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Probability and Statistics Exercise: Normal Distribution and Rayleigh Distribution - Prof., Exámenes de Estadística

Solutions to two statistics exercises from a textbook. The first exercise involves finding the probability that the sample mean of a normally distributed population is greater than a certain value and determining the sample quasivariance such that 5% of the sample variances are less than this value. The second exercise deals with the rayleigh distribution, specifically calculating the bias, mean square error, and consistency of the maximum likelihood estimator for a random variable with a rayleigh distribution. The document also includes some background information on the rayleigh distribution.

Tipo: Exámenes

2014/2015

Subido el 31/10/2015

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Statistics II
30 October 2015
Exercise 1 (5 points)
A random sample of size n = 18 is obtained from a normally distributed population with a population mean of
μ = 46 and a variance of σ2 = 50.
(a) What is the probability that the sample mean is greater than 50?
(b) What is the value of the sample quasivariance such that 5% of the sample variances would be less
than this value?
(From: Newbold, P., W. Carlson and B. Thorne. Statistics for Business and Economics. Pearson-Prentice Hall.)
(a)
P(¯
X>50)=P
(
¯
X−μ
σ2
n
>50−μ
σ2
n
)
=P
(
T>5046
50
18
)
=P(T>2.4)=1P(T2.4 )=10.9918=0.0082
(b) For the sample quasivariance
0.05 =P
(
S2<a
)
=P
(
(n1)S2
σ2<(n1)a
σ2
)
=P
(
T<(181)a
50
)
0.95 =P
(
T(181)a
50
)
(181)a
50 =8.67
a=8.6750
17 =25.5
For the sample variance
0.05 =P
(
s2<a
)
=P
(
n s2
σ2<n a
σ2
)
=P
(
T<18 a
50
)
0.95 =P
(
T18 a
50
)
18 a
50 =8.67
Exercise 2 (5 points)
Let X be a random variable with probability function
f(x ; θ) = x
θ2e
x2
2θ2, x 0, (θ> 0)
such that
E(X)=θ
π
2
and
Var(X)= 4π
2θ2.
Let X = (X1,...,Xn) be a simple random sample. The
application of the method of the moments provides the estimator
^
θM=
2
π¯
X .
For this estimator, calculate
the bias and the mean square error, and study the consistency.
Cultural note: In probability theory and statistics, the Rayleigh distribution is a continuous probability distribution for positive-valued random
variables. A Rayleigh distribution is often observed when the overall magnitude of a vector is related to its directional components. One example
where the Rayleigh distribution naturally arises is when wind velocity is analyzed into its orthogonal 2-dimensional vector components. Assuming
that the magnitudes of each component are uncorrelated, normally distributed with equal variance, and zero mean, then the overall wind speed
(vector magnitude) will be characterized by a Rayleigh distribution. A second example of the distribution arises in the case of random complex
numbers whose real and imaginary components are i.i.d. (independently and identically distributed) Gaussian with equal variance and zero mean.
In that case, the absolute value of the complex number is Rayleigh-distributed. The distribution is named after Lord Rayleigh. (From: Wikipedia.)
Bachelor's Degree in Economics 1 Complutense University of Madrid
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Statistics II

30 October 2015

Exercise 1 (5 points)

A random sample of size n = 18 is obtained from a normally distributed population with a population mean of

μ = 46 and a variance of σ

2

(a) What is the probability that the sample mean is greater than 50?

(b) What is the value of the sample quasivariance such that 5% of the sample variances would be less

than this value?

(From: Newbold, P., W. Carlson and B. Thorne. Statistics for Business and Economics. Pearson-Prentice Hall.)

(a)

P (

X > 50 )= P

X −μ

σ

2

n

50 −μ

σ

2

n

= P

T >

= P ( T > 2.4)= 1 − P ( T ≤ 2.4)= 1 −0.9918=0.

