MATH 5010–001 SUMMER 2003
SOLUTIONS TO ASSIGNMENT 7
Problems from pp. 228–231:
38. The roots (the variable is x) of the quadratic equation 4x2 + 4Y x + (Y + 2) = 0 are
x=
Therefore,
−4Y ±
p
16Y 2 − 16(Y + 2)
−Y ±
=
8
√
Y2−Y −2
.
2
P {Roots are real} = P Y 2 − Y − 2 ≥ 0 .
The roots of y 2 − y − 2 = 0 are y = −1 or y = 2. Therefore (plot the parabola!),
Z
P {Roots are real} = 1 − P {−1 ≤ Y ≤ 2} = 1 −
0
2
3
1
dy = .
5
5
39. If Y = ln X, then because of the fact that ex is increasing, we have
FY (a) = P {ln X ≤ a} = P {X ≤ ea }
Z ea
a
=
e−x dx = 1 − e−e .
0
This is valid for all −∞ < a < ∞. Now differentiate to obtain
a
fY (a) = ea−e .
40. Again, start with the cdf:
FY (a) = P eX ≤ a = P {X ≤ ln(a)} .
If 0 < a ≤ 1, then ln(a) ≤ 0, and this probability is zero. If a ≤ 0, it is also zero because
eX cannot be less than or equal to a. On the other hand, if ln(a) ≥ 1 (i.e., a ≥ e), then
FY (a) = 1. So now consider 1 ≤ a ≤ e. Then, ln(a) ∈ (0, 5), and we have
Z
FY (a) = P {X ≤ ln(a)} =
Therefore,
ln(a)
0
dx = ln(a).

 1 , if 1 ≤ a ≤ e,
fY (a) = a

0,
otherwise.
41. I will assume first that A > 0.
1
Now let us compute the cdf first, viz., FR (r) = P {A sin Θ ≤ r} = 0, if r ≤ −A, and
FR (r) = 1 if r ≥ A because A sin x ≤ A for all x. [We need A ≥ 0 for this to be true, of
course.]
Now, for −A ≤ r ≤ A (need A ≥ 0 here too),
r o
r i
n
n
1 hπ
ro
FR (r) = P sin Θ ≤
= P Θ ≤ sin−1
=
+ sin−1
.
A
A
π 2
A
To understand this inequality best, you need to draw
√ a picture of the sine function. Now
d
differentiate to finish, viz., because dx
sin−1 (x) = 1/ 1 − x2 ,

1


√

 π A2 − r 2 ,
fR (r) =




0,
if −A ≤ r ≤ A,
otherwise.
If A < 0, then by symmetry,
fR (r) =

1



 π √A 2 − r 2 ,




0,
if A ≤ r ≤ −A,
otherwise.
2