.0510
0.9489
Average number of suits
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The average number of suits a player has
at hand is the expected value of the
discrete random variable Y,
 E(Y)=1*1.57^(-12)+2*0.0001+3*0.051+4*0.9489
=3.9488, almost 4.
 That
means you will almost always have all
four suits in your hand.
Variance and standard deviation
y p(y) yp(y)
y-mu
4 0.9489 3.7956 0.0512
3 0.051 0.153 -0.9488
2 0.0001 0.0002 -1.9488
1
0
0 -2.9488
(y-mu)^2
0.00262144
0.90022144
3.79782144
8.69542144
(y-mu)^2P(y)
0.002487484
0.045911293
0.000379782
0
Variance and standard deviation
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The variance and standard deviation of the
number of suits in your hand are:
Var(Y)= 0.04877856 Std. Dev. 0.220858688
Again, they are for “number of suits in your
hand”, not for “average number of suits in your
hand”.
Purpose of the above two
examples
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1. How to put a given problem in probability
terms ( define the random variable of interest,
find its sample space, etc ).
2. How to find the pmf of the random variable.
3. How to find the expected value and variance
given the pmf of the random variable.
The above two examples are for discrete
random variables, the procedure is similar to
continuous random variables.
Another Example
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If you toss a fair coin three time and let X= the
number of heads observed. Find the expected
value and variance of X.
There are different ways to solve this problem.
 From the three tosses, we have a total of 8
outcomes.
 {HHH, HHT, HTH, THH, HTT, THT, TTH,
TTT}
 Each of the above 8 outcomes has a
probability of 1/8.
Coin Toss Example
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One way of finding the mean is to count
the number of heads in each outcome and
take the average.
 {3,
2, 2, 2, 1, 1, 1, 0}
 The mean is therefore 12/8=1.5
 Then we can find the variance of the 8
numbers, which is:
 [(3-1.5)^2+3*(2-1.5)^2+3*(1-1.5)^2+(01.5)^2]/8=0.75
Coin Toss Example
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Another way is to find the pmf.
X
0
1
2
3
P(X)
1/8
3/8
3/8
1/8
E(X)=0*(1/8)+1*(3/8)+2*(3/8)+4*(1/8)=1.5
 Var(X)=(0-1.5)^2*(1/8)+(1-1.5)^2*(3/8)+(21.5)^2*(3/8)+(3-1.5)^2*(1/8)=0.75
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Coin Toss Example
Yet another way.
 In probability, there is something we call
Bernoulli and Binomial trials, which are
repetitions of exactly the same
experiments with two possible outcomes.
 In this case, we repeat the experiment of
tossing a fair coin 3 times, each time with
50% chance of getting head and 50%
chance of getting tail.
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Coin Toss Example
That is a Binomial experiment, or we say
the (discrete) random variable X follows
a Binomial distribution.
 For Binomial distribution, the outcomes
can be summarized with a pmf that does
not have to look like a table, but like a
function instead.
 Use our knowledge:
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Coin Toss Example
There are easier ways to find the expected
value and variance of a Binomial random
variable.
 If X~BIN(n,p)
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 E(X)=np
 Var(X)=np(1-p)
 In
this case, n=3, p=0.5, so
E(X)=np=3*0.5=1.5 and
Var(X)=3*0.5*0.5=0.75
Properties of Expected Values
and Variances
E(X+Y)=E(X)+E(Y)
 E(X+c)=E(X)+c
 E(aX+bY)=aE(X)+bE(Y)
For example, if E(X)=5 and E(Y)=6, then
E(X+5)=5+5=10
E(2X+5)=2*5+5=15
E(3X+2Y)=3*5+2*6=27
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Properties of Expected Values
and Variances
Var(X+Y)=Var(X)+Var(Y), if X and Y are
independent.
 Var(aX)=(a^2)Var(X)
 Var(X+c)=Var(X)
 Var(aX+bY)=(a^2)Var(X)+(b^2)Var(Y), if X
and Y are independent.
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Properties of Expected Values
and Variances
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Example: if Var(X)=4 and Var(Y)=9, X and
Y are independent.
 Var(2X+3Y)=4*Var(X)+9*Var(Y)=4*4+9*9=97
Properties of Expected Value
and Variances
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Example: if you roll a 6-sided fair die and
win $3 based on the number you get, e.g.,
you get $6 if you roll a 2 and $15 if you roll
a $15. Let Y be the money you collect
from playing this game what is the
expected value and variance of Y.
Properties of Expected Values
and Variances
Let X be the outcome of rolling a fair die,
then Y=3*X. Since we know E(X)=3.5,
E(Y)=3*3.5=$10.5 and
Var(Y)=(3^2)Var(X)=9*3.5=31.5
 The standard deviation is sqrt(31.5)=$5.61
 Note: Mean and standard deviation have
the same unit.
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