Stats 242.3(02) – Assignment 3
Solutions
I
Suppose that x1, x2, x3, ..., xn is a sample from the gamma distribution with
parameters 0 and . That is the density of xi is given by:
 0  0 1  x
f ( x|  ) 
x e
  0 
Assume that 0 is a known constant. Find the maximum likelihood estimator of the
unknown parameter .
n
n
   xi
 0  0 1 xi
n 0
 0 1
i


L    f ( xi |  )  
xi e

x
x
x

x
e
1 2 3
n
n
 0 
i 1
i 1  0 
l    ln L   n 0 ln   n ln  0    0  1 ln xi    xi
i
l    n 0
II
1

  xi  0 if ˆ 
i
i
0
x
Suppose that x1, x2, x3, ..., xn is a sample from the density
  1x 
0  x  1,  1

f (x |  )  
 0
elsewhere

Find
i
the method of moments estimator of the unknown parameter .
1

1

0
   xf ( x |  )dx   (  1) x
x
 1

x  2 
 1
dx  (  1)
 
  2o   2

 1
or   2x    1 and  x  1  1  2 x , Thus the method of moments
 2
~ 1  2x
estimator is  
x 1
ii
the maximum likelihood estimator of the unknown parameter .
n
n
i 1
i 1
L    f ( xi |  )     1xi   1  x1 x 2 x3  x n  for   1.
n

l    ln L   n ln   1   ln x1  ln x2  ln x3   ln xn .
n
n
l   
 ln x1  ln x 2  ln x3   ln x n   0 if ˆ  
1
 1
 ln xi
i
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Stats 242.3(02) – Assignment 2
III
Solutions
Suppose that x1, x2, x3, ..., xn is a sample from the density
 1
0  x  2  1
 2  1

f (x |  )  
 0
elsewhere


Find
i
the method of moments estimator of the unknown parameter .

2 1

0
   xf ( x |  )dx  
2 1
 x2 
x
dx  

2  1
 22  1 o

2  12  2  1    1
2
22  1
2
~
x    12 . Thus the method of moments estimator is   x  12
ii
n
L   
i 1
the maximum likelihood estimator of the unknown parameter .
 n
1
0  xi  2  1 for all i
 2  1
i 1

f ( xi |  )  

0
otherwise


 1  n
max xi  2  1


 2  1 


0
otherwise


 n ln 2  1

l    ln L   

0

2n

max xi 1

2
 2  1

l    
 0
  max2xi 1


Hence ˆ  max2xi 1 since l    0 for
max xi  1 / 2  
.
otherwise
max xi 1
2
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
Stats 242.3(02) – Assignment 2
IV
Solutions
Suppose that X and Y are independent unbiased measurements of the angle  and 3
respectively. Namely E(X) =  and E(Y) = 3. In addition assume that X and Y have
the same variance 2.
i) Determine the conditions on a and b that would result in T = aX + bY
being and unbiased estimator of .
ET   EaX  bY   aE X   bEY   a  b3   a  3b   if a + 3b = 1
ii) Determine the values of a and b that would make T = aX + bY the
unbiased estimator of  with the smallest variance.
V  Var T   Var aX  bY   a 2Var  X   b 2Var Y 
 2 1  a 2  
 a  b    a  
 

3

 

dV 
 1  a  1   
V is minimized when
  2a  2
     0
da 
 3  3  
1 a
1
1  a 1  18 8  1 7
That is a 
or 9a = 1 – a and a  . Also b 



9
8
3
3
24
24
7
1
Hence T  8 X  24 Y .
2
2

Page 3