(b) For the sample quasivariance

0.05 = P

S

2

< a

= P

( n − 1 ) S

2

σ

2

( n − 1 ) a

σ

2

= P

T <

( 18 − 1 ) a

→ 0.95 = P

T ≥

( 18 − 1 ) a

( 18 − 1 ) a

= 8.67 → a =

For the sample variance

0.05 = P

s

2

< a

= P

n s

2

σ

2

n a

σ

2

= P

T <

18 a

→ 0.95 = P

T ≥

18 a

18 a

= 8.67 → a =

Exercise 2 (5 points)

Let X be a random variable with probability function

f ( x ; θ) =

x

θ

2

e

x

2

2 θ

2

, x ≥ 0, (θ> 0 )

such that E ( X )=θ

π

and Var ( X )=

4 −π

θ

2

Let X = ( X 1

,...,X

n

) be a simple random sample. The

application of the method of the moments provides the estimator

^

θ

M

π

X. For this estimator, calculate

the bias and the mean square error, and study the consistency.

Cultural note: In probability theory and statistics, the Rayleigh distribution is a continuous probability distribution for positive-valued random

variables. A Rayleigh distribution is often observed when the overall magnitude of a vector is related to its directional components. One example

where the Rayleigh distribution naturally arises is when wind velocity is analyzed into its orthogonal 2-dimensional vector components. Assuming

that the magnitudes of each component are uncorrelated, normally distributed with equal variance, and zero mean, then the overall wind speed

(vector magnitude) will be characterized by a Rayleigh distribution. A second example of the distribution arises in the case of random complex

numbers whose real and imaginary components are i.i.d. (independently and identically distributed) Gaussian with equal variance and zero mean.

In that case, the absolute value of the complex number is Rayleigh-distributed. The distribution is named after Lord Rayleigh. (From: Wikipedia.)

Bachelor's Degree in Economics 1 Complutense University of Madrid

Mean or expectation : E

^

θ

M

= E

(

π

X

)

π

E

X

π

E

X

π

θ

π

Bias : b

^

θ

M

= E

^

θ

M

−θ=θ−θ= 0 →

^

θ

M

is an unbiased estimator of θ.

Variance : Var

^

θ

M

= Var (

π

X )=

π

Var (

X )=

π

Var ( X )

n

π

( 4 −π)

2 n

θ

2

( 4 −π)

π n

θ

2

Mean square error : ECM

^

θ

M

= b

^

θ

M

2

  • Var

^

θ

M

( 4 −π)

π n

θ

2

( 4 −π)

π n

θ

2

Consistency : lim

n →∞

MSE

^

θ

M

=lim

n →∞

( 4 −π)

π n

θ

2

= 0 and therefore σ^

M

is consistent (for θ).

Additional Exercise

In exercise 1,

(c) What is the value of the sample quasivariance such that 5% of the sample variances would be greater

than this value?

(d) Calculate E (S

2

Hint: In the last section, select the proper statistic T and make it appear; then apply the basic properties of the mean and the variance

as well as the information of following probability distribution:

(c) For the sample quasivariance

0.05 = P

S

2

> a

= P

(

( n − 1 ) S

2

σ

2

( n − 1 ) a

σ

2 )

= P

(

T >

( 18 − 1 ) a

)

( 18 − 1 ) a

= 27.59 → a =

For the sample variance

0.05 = P

s

2

> a

= P

(

n s

2

σ

2

n a

σ

2 )

= P

(

T >

18 a

)

18 a

a =

(d) For the sample quasivariance we can do

E

S

2

= E

(

σ

2

( n − 1 )

( n − 1 ) S

2

σ

2

)

σ

2

( n − 1 )

E ( T )=

σ

2

( n − 1 )

κ=

σ

2

( n − 1 )

( n − 1 )=σ

2

or use that the sample quasivariance is unbiased for the population variance. For the sample variance

E

s

2

= E

(

σ

2

n

n s

2

σ

2 )

σ

2

n

E ( T ) =

σ

2

n

κ=

σ

2

n

( n − 1 )=

Bachelor's Degree in Economics 2 Complutense University of Madrid