Partial solutions to assignment 1
Math 5080-1, Spring 2011
p. 88, #34.
(a) Because there is only one distribution for a given MGF, it must be that X
is a discrete random variable with distribution
P {X = 1} =
1
,
8
P {X = 2} =
1
,
4
P {X = 5} =
5
.
8
(b) 1/4.
p. 89, #40.
(a) By the chain rule,
0
ψX
(t) =
0
(t)
MX
,
MX (t)
2
00
ψX
(t) =
00
(t) − [MX (t)]
MX (t)MX
2
[MX (t)]
.
00
0
(0) = E(X 2 ).
(0) = E(X) = µ, and MX
(b) We know that MX (0) = 1, MX
Therefore,
00
ψX
(0) = E(X 2 ) − µ2 = Var(X) = σ 2 .
(c) First of all, let us recall that the pdf is
(
e−(x+2) if x ≥ −2,
f (x) =
0
if x < −2.
Therefore, the MGF is
Z
∞
M (t) =
etx e−(x+2) dx
−2
 −2t
e
= 1−t

∞
if t < 1,
if t ≥ 1.
Therefore, whenever t < 1,
−2t e
ψ(t) = ln
= ln e−2t − ln(1 − t) = −2t − ln(1 − t).
1−t
Differentiate to find that, for t < 1,
ψ 0 (t) = −2 +
1
,
1−t
and therefore
µ = ψ 0 (0) = −1.
Also, if t < 1, then
ψ 00 (t) =
1
(1 − t)2
and therefore
1
σ 2 = ψ 00 (0) = 1.
(d) Example 2.5.5 is on page 83 and refers to Example 2.5.3 on page 80, where
it is shown that the MGF is

 1
if t < ln 2,
MX (t) = 2 − et
∞
if t ≥ ln 2.
It follows that for t < ln 2,
ψX (t) = ln
whence
0
ψX
(t) =
1
2 − et
1
2 − et
= − ln 2 − et ,
0
µ = ψX
(0) = 1;
and therefore
0
(0). And similarly,
this quantity was computed on page 80 by inspecting MX
00
ψX
(t) =
1
(2 − et )2
and therefore
00
σ 2 = ψX
(0) = 1.
p. 227, #14.
(a) Note that X and Y are independent random variables, both with the
EXP( 21 ) distribution. Therefore, we compute the MGF of W = X + Y as
follows:
2
1
MX+Y (t) = MX (t) · MY (t) =
.
1 − 2t
That is, W has a GAMMA( 12 , 2) [because its MGF is that of a GAMMA( 21 , 2);
see the back of the front cover of your text for a table of MGFs]. By using the
gamma density with θ = 21 and κ = 2, we obtain
(
4we−2w if w > 0,
fW (w) =
0
if w ≤ 0.
[Again, see the table on the back of the front cover of your text for this pdf.]
Consequently, FW (w) = 0 if w ≤ 0, and for w > 0 we have
Z w
Z w
FW (w) =
fW (z) dz =
4ze−2z dz.
0
0
Integrate by parts to find that for all w > 0,
w
Z w
−2z FW (w) = −2ze
e−2z dz = 1 − e−2w − 2we−2w .
+2
0
0
2
(b) If u = x/y and v = x, then

∂x
 ∂u

J = det 
 ∂y
∂u
x = v and y = v/u.


∂x
0

∂v 

 = det 
v
∂y 
− 2
u
∂v
Therefore
1
1
u

v

 = 2.
u
We are interested in positive x and y, and hence positive u and v. Therefore,
v v
4v
1
fU,V (u , v) = fX,Y (x , y)|J| = fX,Y v ,
=
exp
−2v
1
+
.
u u2
u2
u
(c) We work directly: Fix u > 0 and note that
Z ∞
fU (u) =
fU,V (u , v) dv
0
Z ∞
1
4v
exp
−2v
1
+
dv
=
u2
u
0
Z ∞
4
= 2
ve−vc dv,
u 0
where
1
2(1 + u)
.
c := 2 1 +
=
u
u
Now,
Z
∞
−cv
ve
0
1
dv = 2
c
Z
∞
xe−x dx =
0
Γ(2)
1
u2
=
=
.
c2
c2
4(1 + u)2
Therefore,
fU (u) =
1
(1 + u)2

