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STAT 450
Problems: Assignment 4
1.
2. Suppose X, Y have joint density
⎧
⎪⎨ 3x 0 ≤ x ≤ y ≤ 1
f(x, y) = 3y 0 ≤ y ≤ x ≤ 1
⎪⎩
0 otherwise
(a) Find the mean of Y .
∫ ∫
E(Y ) = yf(x, y)dxdy
∫ 1
{∫ y
= 3xydx +
=
= 3
0
∫ 1
0
∫ 1
0
= 5/8
(b) Find the variance of Y .
Replace y above by y 2 to get
0
∫ 1
y
}
3y 2 dx dy
{
3y 3 /2 + 3y 2 (1 − y) } dy
{y 2 − y 3 /2}dy
E(Y 2 ) = 3(1/4 − 1/10) = 0.45
so that
Var(Y ) = 45/100 − 25/64 = 0.059375
(c) Find the moment generating function of Y .
Now replace y by e ty to get
∫ 1
M y (t) = 3 e ty (y − y 2 /2)dy
0
= 3(t2 e t − 2e t + 2t + 2)
2t 3
(I used Maple.)
1

3. Find the moment generating function of the number of tails before the first head when
you toss a coin with probability p of landing heads.
We have
so that
P (X = k) = p(1 − p) k k = 0, 1, · · ·
M X (t) = p
=
∞∑ { } (1 − p)e
t k
0
p
1 − (1 − p) exp(t)
4. Suppose X is a random variable with mean µ and variance σ 2 . Find the value of a
which minimizes
E{(X − a) 2 }.
Define h(b) by
h(a) = E [ (X − a) 2]
= E [ (X 2 − 2aX + a 2 ) ]
= E(X 2 ) − 2aE(X) + a 2
To minimize h set
h ′ (a) = 2a − 2E(X) = 0
and solve to find a = E(X). That this is a minimum follows from h ′′ = 2 > 0 for all
a.
5. Find the mean and variance of the density
{
|1 − x| 0 ≤ x ≤ 2
f(x) =
0 otherwise
E(X) =
=
∫ 2
0
∫ 1
0
xf X (x) dx
x(1 − x) dx +
∫ 2
1
x(x − 1) dx
= ( x 2 /2 − x 3 /3 )∣ ∣ 1 0 + ( x 3 /3 − x 2 /2 )∣ ∣ 2 1
= 1/6 + [(8/3 − 2) − (1/3 − 1/2)]
= 1
2

(which is to be expected because the density is symmetric about x = 1.) Also:
E(X 2 ) =
=
∫ 2
0
∫ 1
0
x 2 f X (x) dx
x 2 (1 − x) dx +
∫ 2
1
x 2 (x − 1) dx
= ( x 3 /3 − x 4 /4) )∣ ∣ 1 0 + ( x 4 /4 − x 3 /3 )∣ ∣ 2 1
= 1/(12) + [(16/4 − 8/3) − (1/4 − 1/3)]
= 3/2
Hence
Var(X) = E(X 2 ) − [E(X)] 2 = 1/2 .
6. Find the mean and variance of the Rayleigh density
{
λ 2 xe −λ2 x 2 /2
x > 0
f(x) =
0 otherwise
E(X r ) =
=
∫ ∞
0
∫ ∞
0
x r f X (x) dx
x r λ 2 xe −(λx)2 /2 dx
Make the substitution y = (λx) 2 /2 and dy = λ 2 xdx to see that
E(X r ) =
∫ ∞
0
(2y) r
e −y dy
λ r
∫ ∞
= 2 r/2 λ −r y r/2+1−1 e −y dy
The integral matches the definition of the Gamma function precisely so
In particular
E(X r ) = 2r/2 Γ(1 + r/2)
λ r
E(X) = 2 1/2 Γ(3/2)/λ
(In fact Γ(3/2) = (1/2)Γ(1/2) = π 1/2 /2 but I don’t really expect you to know this!)
Moreover
E(X 2 ) = 2Γ(2)/λ 2 = 2/λ 2
and
Var(X) = 2 λ 2 − ( 2 1/2 Γ(3/2)/λ ) 2
= (2 − π/2)/λ
2
0
3

7. Find the mean and variance of the Maxwell density
{
4λ 3 x 2 e −λ2 x 2 / √ π
f(x) =
0 otherwise
E(X r ) =
=
∫ ∞
0
∫ ∞
0
x r f X (x) dx
4x r λ 3 x 2 e −(λx)2 dx/ √ π
Make the substitution y = (λx) 2 and dy = 2λ 2 xdx to see that
E(X r ) =
∫ ∞
2y (r+1)/2
0
= 2
λ r√ π
λ r√ π e−y dy
∫ ∞
0
= 2
λ r√ Γ((r + 3)/2)
π
y (r+1)/2+1−1 e −y dy
In particular
and
E(X) = 2
λ √ π Γ(2) = 2
λ √ π
E(X 2 ) = 2
λ 2√ π Γ(5/2) = 3
2λ 2√ π Γ(1/2) = 3
2λ 2
Var(X) = (3/2 − 4/π)/lambda 2 .
8. Suppose X and Y are random variables such that X is Uniform[0,1] and
{
1 x < y < x + 1
f Y |X (y|x) =
0 otherwise
(a) Find E(Y ).
4

∫ ∫
E(Y ) =
yf XY (x, y) dy dx
=
=
=
=
∫ 1 ∫ x+1
0
∫ 1
0
∫ 1
0
∫ 1
0
x
y 2
2
∣
x+1
x
y dy dx
dx
[(x + 1) 2 − x 2 ]/2 dx
(x + 1/2) dx
= [(x 2 + x)/2]| 1 0
= 1
(b) Find Cov(X, Y ).
We need to compute E(XY ) − E(X)E(Y ). Note that E(X) = 1/2 and
∫ ∫
E(XY ) = xyf(x, y)dydx
=
=
=
∫ 1
∫ x+1
0
∫ 1
0
∫ 1
0
x
= 1/3 + 1/4
xydydx
x { (x + 1) 2 − x 2} dx/2
x(2x + 1)dx/2
Thus
Cov(X, Y ) = 7/12 − (1/2)(1) = 1/12.
9. Derive the moment generating function of Poisson(λ) distribution and use it to compute
the first four moments about the origin of this distribution.
∞∑
M X (t) = e tk e −λ λ k /k!
0
∑ ∞
{
= e −λ e t λ } k
/k!
0
= exp{−λ + λ exp(t)}
5

The first 4 derivatives are
M ′ X(t) = λ exp(t)M X (t)
M ′′
X (t) = M ′ X (t) + λ exp(t)M ′ X (t)
M (3)
′′
X
(t) = M X (t) + λ exp(t)M X ′ ′′
(t) + λ exp(t)M X (t)
M (4)
(3)
X
(t) = M
X (t) + λ exp(t)M X ′
(t) + λ exp(t)M
′′
X
Now plug in t = 0 in each formula to find:
M ′ X (0) = µ X = λ
M ′′
X (0) = E(X 2 ) = λ + λ 2
M (3)
X (0) = µ′ 3 = λ + λ2 + λ 2 + λ(λ + λ 2 ) = λ + 3λ 2 + λ 3
′′
(3)
(t) + λ exp(t)M X (t) + λ exp(t)M
X (t)
M (4)
X (0) = µ′ 4 = λ + 3λ2 + λ 3 + λ 2 + λ(λ + λ 2 ) + λ(λ + λ 2 ) + λ(λ + 3λ 2 + λ 3 )
= λ + 7λ 2 + 6λ 3 + λ 4 6