Solutions for Q6-Q10

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Solutions
Q6)
Moment generating function of Y = ∑
i=1
n 1
i X I is equivalent to;
MY(t) = E(etY ) = E(et
∑
i=1
n 1
i Xi) = E(∏i=1
n
e
1
i tXi
)
Therefore,
E( ∏i=1
n
e
1
i tXi
) =∏i=1
n
E ( et i¿ Xi ) since Xi’is independent and ti* = t
i
Hence,
=∏
i=1
n
∫
0
∞
et i¿ x e− x dx=∏
i=1
n
( 1
1−t i¿ )= 1
( 1−t ) (1− t
2 )(1− t
3 )… . (1− t
n )
→ et as n → ∞
PDF of Vn is
fVn(v) = n(1- e− x ¿n-1
e− x , x>0, 0 otherwise
Moment generating function of Vn is
MV n(t) = E ( eVt ) =∫
0
∞
ext n ( 1−e−x ) n−1
dx
Take y = e-x
= n∫
0
1
yt ( 1− y ) n −1 dy
= n Г (t+ 1) Г (n)
Г (n+ t+1) = n !
( n+t ) ( n+t −1 ) …(t+1) = 1
(1+ t
n )(1+ t
n−1 )…(t +1)
→ et as n → ∞

Therefore, the distribution of Vn and Y are same
Q7)
The characteristic function of a random variable X is defined as ϕ(t) = E(eitX)
φ(t) = ∫
0
∞
( eitu ) dF(u)
φ(t) = ∫
0
∞
( eitu ) f (u)du
Given X > 0, the characteristic properties function will be;
φ ( 0 )=1
dn φ(t )
d tn∨t=0 =E ( Xn )
φn ( t )=E ( Xn )
Let’s take the new characteristic function to be represented as;
φn(t) = ∫
0
∞
φ ( tu ) dF (u)
φn(t) = ∫
0
∞
φ ( tu ) f (u)du
Showing the characteristic function satisfies all the conditions of non-negative random variable
φn(t) = ∫
0
∞
φ(tu)dF (u)
φn(t) = ∫
0
∞
φ(tu)f (u)du

Differentiating the second new characteristic function above
φ’n(t) = d
dt ∫
0
∞
φ(tu )f (u)du
φ’n(t) = d
dt ∫
0
∞
φ(tu )uf (u) du
Therefore, the derivative of the characteristic function at any order ‘m’ will be represented as;
φnm(t) = dm
d tm ∫
0
∞
φ ( tu ) f ( u ) du
φnm(t) = ∫
0
∞
φ ( u ) um f ( u ) du
At t = 0, the characteristic function will be equivalent to;
φ ( 0 )=∫
0
∞
φ ( 0 ) dF (u)
φ ( 0 ) =∫
0
∞
(1)f (u)du
φ ( 0 )=1
At t = 0, at derivative order ‘m’ the characteristic function be equivalent to;
φnm(0) = ∫
0
∞
φ ( 0 ) um f ( u ) du
Substituting, φ ( 0 )=1
φnm(0) = ∫
0
∞
(1)um f ( u ) du
φnm(0) = ∫
0
∞
um f ( u ) du
φnm(t) = E ( Um )
It should be noted that Um >0

Therefore, φ ( t ) =∫
0
∞
etu f ( u ) du will satisfies all the function of a random variable
Q8)
The probability distribution
Xn -n 0 n
P(Xn) 1
2nln (n) 1 - 1− 1
nln(n)
1
2nln(n)
E(Xn) = ∑ xnP(Xn)
Sn = 1
n ∑
k=l
n
Xk = 1
n ∑
k=2
n
Xk
= −n ( 1
2 nln ( n ) )+o x (1− 1
ln ( n ) )+ n( 1
2 nln(n))
E(Xn) = −1
2 ln ( n ) + 1
2 ln ( n)
E(Xn) = 0
V(xn) = E(xn2) – (E(xn))2
= ∑ xn2P(xn)
= (-n)2( 1
2nln(n) ¿+02
(1− 1
nln ( n ) )+n2 ( 1
2 ln ( n ) )
= n2 1
2nln(n)+ n2 ( 1
2nln (n) )
= n
2 ln ( n ) + n
2 ln ( n )
E(xn2) = 2 n
2 ln ( n ) = n
ln ( n )

