School of Mathematical Sciences
MTH5122 Statistical Methods, 2016
Assignment 2
1. Let X have the pdf f (x) = 4x3 , 0 < x < 1 and zero otherwise. Find
the pdf of Y = 2X − 1.
For y = 2x − 1 we have x = (y + 1)/2, dx/dy = 1/2, hence
3
y+1 1 1
= (y + 1)3 , y ∈ (−1, 1).
fY (y) = 4
2
2 4
2. (a) This is just bookkeping:
y 1
λα −λ y y α−1 1 (λ/a)α −(λ/a)y α−1
=
e a
=
e
y
faX (y) = fX
a a Γ(α)
a
a
Γ(α)
which is the Gamma(α, λ/a) density.
(b) The supports of X and Y are both (0, ∞), where the function y =
x2 is an increasing bijection. The inverse function is x = y 1/2 with
dx/dy = y −1/2 /2. Thus substituting in the formula for the density
transform we get after some simplification
fY (y) = fX (y
1/2
)y
−1/2
λα α −1 −λy1/2
/2 =
y2 e
.
2Γ(α)
3. (a) , (b) Not independent, since the joint support is a triangle. The
support must be rectangular when the marginal supports of X and
Y are intervals.
(b) c = 8/9 from
Z 2Z
x
xydydx = 8/9.
1
1
(c) For 1 < x < 2 and 1 < y < 2
Z x
Z 2
4 3
4
fX (x) = c
xydy = (x −x), fY (y) = c
xydx = (4y−y 3 ) .
9
9
1
y
(d) For 1 < y < x < 2
fX (x|Y = y) =
8
9 xy
4
9 (4y
− y3)
=
1
2x
2y
,
f
(y|X
=
x)
=
.
Y
4 − y2
x2 − 1
(e)
x
Z
E(Y |X = x) =
1
2
Z
2
2y
2(x3 − 1)
y 2
dy =
.
x −1
3(x2 − 1)
x2
E(X |Y = y) =
y
16 − y 4
2x
dx
=
.
4 − y2
2(4 − y 2 )
(f)
2 · (1.53 − 1)
E(Y |X = 1.5) =
,
3 · (1.52 − 1)
E(Y |Y = 1.5) = 1.5,
E(Y |X) =
2(X 3 − 1)
3(X 2 − 1)
4. FEEDBACK QUESTION
(a) (0, ∞) × (0, 1)
(b) Changing the variable of integration for x/y = t (where y ∈ (0, 1)
is fixed) yields
∞
Z
fY (y) =
0
1 2 −3 −x/y
1
xy e
dx =
2
2
Z
∞
t2 e−t dt =
0
Γ(3)
= 1,
2
implying that Y ∼ U nif orm[0, 1].
(c)
1 2 −3 −x/y
xy e
fX,Y (x, y)
1
fX (x|Y = y) =
= 2
= x2 y −3 e−x/y
fY (y)
1
2
which is the Gamma(3, y −1 ) density.
(d)
Z
E(X|Y = y) =
0
∞
Z
∞
1 3 −3 −x/y
xy e
dx =
xfX (x|Y = y)dx =
2
0
Z
y ∞ x3 − xy dx yΓ(4)
e
=
= 3y.
2 0 y3
y
2
Therefore E(X|Y ) = 3Y . Using the iterated expectation identity
and (b)
Z 1
3
E X = E(E(X|Y )) = E(3Y ) =
3ydy =
2
0
2
(e) Keeping x constant we use substitution
z=
xdz
x
, dy = − 2
y
z
to obtain
Z
fX (x) =
0
Z
1 2 −3 −x/y
1 ∞ −z
xy e
dy =
ze dx =
2
2 x
∞ 1
1
− ze−z − e−z x = (1 + x)e−x ,
2
2
1
where the last integral was evaluated via the integration by parts.
(f) Evaluating the expectation as
Z ∞
Z ∞
1
3
1
EX =
xfX (x)dx =
x (1+x)e−x dx = (Γ(2)+Γ(3)) =
2
2
2
0
0
we see that this is the same value as the one computed in (d).
(g) Not independent. E(X|Y = y) depends on y, and is not equal to
constant E X. The same conclusion can be made looking at the
conditional pdf fX (x|Y = y) 6= fX (x).
3