P_PracticeExam6_05-1..

PRACTICE EXAMINATION NO. 6
Let z0.6 be the 60-th percentile of the standard normal distribution. Let Z be a standard normal
ln X ! µ
random variable. Then Pr( Z ! "z0.6 ) = 0.40. But Z = is standard normal, thus
1
ln2 ! µ = !z0.6 , and µ = ln 2 + z0.6 . From the table, ! ( 0.25)
= 0.5987 and ! ( 0.26)
= 0.6026.
By linear interpolation,
0.6 " 0.5987
1
z0.6 ! 0.25 + # ( 0.26 " 0.25)
= 0.25 + # 0.01 ! 0.2533.
0.6026 " 0.5987 3
This gives
µ = ln 2 + z0.6 ! 0.9464805.
The mean of the log-normal distribution is E( X)
= e
E X
Answer A.
1
0.9464805+
( ) = e
2 ! 4.2481369.
ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2007 by Krzysztof Ostaszewski - 279 -
1
µ +
2
23. You are given a continuous random variable with the density
fX ( x)
= 1
2
x + 1
2 ,
! 2
, so that in this case
for !1 " x " 1, and 0 otherwise. Find the density of Y = X 2 , for all points where that density is
nonzero.
A.
1 3
B. 2y C.
2 y
2y
4
D.
3 y2 E. 1
yln2
Solution.
This is an unusual problem because you cannot use the direct formula for the density, as the
transformation Y = X 2 is not one-to-one for !1 " x " 1. So the approach using the cumulative
distribution function will work better. We have
FX ( x)
=
x
! 1 1$
!
( t +
"
#
2 2%
&dt
=
"
#
'1
t 2
4
t $
+
2%
t = x
& t ='1
=
( )2
= x2 ! x $ ! 1 1$
+
"
#
4 2%
& ' '
"
#
4 2%
& = x2 x 1
+ +
4 2 4 = x2 + 2x + 1
=
4
x + 1
4
,
for !1 " x " 1, 0 for x < !1, and 1 for x > 1. Note that Y takes on only non-negative values.
Therefore,
FY ( y)
= Pr( Y ! y)
= Pr X 2 ( ! y)
= Pr " y ! X ! y
= FX ( y ) " FX " y
( ) =
( )
(
=
y + 1)
2
" " y + 1 ( ) 2
= y + 2 y + 1 " y + 2 y " 1
4
4
= y.
4