4MA501 Foundations of Probability: Assignment Solutions

This document presents solutions to a probability assignment from the course 4MA501 Foundations of Probability. The solutions cover a range of problems, including those related to bivariate normal density functions, moment generating functions, and characteristic functions. The assignment explores the independence of random variables, the weak law of large numbers, and the central limit theorem. Detailed step-by-step solutions are provided for each problem, including calculations of expected values, variances, and probabilities. The problems involve concepts such as probability distributions, conditional probabilities, and the analysis of random variables. The solutions demonstrate the application of key statistical principles and techniques to solve complex probability problems. The document is a valuable resource for students studying probability and statistics, offering comprehensive explanations and insights into the subject matter.

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