V(xn) = n
ln ( n ) − ( 0 )2
V(xn) = n
ln ( n )
Let Sn = n and Bn = V(Sn)
E(Sn/n) = 1
n E ¿n) = 1
n E ¿k)
= n
n E ¿k)
E(Sn/n) = 0
Bn = V(Sn) = V(∑
k =2
n
Xk) = nV(Xk) = n∗n
ln ( n ) = n2
ln ( n )
Therefore if,
P
[|Sn
n − ∈ ( Sn )
n |> ∈ ]→ 0 as n → ∞
Provided Bn/n2 → 0 as n → ∞
Therefore,
B n/n2 = n2
n2 ln ( n ) = 1
ln ( n )
1
ln ( n ) → 0 as n → ∞
Therefore,
P
[|Sn
n − ∈(Sn)
n |>∈ ]→ 0 as n→ ∞
p[|Sn
n −0|>∈ ]→ 0 as n→ ∞
if Sn/n p
→
0

Sn/n = 1
n ∑
k=2
n
Xk
1
n ∑
k=2
n
Xk0
→
0
Therefore, Show that this sequence satisfies the weak law of large numbers but not the strong
Q9)
For n ≥ 2
E(x1) = 1
Are bounded real numbers
Therefore,
E(Xk) = 1 ⍱k = 1,2
Xk p
→
1
If Xk p
→
1 then Xk as
→
1
E(Xk) is constant
Now
w| ∑
k =2
n ak Xk (w)
n →
∑
k=2
n
ak
n ∨¿ since xk(w) → 1 as k → ∞
w¿ ∑
k=2
n
ak
Xk (w)
n → ∑
k=2
n ak
n → 1
The probability of this expression will be represented as;
P{W ∨∑
k=2
n akXk (N )
n → 1}=1

Therefore, ∑
k =2
n
akXk as
→
1
Making 1
n the subject of the equation, then;
1
n =∑
k=2
n
akXk → 1
Q10)
Given that = f ( x ) =
{ 1
2 ,∧0≤ x ≤ 1
1− 1
2 x∝ ,∧x ≥ 1
For all real values x except x = 0, therefore the pdf of x can be obtained by just differentiating
cdf other than x = 0, hence
Fx(x) = ¿
Fx(x) =
{ 1
2 −0 x=0
∝
2 x− ( ∝+ 1 ) x ≥1
0 otherwise

=
{ 1
2 ; x=0
∝
2 x∝+1 ; x ≥ 1
E ( x2 ) = ( 0 ) 2∗1
2 + ∫
−∞
∞
x2∗¿ ∝
2 x− ( ∝+1 ) dx ¿
= ∫
1
∞
∝
2 x− ( ∝+1 ) dx= ∝
2 [ x−∝+1+1
−∝+1+1 ] at the limits of ∞ ¿ 1
= ∝
2(2−∝) [ 1
x2 −∝ ] at the limit of ∞ ¿ 1
= ∝
2(2−∝) [ 1
∝2−∝ − 1
12−∝ ]
= ∝
2(2−∝) (0− 1
1 )
= −∝
2(2−∝) , if 2 - ∝>0 ,∧if 2> ∝
E ( x )= 0∗1
2 +∫
1
∞
x∗∝
2 x−∝−1 dx= ∝
2 ∫
1
∞
xdx
= ∝
2 [ x−∝+1
(−∝+1) ] for the limit ∞ ¿ 1
= ∝
2(1−∝) [0− 1
11−∝ ] for 1 - ∝>0 , 1> ∝=1 ,∝<1
Therefore, E ( x ) ∃if ∝b <1∧∈ ( x2 ) exist if ∝<2
Then V(x) = E ( x2 )− ( E ( x ) )2 exist when ∝<1 ,∝<2 if ∝<3
Hence the central limit exist, when random variable have finite mean and finite variance which
implies CLT exists when 0 < ∝<1