Download Continuous Random Variables

Continuous Random
Variables
Lecture 26
Section 7.5.4
Mon, Mar 5, 2007
Uniform Distributions

The uniform distribution from a to b is
denoted U(a, b).
1/(b – a)
a
b
Hypothesis Testing (n = 1)

An experiment is designed to determine
whether a random variable X has the
distribution U(0, 1) or U(0.5, 1.5).
 H0:
X is U(0, 1).
 H1: X is U(0.5, 1.5).

One value of X is sampled (n = 1).
Hypothesis Testing (n = 1)

An experiment is designed to determine
whether a random variable X has the
distribution U(0, 1) or U(0.5, 1.5).
 H0:
X is U(0, 1).
 H1: X is U(0.5, 1.5).
One value of X is sampled (n = 1).
 If X is more than 0.75, then H0 will be
rejected.

Hypothesis Testing (n = 1)

Distribution of X under H0:
1
0

0.5
1
1.5
1
1.5
Distribution of X under H1:
1
0
0.5
Hypothesis Testing (n = 1)

What are  and ?
1
0
0.5
1
1.5
0
0.5
1
1.5
1
Hypothesis Testing (n = 1)

What are  and ?
1
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
1
Hypothesis Testing (n = 1)

What are  and ?
1
0
0.5
0.75
Acceptance Region
1
1.5
Rejection Region
1
0
0.5
0.75
1
1.5
Hypothesis Testing (n = 1)

What are  and ?
1
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
1
Hypothesis Testing (n = 1)

What are  and ?
 = ¼ = 0.25
1
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
1
Hypothesis Testing (n = 1)

What are  and ?
 = ¼ = 0.25
1
0
1
0.5
0.75
1
1.5
0.5
0.75
1
1.5
 = ¼ = 0.25
0
Example
Now suppose we use the TI-83 to get two
random numbers from 0 to 1, and then
add them together.
 Let X2 = the average of the two random
numbers.
 What is the pdf of X2?

Example

The graph of the pdf of X2.
f(y)
?
y
0
0.5
1
Example

The graph of the pdf of X2.
f(y)
2
Area = 1
y
0
0.5
1
Example

What is the probability that X2 is between
0.25 and 0.75?
f(y)
2
y
0
0.25
0.5
0.75
1
Example

What is the probability that X2 is between
0.25 and 0.75?
f(y)
2
y
0
0.25
0.5
0.75
1
Example

The probability equals the area under the
graph from 0.25 to 0.75.
f(y)
2
y
0
0.25
0.5
0.75
1
Example

Cut it into two simple shapes, with areas
0.25 and 0.5.
f(y)
2
Area = 0.25
0.5
Area = 0.5
y
0
0.25
0.5
0.75
1
Example

The total area is 0.75.
f(y)
2
Area = 0.75
y
0
0.25
0.5
0.75
1
Verification
Use Avg2.xls to generate 10000 pairs of
values of X.
 See whether about 75% of them have an
average between 0.25 and 0.75.

Hypothesis Testing (n = 2)

An experiment is designed to determine whether
a random variable X has the distribution U(0, 1)
or U(0.5, 1.5).
 H0:
X is U(0, 1).
 H1: X is U(0.5, 1.5).



Two values of X are sampled (n = 2).
Let X2 be the average.
If X2 is more than 0.75, then H0 will be rejected.
Hypothesis Testing (n = 2)

Distribution of X2 under H0:
2
0

0.5
1
1.5
Distribution of X2 under H1:
2
0
0.5
1
1.5
Hypothesis Testing (n = 2)

What are  and ?
2
0
0.5
1
1.5
0
0.5
1
1.5
2
Hypothesis Testing (n = 2)

What are  and ?
2
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
2
Hypothesis Testing (n = 2)

What are  and ?
2
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
2
Hypothesis Testing (n = 2)

What are  and ?
2
 = 1/8 = 0.125
0
0.5
0.75
1
1.5
0
0.5
0.75
1
1.5
2
Hypothesis Testing (n = 2)

What are  and ?
2
 = 1/8 = 0.125
0
2
0.5
0.75
1
1.5
0.75
1
1.5
 = 1/8 = 0.125
0
0.5
Conclusion

By increasing the sample size, we can
lower both  and  simultaneously.
Example
Now suppose we use the TI-83 to get
three random numbers from 0 to 1, and
then average them.
 Let X3 = the average of the three random
numbers.
 What is the pdf of X3?

Example

The graph of the pdf of X3.
3
y
0
1/3
2/3
1
Example

The graph of the pdf of X3.
3
Area = 1
y
0
1/3
2/3
1
Example

What is the probability that X3 is between
1/3 and 2/3?
3
y
0
1/3
2/3
1
Example

What is the probability that X3 is between
1/3 and 2/3?
3
y
0
1/3
2/3
1
Example

The probability equals the area under the
graph from 1/3 to 2/3.
3
Area = 2/3
y
0
1/3
2/3
1
Verification
Use Avg3.xls to generate 10000 triples of
numbers.
 See if about 2/3 of the averages lie
between 1/3 and 2/3.

Hypothesis Testing (n = 3)

An experiment is designed to determine whether
a random variable X has the distribution U(0, 1)
or U(0.5, 1.5).
 H0:
X is U(0, 1).
 H1: X is U(0.5, 1.5).


Three values of X3 are sampled (n = 3). Let X3
be the average.
If X3 is more than 0.75, then H0 will be rejected.
Hypothesis Testing (n = 3)

Distribution of X3 under H0:
0

1/3
2/3
1
4/3
1
4/3
Distribution of X3 under H1:
1
0
1/3
2/3
Hypothesis Testing (n = 3)

Distribution of X3 under H0:
 = 0.07
0

1/3
2/3
1
4/3
1
4/3
Distribution of X3 under H1:
1
 = 0.07
0
1/3
2/3
Example
Suppose we get 12 random numbers,
uniformly distributed between 0 and 1,
from the TI-83 and get their average.
 Let X12 = average of 12 random numbers
from 0 to 1.
 What is the pdf of X12?

Example

It turns out that the pdf of X12 is nearly
exactly normal with a mean of 1/2 and a
standard deviation of 1/12.
N(1/2, 1/12)
x
1/3
1/2
2/3
Example
What is the probability that the average
will be between 0.45 and 0.55?
 Compute normalcdf(0.45, 0.55, 1/2, 1/12).
 We get 0.4515.

Experiment
Use the Excel spreadsheet Avg12.xls to
generate 10000 values of X, where X is
the average of 12 random numbers from
U(0, 1).
 Test the 68-95-99.7 Rule.
