Transcript Hatfield.Topic 5

Topic 5 - Joint distributions and the CLT
• Joint distributions
– Calculation of probabilities, mean and variance
– Expectations of functions based on joint distributions
• Central Limit Theorem
– Sampling distributions
• Of the mean
• Of totals
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• Often times, we are interested in more than one random
variable at a time.
• For example, what is the probability that a car will have
at least one engine problem and at least one blowout
during the same week?
• X = # of engine problems in a week
• Y = # of blowouts in a week
• P(X ≥ 1, Y ≥ 1) is what we are looking for
• To understand these sorts of probabilities, we need to
develop joint distributions.
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Discrete distributions
• A discrete joint probability mass function is given by
f(x,y) = P(X = x, Y = y)
where
1. f (x , y )  0 for all x , y
2.

all (x ,y )
f (x , y )  1
3. P (( X ,Y )  A ) 
4. E (h ( X ,Y )) 


all ( x ,y )A
all (x ,y )
f (x , y )
h (x , y ) f (x , y )
3
Return to the car example
• Consider the following joint pmf for X and Y
X\Y
0
1
2
3
4
0
1/2
1/16 1/32 1/32 1/32
1
1/16 1/32 1/32 1/32 1/32
2
1/32 1/32 1/32 1/32 1/32
• P(X ≥ 1, Y ≥ 1) =
• P(X ≥ 1) =
• E(X + Y) =
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Joint to marginals
• The probability mass functions for X and Y individually
(called marginals) are given by
f X (x )   all y f (x , y ), fY (y )   all x f (x , y )
• Returning to the car example:
fX(x) =
fY(y) =
E(X) =
E(Y) =
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Continuous distributions
• A joint probability density function for two continuous
random variables, (X,Y), has the following four properties:
1. f (x , y )  0 for all x , y
 
2.
  f (x , y )dxdy  1
- -
3. P (( X ,Y )  A ) 

A
f (x , y )dxdy
 
4. E (h ( X ,Y )) 
  h (x , y ) f (x , y )dxdy
- -
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Continuous example
• Consider the following joint pdf:
x (1  3y 2 )
f (x , y ) 
4
0  x  2, 0  y  1
• Show condition 2 (total volume is 1) holds on your own.
• Show P(0 < X < 1, ¼ < Y < ½) = 23/512
x(1  3 y 2 )
P(0  x  1,1/ 4  y  1/ 2)   
dydx
4
0 1/ 4
1 1/ 2
1
 1/ 4 x[ y  y 3 ]
0
1
y 1/ 2
y 1/ 4
1
dx 1/ 4  x[5 / 8  17 / 64]dx
0
 23 / 256  xdx  23 / 256[ x 2 / 2]
0
x 1
x 0
 23 / 256[1/ 2  0]  23 / 512
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Joint to marginals
• The marginal pdfs for X and Y can be found by

f X (x ) 
 f (x , y )dy,


fY (y ) 
 f (x, y )dx

• For the previous example, find fX(x) and fY(y).
x(1  3 y 2 )
f x ( x)  
dy = x / 4[ y  y 3 ]
4
0
1
y 1
y 0
= x / 4[2  0]  x / 2
x(1  3 y 2 )
(1  3 y 2 )
(1  3 y 2 ) 2
f y ( y)  
dx =
xdx =
[ x / 2]

4
4
4
0
0
2
1 3y2

2
2
x2
x 0
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Independence of X and Y
• The random variables X and Y are independent if
– f(x,y) = fX(x) fY(y) for all pairs (x,y).
• For the discrete clunker car example, are X and Y
independent?
• For the continuous example, are X and Y independent?
x(1  3 y 2 )
x (1  3 y 2 ) x(1  3 y 2 )
f ( x, y ) 
 f x ( x) f y ( y )  (
)

