University of Wollongong STAT333/STAT833 Assignment 1 Solution

This document presents a detailed solution to Assignment 1 for STAT333 and STAT833, focusing on statistical inference. The solution includes finding the marginal probability density function (pdf) of Y, calculating the mean and variance of Y, determining the conditional pdf of X given Y, and computing the expected value of X. It also covers finding the joint moment generating function, the likelihood function, and the maximum likelihood estimate. Additionally, the solution addresses the score function, observed information, and confirms that certain estimates are unbiased and consistent. The assignment utilizes concepts from probability models and statistical analysis, providing step-by-step explanations and derivations to guide the reader through complex statistical calculations.

Let X, Y be continuous random variables with joint probability density function
f X , Y ( x , y ) = {α βα y− ( α +1 ) e− ( x− y ) 0 <β< y< x< ∞
0 Otherwise
With parameters α >2 and β >0.
a) Find the marginal pdf (probability density function) f Y ( y) of Y.
Solution
We compute the marginal pdf:
f Y ( y ) =∫
−∞
∞
f ( x , y ) dx
¿
{∫
y
∞
( α βα y− (α+ 1 ) e−( x− y ) ) dx 0< y < x
0 Otherwise
b) Find the expressions for the mean and variance of Y
Solution
Mean M ( t )=E ( ety )=∫
−∞
∞
ety f X , Y ( y ) dy
M ( t ) =∫
0
∞
ety α βα y− ( α +1 ) e−(x− y) dy
Variance μ= E (Y )=∫
−∞
∞
y f Y ( y ) dy
μ=∫
0
∞
yα βα y− ( α +1 ) e−(x− y) dy
c)
i) Find the conditional pdf of X given Y = y, and hence find E( X∨Y ).
Solution
f X∨Y ( x|y ) = f XY (x , y)
f Y ( y )
¿ α βα y− (α+1 ) e− ( x− y )
α βα y− ( α+1 ) e y
E ( X |Y ) =∫
−∞
∞
x f X ∨Y ( x| y ) dx

¿∫ x α βα y− ( α+ 1 ) e− ( x− y )
α βα y− ( α +1 ) e y dx
¿−e−x x−e− x
ii) Hence calculate the expected value of X.
Solution
E [ X |Y = y ] = [ −e− x x−e− x ]2
∞
¿ 3
e2 ≈ 0.406
d) Find the joint moment generating function M(s, t) of (X, Y) for s<0 , t<0.
Solution
M ( s , t )=E ( est ) =∫
x1
x2
est f X ,Y ds(Ross, 2014)
¿∫
0
∞
est α βα y− ( α+1 ) e−( x− y ) ds
e) Determine the likelihood function L ( α , β|y )=∏
i=1
n
f Y ( yi∨α , β)
Solution
L ( α , β| y ) =∑
i=1
n
log f ( Xi ∨θ)
¿ ∑
i=1
n
(¿ log α + log βα−log Γ +log y− ( α +1 ) +log e− ( x− y ) )¿
¿ n log α +nα log β−n log Γ −n ( α + 1 ) log y − ( x− y ) log e
f) Show that ( T α ( y ) , T β ( y ) ) =
( ∏
i=1
n
yi , y ( 1 ) ) is minimally sufficient for ( α , β ) .
Solution
For every set of nonnegative integers x1 , … . xn, the joint probability mass function
f n ( X ∨θ) of X1 … Xn is as follows (Grace, 2017);
f n ( X |θ ) =∏
i=1
n e−θ θxi
xi !
¿ ∏
i=1
n
yi , y(1)

g) Assume that β is known:
i) Determine the score function of α given β.
Solution
From (e) above, we have;
L ( α , β|y )=n logα +nα log β−n log Γ −n ( α +1 ) log y − ( x− y ) log e
The score function will be a vector;
( ∂
∂ α ln L ( α , β| y ) , ∂
∂ β ln L ( α , β | y ) )
ii) Show the maximum likelihood estimate of α given α = 1
1
n ∑
i=1
n
ln yi−ln β
Solution
∂
∂ α ln L ( α , β |y ) =n ( d
dα ln α+ ln β−α ln y− d
dα ln e )=0(Garg, 2017)
Solving, we obtain;
α= 1
1
n ∑
i=1
n
ln yi−ln β
iii) Find the observed information J ( α) and confirm that α is a maximum.
Solution
J ( α ) =∑
i=1
n
α 2−λ ( ∑
i=1
n
α i−1 )
¿
( 1
1
n ∑
i=1
n
ln yi−ln β )
2
−λ
( 1
1
n ∑
i=1
n
ln yi−ln β )
Checking if α is maximum;
∂ J
∂ α =0
Hence α is maximum.
iv) Confirm that T α ( y ) can be expressed as a function of α .
Solution
T α ( y ) =n log α +nα log β−n log Γ−n ( α +1 ) log y− ( x− y ) log e

v) Show that Zi Gamma(1 ,α ) where Zi =ln ( Y i )−ln(β ) for i=1 , … . ,n.
Solution
P ( λ|Y ) ∝ P ( Y |λ ) P( λ)
∝ λn e−λ
∑ y λ− (α+ 1 ) α βα e− ( x−λ )
Hence, Zi Gamma(1 ,α )
h) The cumulative distribution function of Y is given by:
P ( Y ≤ y )=FY ( y ) =
{1−( β
y )α
β< y< ∞
0 Otherwise
With the maximum likelihood estimate ^β of β given by ^β= y(1).
i) Show that: P ( β ≤ b ) =F ^β ( b ) = {1− ( β
b )
nα
0< β <b< ∞
0 Otherwise
Solution
P ( β ≤ b ) =F ^β ( b ) =∫
−∞
b
¿ ¿
¿
{1−( β
y ) nα
0< β <b< ∞
0 Otherwise
ii) Hence find the pdf of ^β.
Solution
f ( y ) =F ( y )'
¿ d
dy ( 1−( β
y ) α
)
¿ α βα
yα+1
iii) Determine whether ^β is an unbiased estimate β.
Solution
E ( X ) =1
n ∑
i=1
n
(1−( β
y )α
)

¿ 1− ( β
y )
α
Thus, ^β is an unbiased estimate β.
iv) Show that ^β is a consistent estimator of β
Solution
^β=β + ( X T X )−1
XT u
(lim
n → ∞
1
n X T X )−1
lim
n → ∞
1
n X T u= ( SX T X )−1 lim
n → ∞
1
n X T u
¿ 0
References
Garg, H., 2017. A new improved score function of an interval-valued Pythagorean fuzzy
set based TOPSIS method. International Journal for Uncertainty Quantification, 7(5).
Grace, Y.Y., 2017. Multi-State Models with Error-Prone Data. In Statistical Analysis
with Measurement Error or Misclassification (pp. 257-300). Springer, New York, NY.
Ross, S.M., 2014. Introduction to probability models. Academic press, Pp. 44-47.