School of Mathematical Sciences
MTH5122 Statistical Methods, 2016
Assignment 3
1. The cdf of X ∼ Exp(λ) is FX (x) = 1 − e−λx for x > 0. By the
probability integral transform theorem U := FX (X) = 1 − e−λX ∼
U nif orm(0, 1). But then for Y = 1−U we also have Y ∼ U nif orm(0, 1).
That is, fY (y) = 1 for y ∈ (0, 1). Alternatively, the density transform
formula can be used to obtain fY .
2. We can write X1 = Z1 + µ1 , X2 = Z2 + µ2 where Z1 , Z2 are iid standard
normal rv’s. Also, since X1 + X2 = (µ1 + µ2 ) + (Z1 + Z2 ) it is enough
to show that Z1 + Z2 ∼ N (0, 2). The joint pdf of Z1 , Z2 is
fZ1 ,Z2 (z1 , z2 ) =
1 −(z12 +z22 )/2
e
.
2π
Let Y1 = Z1 + Z2 , Y2 = Z2 , which is a linear transfromation of rv’s with
the Jacobian equal to 1. The inverse is z1 = y1 − y2 , z2 = y2 , hence the
density transform formula yields
fY1 ,Y2 (y1 , y2 ) =
1 − 1 ((y1 −y2 )2 +y22 )
e 2
.
2π
To integrate this in the variable y2 we complete a square
1
1
(y1 − y2 )2 + y22 = 2(y2 − y1 )2 + y12 ,
2
2
and so the joint pdf of Y1 , Y2 becomes
fY1 ,Y2 (y1 , y2 ) =
Using that
Z
1 −(y2 −y1 /2)2 −y12 /4
e
e
.
2π
∞
2
e−(z−a) dz =
√
π
−∞
we obtain
Z
fY1 (y1 ) =
∞
1
2
fY1 ,Y2 (y1 , y2 )dy2 = √ e−y1 /4 ,
2 π
−∞
whence Y1 ∼ N (0, 2), as wanted.
1
3. Let Y1 = X1 /X2 , Y2 = X2 . The inverse transform is x1 = y1 y2 , x2 = y2
with the Jacobian |y2 |. By the density transformation formula
fY1 ,Y2 (y1 , y2 ) =
1 −y12 y22 /2 −y22 /2
e
e
|y2 |.
2π
We need to integrate this in the variable y2 from −∞ to ∞, but the
function is even, hence
Z ∞
Z ∞
1 −y12 y22 /2 −y22 /2
fY1 (y1 ) =
fY1 ,Y2 (y1 , y2 )dy2 = 2
e
e
y2 dy2 .
2π
−∞
0
With the substitution t = y22 this becomes
Z
∞
1
1 − y12 +1 t
e 2 dt =
,
fY1 (y1 ) =
2π
π(1 + y12 )
0
which is the Cauchy pdf with support (−∞, ∞).
4. (a) The halfplane above the line y1 = y2 .
(b) For y1 < y2 there are two pairs x1 = y1 , x2 = y2 and x1 = y2 , x2 =
y1 and in both case |∂(x1 , x2 )/∂(y1 , y2 )| = 1. Since by the iid property fX1 ,X2 (x1 , x2 ) = f (x1 )f (x2 ) it follows that
fY1 ,Y2 (y1 , y2 ) = f (y1 )f (y2 )+f (y2 )f (y1 ) = 2f (y1 )f (y2 ),
y1 < y2
(and the pdf is 0 for y1 > y2 ).
(c) In the case of Xi ∼ U nif orm(0, 1)
fY1 ,Y2 (y1 , y2 ) = 2,
y1 < y2 .
Integrating
fY1 (y) = 2(1 − y), fY2 (y) = 2y,
(d) fZ1 ,Z2 ,Z3 (z1 , z2 , z3 ) = 6f (z1 )f (z2 )f (z3 ),
y ∈ (0, 1).
z1 < z2 < z3 .
5. FEEDBACK QUESTION
(a) , (b)We consider the change of variables
u = x/(x + y), v = x + y
where the range of u, v is u ∈ (0, 1), v ∈ (0, ∞), so the support
of the pdf fU,V is (0, 1) × (0, ∞). Solving for x, y we obtain the
inverse relation
2
x = uv, y = v(1 − u)
and the Jacobian is computed as
v
u
det
= v.
−v (1 − u)
The joint density of (X, Y ) is by independence the product form
1
xα−1 e−x y β−1 e−y
Γ(α)Γ(β)
where x > 0, y > 0. By the density transformation theorem we
need to substitute x = uv, y = v(1 − u) and multiply by v:
fX,Y (x, y) =
fU,V (u, v) = fX,Y (uv, v(1 − u))v =
1
(uv)α−1 e−v v β−1 (1 − u)β−1 v =
Γ(α)Γ(β)
1
uα−1 (1 − u)β−1 v α+β−1 e−v =
Γ(α)Γ(β)
Γ(α + β) α−1
1
u (1 − u)β−1
v α+β−1 e−v
Γ(α)Γ(β)
Γ(α + β)
Integrating out v we get applying the Gamma integral formula for
the second factor in brackets
fU (u) =
Γ(α + β) α−1
u (1 − u)β−1
Γ(α)Γ(β)
Z
0
∞
1
v α+β−1 e−v dv =
Γ(α + β)
Γ(α + β) α−1
u (1 − u)β−1
Γ(α)Γ(β)
so U has the asserted beta distribution on [0, 1].
Likewise, integrating out u we confirm that V ∼ Gamma(α+β, 1).
Factorisation
fU,V (u, v) = fU (u)fV (v)
for u ∈ [0, 1], v ∈ [0, ∞) implies that the random variables U and
V are independent.
(c) We have X = U V where U , V independent. Thus
E X = E U E V.
3
From the properties of Gamma distribution (LN Ch1 page 13) we
know that
E X = α, E V = α + β
whence from the above
EU =
α
EX
=
EV
α+β
(b)
Z
EU =
0
1
Z
Γ(α + β) 1 α
xfU (x)dx =
x (1 − x)β−1 dx =
Γ(α)Γ(β) 0
α
Γ(α + β) Γ(α + 1)Γ(β)
=
Γ(α)Γ(β) Γ(α + β + 1)
α+β
using the beta integral and that Γ(a + 1) = aΓ(a).
4