4
2
2
4
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Sampling distributions
• We assume that each data value we collect represents a
random selection from a common population distribution.
• The collection of these independent random variables is called
a random sample from the distribution.
• A statistic is a function of these random variables that is
used to estimate some characteristic of the population
distribution.
• The distribution of a statistic is called a sampling
distribution.
• The sampling distribution is a key component to making
inferences about the population.
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Statistics used to infer parameters
We take samples and calculate statistics to make
inferences about the population parameters.
Sample
Mean
x
Std. Dev.
s
2
Variance
s
Proportion
p̂




Population



p
2
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StatCrunch example
• StatCrunch subscriptions are sold for 6 months ($5) or 12
months ($8).
• From past data, I can tell you that roughly 80% of
subscriptions are $5 and 20% are $8.
• Let X represent the amount in $ of a purchase.
• E(X) =
• Var(X) =
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StatCrunch example continued
• Now consider the amounts of a random sample of two
purchases, X1, X2.
• A natural statistic of interest is X1 + X2, the total amount
of the purchases.
Outcomes X1 + X2 Probability
5,5
X1 + X2 Probability
5,8
8,5
8,8
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StatCrunch example continued
• E(X1 + X2) =
• E([X1 + X2]2) =
• Var(X1 + X2) =
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StatCrunch example continued
• If I have n purchases in a day, what is
– my expected earnings?
– the variance of my earnings?
– the shape of my earnings distribution for large n?
• Let’s experiment by simulating 10,000 days with 100
purchases per day using StatCrunch.
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Simulation instructions
•
Data > Simulate data > Binomial
•
Specify Rows to be 10000, Columns to be 1, n to be 100 and p
to be .2. This will give you a new column called Binomial1
•
To compute the total for each day, go to Data > Transform data
and enter the expression, 8*Binomial1+5*(100-Binomial1). This
will add a new column to the data table.
•
Make a histogram and set the bin width to 1 for best results.
•
For the new sum column, do a histogram and a QQ plot. Both
should verify normality!
StatCrunch
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Should result in a dataset like this
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Central Limit Theorem
• We have just illustrated one of the most important
theorems in statistics.
• As the sample size, n, becomes large the distribution of
the sum of a random sample from a distribution with
mean  and variance 2 converges to a Normal
distribution with mean n and variance n2.
• A sample size of at least 30 is typically required to use
the CLT (arguable in the general statistics community).
• The amazing part of this theorem is that it is true
regardless of the form of the underlying distribution.
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Airplane example
• Suppose the weight of an airline passenger has a mean of
150 lbs. and a standard deviation of 25 lbs.
• What is the probability the combined weight of 100
passengers will exceed the maximum allowable weight of
15,500 lbs?
• How many passengers should be allowed on the plane if
we want this probability to be at most 0.01?
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What are the probabilities at n = 99?
The mean is 99*150  14850
The variance is 99 * 252  61850
The standard deviation is 61875  248.75
P( X TOT  15500)
 0.004487
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The distribution of the sample means
• For constant c, E(cY) = cE(Y) and Var(cY) = c2Var(Y)
•
2
1
1
1

2
• Var ( X )  Var (  x )  Var (  x ) 
n 
2
2
n
n
n
n
• The CLT says that for large samples, X is approximately
normal with a mean of  and a variance of 2/n.
• So, the variance of the sample mean decreases with n.
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What are the probabilities we get a
sample average at some level?
If the parent population is assumed with a mean of 150 lbs.
and a standard deviation of 25 lbs., what’s the probability
we get a sample average below 141 with a sample size of 30?
Talking about the
sampling distribution,
the mean is 150 and the
standard deviation is
25
 4.5644
30
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Sampling distribution applet
• In StatCrunch, go to the “Applets” tab and click
on “sampling distributions”. It will demonstrate
how any parent distribution will converge to
normal with larger, repeated samples.
• The closer the parent is to symmetrical, the
quicker the sampling distribution will converge.
The additional file for Topic 5 has discussion and examples
on both sampling distributions and joint probability
distributions. There are also additional examples of double
integration.
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