STAT 610 Homework #10 solution 5.12. Note that for any n and for ...

STAT 610 Homework #10 solution
5.12. Note that for any n and for any X i , i = 1, ..., n one has
∣ 1
n∑ ∣∣∣∣
Y 1 =
X i ≤ 1 n∑
|X i | = Y 2
∣n
n
i=1
so, the same ineqaulity holds for expectations.
Thus,
E(Y 1 ) =
=
∫ ∞
−∞
∫ 0
−∞
∫ ∞
√ n
{
|z| √ exp − n 2π 2 z2} dz
√ n
−z √ exp 2π
{
− n 2 z2} dz +
√ n
= 2 z √ exp
0 2π
√ ∫ n 1 ∞
= 2√ exp{−u}du
2π n 0
√
2
= √ nπ
∫ ∞
0
i=1
1
n
n∑
X i ∼ N(0, 1 n ) since X i ∼ N(0, 1).
i=1
√ n
{
z √ exp − n 2π 2 z2} dz
{
− n 2 z2} dz ( Let n 2 x2 = u, nxdx = du)
5.15. (a)
E(Y 2 ) = 1 n
=
= 2
= 2
=
i=1
∫ ∞
n∑
E(|X i |) = E(|X 1 |)
−∞
∫ ∞
0
∫ ∞
0
√
2
√ π
.
1
|x| √ exp 2π
Thus, E(Y 1 ) < E(Y 2 ).
(b)
{− 1 2 x2 }
dx
x√ 1 exp
{− 1 }
2π 2 x2 dx ( Let 1 2 x2 = u, xdx = du)
1
√
2π
exp −udu
¯X n+1 =
∑ n+1
i=1 X i
n + 1
= X n+1 + ∑ n
i=1 X i
n + 1
= X n+1 + n ¯X n
.
n + 1
nS 2 n+1 =
=
=
n
n∑
(X i −
(n + 1) − 1
¯X n+1 ) 2
i=1
n+1
∑
(
X i − X n+1 + n ¯X ) 2
n
( use (a) )
n + 1
i=1
n+1
∑
(
X i − X n+1
n + 1 − n ¯X ) 2
n
n + 1
i=1
= ∑ i = 1 n+1 [(X i − ¯X n ) −
1
(
Xn+1
n + 1 −
¯X )] 2
n
(±
n + 1
¯X n )

=
=
=
n+1
∑
i=1
[
(X i − ¯X n ) 2 − 2(X i − ¯X n )
(
Xn+1 − ¯X )
n
+
n + 1
n∑
(X i − ¯X n ) 2 + (X n+1 − ¯X n ) 2 − 2(X n+1 − ¯X n )
i=1
n∑
(X i − ¯X n ) 2 + n
n + 1 (X n+1 − ¯X n ) 2
i=1
= (n − 1)S 2 n + n
n + 1 (X n+1 − ¯X n ) 2
The sixth equlity holds since ∑ n
i=1 (X i − ¯X n ) = 0.
3∑
( ) 2 Xi − i
5.16. (a)
∼ χ 2
i
3.
i=1
/ √ √√ 3∑
( ) 2 /
Xi − i
(b) (X 1 − 1)
2 ∼ t 2 .
i
i=2
/ ( 3∑
( ) 2 / )
(c) (X 1 − 1) 2 Xi − i
2 ∼ F 1,2 .
i
i=2
]
1
(n + 1) 2 (X n+1 − ¯X n ) 2
(
Xn+1
n + 1 −
5.18. If X ∼ t p , then X = Z/ √ V/p where Z ∼ N(0, 1), V ∼ χ 2 p and Z and V are independent.
(a) E(X) = E(Z)/E(1/ √ V/p) = 0, since E(Z) = 0, as long as the other expectation is
finite. This is so if p > 1. Var(X) = E(X 2 ) = p/(p − 2) if p > 2 since X 2 ∼ F 1,p .
(b) X 2 = Z 2 /(V/p). Z 2 ∼ χ 2 1, so the ratio is distributed F 1,p .
(c) The pdf of X is
f X (x) =
[ ]
Γ((p + 1)/2)
Γ(p/2) √ 1
pπ (1 + p/x 2 ) . (p+1)/2
Denote the qunatity in square brackets by C p . Using Stirling’s formula
lim C p = lim
p→∞ p→∞
= e−1/2
√ π
= e−1/2
√
2π
= e−1/2
√
2π
√
2π
( p−1
2
√
2π
( p−2
lim
p→∞
lim
p→∞
lim
p→∞
) p−1
2 + 1 2
e − p−1
2
) p−2
2 + 1 2
e − p−2
2
2
( p−1
2
( p−2
) p−2
2
) p−1
2 + 1 2
1
√ pπ
2 + 1 2
√ p
( ) p−2 ( p − 1
2 p − 1
p − 2 p − 2
(
) p−2
2
1 −
1/2
(p − 2)/2
Also we can show that for each x
lim
(1 + x2
p→∞ p
) (p+1)/2
= lim
p→∞
) 1
( 2
p − 1
p
( ) 1
p − 1
p − 2
(
1 + x2 /2
p/2
) 1
2
2 ( p − 1
p
¯X )
n
+ 1
n + 1 n + 1 (X n+1 − ¯X n ) 2
) 1
2
=
e −1/2 e 1/2
√
2π
= 1 √
2π
.
) p/2 ) 1/2 (1 + x2
= e x2 /2 .
p
(d) As the random variable F 1,p is the square of a t p , we conjecture that it would converge
to the square of N(0, 1) random variable, a χ 2 1.
(e) The random variable qF q,p can be thought of as the sum of q random variables, each a
t p squared. Thus, by all of the above, we expect it to converge to a χ 2 q rondom variable
as p → ∞.
2