0


if u > 0,
if u ≤ 0.
p. 227, #17. There isn’t a unique way of proceeding. I will follow the method
of the previous question because it has other uses in this course.
(a) Let us write Y1 instead of Y , and define Y2 = X2 . If y1 =
y2 = x2 , then
x1 = y12 − y2 and x2 = y2 .
And the Jacobian of this transformation leads us to


∂x1 ∂x1
 ∂y1 ∂y2 
2y1 −1


J = det 
=
det
= 2y1 .

0
1
 ∂x
∂x2 
2
∂y1 ∂y2
3
√
x1 + x2 and
Therefore,
fY1 ,Y2 (y1 , y2 ) = fX1 ,X2 (x1 , x2 )|J| = 2|y1 |fX1 ,X2 y12 − y2 , y2 .
Because X1 and X2 are independent gamma variables with θ = 2 and κ = 1/2,
fX1 ,X2 (x1 , x2 ) = fX1 (x1 )fX2 (x2 ) =
1
1
−1/2
−1/2 −x2 /2
x1 e−x1 /2 1/2
x
e
,
1/2
2 Γ(1/2)
2 Γ(1/2) 2
for x1 , x2 > 0; otherwise fX1 ,X2 (x1 , x2 ) = 0.
Remark. The Γ function [the “gamma” function] is simply chosen to make
the integral of the gamma pdf equal to one. That is,
Z ∞
Γ(α) =
xα−1 e−x dx
for α > 0.
0
One can integrate by parts to see that Γ(α) = (α − 1)Γ(α − 1) for every α > 1;
in particular, Γ(n) = (n − 1)! for integers n ≥ 1 [where 0! = 1, as usual]. Also
Z ∞
√
√ Z ∞ −y2 /2
√
Γ(1/2) =
[y = 2x].
x−1/2 e−x dx = 2 ·
e
dy = π
0
Because Γ(1/2) =
0
√
π, the preceding reduces to
fX1 ,X2 (x1 , x2 ) = fX1 (x1 )fX2 (x2 ) =
1 −1/2 −1/2 −(x1 +x2 )/2
x
x2 e
,
2π 1
for x1 , x2 > 0; otherwise fX1 ,X2 (x1 , x2 ) = 0. Plug the y’s in place of the x’s to
find that
fY1 ,Y2 (y1 , y2 ) = fX1 ,X2 (x1 , x2 )|J| =
2
y1
1
p
e−y1 /2 ,
π y2 (y12 − y2 )
for y2 > 0 and y12 > y2 ; otherwise fY1 ,Y2 (y1 , y2 ) = 0.
Recall that we want fY1 (y1 ). This is found as follows: Fix y1 > 0 and
compute
Z
y12
2
y1
1
p
e−y1 /2 dy2
2
π y2 (y1 − y2 )
0
Z 2
2
y1 e−y1 /2 y1
1
p
=
dy2
π
y2 (y12 − y2 )
0
Z
2
y1 e−y1 /2 1
1
p
=
dt
[t := y2 /y12 ].
π
t(1 − t)
0
fY1 (y1 ) =
Recall from lectures that
Z 1
Γ(α)Γ(β)
tα−1 (1 − t)β−1 dt =
Γ(α + β)
0
4
for every α, β > 0.
Apply the preceding with α = β = 21 , and recall that Γ(1/2) =
Z
0
And hence
1
1
p
t(1 − t)
dt =
Γ(1/2)Γ(1/2)
= π.
Γ(1)
(
2
y1 e−y1 /2
fY1 (y1 ) =
0
5
if y1 > 0,
if y1 ≤ 0.
√
π, to see that