STA331 TEST 1 – 29 February 2012
Time: 1 hour
Marks: 35
Question 1 (7)
(i)
Define (p) .
(ii)
1
(ii) Show that ( ) 
2
(iii)
Define μ r and μ r for a continuous random variable X .
π.
Question 2 (18)
If the probability density function (p.d.f.) of X is f x ( x )  e  x , x  0
 0 otherwise
(i) Show that M x ( t )  (1  t ) 1
(ii) Hence, obtain μ′
r for this distribution
Using the transformation theorem show that the probability density function ( p.d.f .) of y  x 2 (a
monotonic function over the range x  0) and x having p.d.f. in (i)) is:
f y ( y) 
1
.e
 y
, 0  y
2 y
0
otherwise
(iii) Hence show that
E[Y]  2E[X]
Question 3 (10 marks)
A random variable X has a chi-square distribution with n degrees of freedom with p.d.f .
f (x) 
1
n
2 2 ( n2 )
n 1
x2 e
0
x
2
0x
otherwise
n
Show that the cumulant generating function (c.g.f.) of X is K X ( t )   n (1  2t )
2
Hence, show that for the above chi-square distribution μ1  n ; μ 2  var( X)  2n
Hint: n (1  x)  x 
x 2 x3


2
3
for x  1
Memorandum
Question 1 [7]
(i) Define (p) .
1
(ii) Show that ( ) 
2
π.
(ii) Define μ r and μ r for a continuous random variable X .
Solution:

(i) (p)   x p1e  x dx
for p  0
0
1
 2
(1)
 1
 2
(ii)      !  π

1

1
    x 2 e  x dx
2 0
put x =
1 2

2 2y
 
e
y dy 
y
0
1 2
2  2y

dy
e
2 0

1 2
y then dx = y dy
2
2
  1 y2
e 2 dy

0
and as e
1
 y2
2
is an even function
2
. 2π  π
2
(4)
(iii) If X has pdf f x (x) μ r   x r f x (x)dx ; μ r   (x  μ1 ) r f x (x)dx
x
x
(2)
Question 2 (18)
If the probability density function (p.d.f.) of X is f x ( x )  e  x , x  0
 0 otherwise
(i) Show that M x ( t )  (1  t ) 1
(ii) Hence, obtain μ′
r for this distribution
Using the transformation theorem show that the probability density function ( p.d.f .) of y  x 2 (a
monotonic function over the range x  0) and x having p.d.f. in (i)) is:
f y ( y) 
1
2 y
.e
 y
, 0  y
0
otherwise
(iii) Hence show that
E[Y]  2E[X]
Solution:


0
0
(i) M x ( t )  E[e xt ]   e xt e  x dx   e x (1 t ) dx

 0
(ii) Now
e  x (1t ) 
] If t  1
 (1  t ) 0
1
 (1  t ) 1
(1  t )
1
 1  t  t 2  ...  t r  ...
(1  t )
t 1
 1  t  2!
Coefficient of
Alternatively
dt
2
d3
dt
3
dr
dt
μ r 
t2
t2
 ...  r!  ...
2!
r!
tr
 μr  r! r  1, 2, ...
r!
d
d
μ r (t )  (1  t ) 1  1(1  t ) 2 (1)  1(1  t ) 2
dt
dt
d2
In general
(3)
r
(1  t ) 1  2(1  t ) 3 (1)  1.2(1  t ) 3
(1  t ) 1  1.2(3)(1  t ) 4 (1)  1.2.3(1  t ) 4
(1  t ) 1  1.2. ... .r (1  t ) (r 1) (1)

dr
M x ( x )

r!(1  t ) ( r 1)
 r!
r
r
dt
 t 0 dt
t 0
dr
(iii) For Y  0 Fr ( y)  P[Y  y]  P[ x 2  y]  0
(4)
 f r ( y) 
dFy ( y)
dy
 0 ,y  0
For Y  0 pdf of Y is
f y ( y)  f x ( x )
dx
dy
now Y  x 2 x   Y  x  0
1
 e x

2 y
dx
1

dy 2 y
0
=e
y
.
1
y>0
2 y
y0
(4)

(iv) E(X)   xf x ( x )dx
0


0
0
  xe  x dx   x 21e  x dx  (2)  1!  1


and E(Y)   yf y ( y)dy   y
0
Now put Z  y
0
 y  z2
1
e
 y
dy
2 y
and dy  2zdz

 E( Y)   z 2 .
0

1 z
.e 2zdz   z 31e z dz  (3)  2! 2
2z
0
E(Y)  2E(X)  2.1  2


0
0
Alternatively E(Y)  E(X 2 )   x 2 f x ( x )dx   x 2 e x dx

  x 31e  x dx  (3)  2!  2
0
(7)
Question 3 (10 marks)
A random variable X has a chi-square distribution with n degrees of freedom with p.d.f .
f (x) 
1
n
2 2 ( n2 )
0
n 1
x2 e
x
2
otherwise
0x
(i)
Show that the cumulant generating function (c.g.f.) of X is
n
K X ( t )   n (1  2t )
2
Hence, show that for the above chi-square distribution
μ1  n ; μ 2  var( X)  2n
(ii)
x 2 x3


2
3
Hint: n (1  x)  x 
for x  1
Solution:


(i) M X ( t )  E[e xt ]   e xt f X ( x )dx   e xt 
0
0



2 n2 ( n2 ) 0
 dx 
1
2
 n 1
2 n2 ( n2 ) 0 1
( 2  t) 2
1
n
dx
t
y2

2 dx
dy
n 1

1
[2( 12  t )] 2
i)
2 n2 ( n2 )
x2 e
 n 1 ( 1  t ) x
 x2 e 2
1
1
Put y  (  t ) x
2
 M X (t) 
n 1  x
1
e  yt 
dy
( 12  t )
1

n

( n2 )
n
2 2 ( n2 ) ( 12  t ) 2
1

n
(1  2t ) 2
(5)
(Use K or C for cumulant referral)
n
 K X (t )  n M X (t )   n(1  2t )
2
(2)
n
ii)  K X (t )  n M X (t )   n(1  2t )
2
n
(2t ) 2 (2t ) 3
  (2t 

 )
2
2
3
 nt  2n
t
1
2
t2

2
 Coefficient of t in expression of K x ( t ) : μ1  n
Coefficient of
t2
in expression of K x ( t ) : μ 2  2n
2!
(3)