5.22. Calculating the cdf of Z 2 , we obtain
F Z 2(z) = P ((min(X, Y )) 2 ≤ z) = P (− √ z ≤ min(X, Y ) ≤ √ z)
= P (min(X, Y ) ≤ √ z) − P (min(X, Y ) ≤ − √ z)
= [1 − P (min(X, Y ) > √ z)] − [1 − P (min(X, Y ) > − √ z)]
= P (min(X, Y ) > − √ z) − P (min(X, Y ) > √ z)
= P (X > − √ z)P (Y > − √ z) − P (X > √ z)P (Y > √ z) X and Y are independent
= (1 − F X (− √ z)) 2 − (1 − F X ( √ z)) 2 X and Y are identically distributed
= 1 − 2F X (− √ z) since 1 − F X ( √ z) = F X (− √ z)
Differentiating and substituting gives
the pdf of χ 2 1 random variable.
f Z 2(z) = d dz F Z 2(z) = f X(− √ z) 1 √ z
= 1 √
2π
e −z/2 z −1/2
5.23.
∞∑
∞∑
P (Z > z) = P (Z > z|x)P (X = x) = P (U 1 > z, ..., U x > z|x)P (X = x)
=
x=1
∞∑
x=1 i=1
x=1
x∏
P (U i > z)P (X = 1) (by independence of theU is)
′
∞∑
∞∑
= P (U i > z) x P (X = x) = (1 − z) x 1
(e − 1)x!
x=1
x=1
= 1 ∞∑ (1 − z) x
= e1−z − 1
0 < z < 1
e − 1 x! e − 1
x=1
5.24. F X (x) = x/θ, 0 < x < θ. Let Y = X (n) , Z = X (1) . Then from Theorem 5.4.6.
( ) n−2
n! y − z n(n − 1)
f Y,Z (y, z) =
=
(n − 2)! θ
θ n (y − z) n−2 , 0 < z < y < θ.
Now let W = Z/Y, Q = Y. Then Y = Q, Z = W Q, and |J| = q. Therefore
f W,Q (w, q) =
n(n − 1)
θ n (q − wq) n−2 n(n − 1)
q =
θ n (1 − w) n−2 q n−1 , 0 < w < 1, 0 < q < θ.
5.35. (a) µ X = 1 and σX 2 = 1 since X i ∼ exponential(1). By CLT, ¯Xn is approximately distributed
as N(0, 1/n). So
( )
¯X n − 1
¯Xn
1/ √ n → Z ∼ N(0, 1) and P − 1
1/ √ n ≤ x → P (Z ≤ x).
(b)
d
d
P (Z ≤ x) =
dx dx F Z(x) = f Z (x) = √ 1 e −x2 /2 . 2π
( )
d ¯Xn
dx P − 1
1/ √ n ≤ x = d
n dx P ( ∑
X i ≤ x √ n + n) , (W =
i=1
n∑
X i ∼ gamma(n, 1))
i=1
= d
dx F W (x √ n + n) = f W (x √ n + n) √ n
= 1
Γ(n) (x√ n + n) n−1 e −(x√ n+n) √ n.
3

1
Therefore,
Γ(n) (x√ n + n) n−1 e −(x√ n+n) √ n ≈ √ 1 −x2 /2
e as n → ∞. Substituting
2π
x = 0 yields n! ≈ n n+1/2 e −n√ 2π.
5.36. (a)
E(X) = E(E(X|N)) = E(2N) = 2θ
Var(X) = E(Var(X|N)) + Var(E(X|N)) = E(4N) + Var(2N) = 4θ + 4θ = 8θ
(b) The moment generating function of Y is
[ ( ) ] N 1
∞∑
( ) n 1
M Y (t) = E[E(e tY e −θ θ n ∑ ∞
|N)] = E
=
= e −θ [θ(1 − 2t) −1 ] n
1 − 2t
1 − 2t n!
n!
n=0
= e 2θt/(1−2t) .
Using M aY +b (t) = e bt M Y (at), the moment generating function of Y − EY/ √ Var(Y ) is
√ θ/2t
M √
Y −EY/ Var(Y )
(t) = e −√θ/2t 1−
e
√ 1/2θt
= e −√ √
e√ θ/2t θ/2t(1+ 1/2θt+O(t 2 /θ)) → e t2 /2 .
n=0
The last equality is obtained by taylor expansion of 1/(1 − √ 1/2θt).
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