Please see attached mathematical statistics question attached along with attempt of proof. part(b)How to prove the convergence in distribution to a normal distribution with mean zero and variance v(theta) and nd v(theta) as a function in terms of theta? Is my solution correct? Page-10-1) ECx)%3D%3D1.0+2%3DO+20+2.'. Vas (x) = E (x³) - E²(x) = _©2.0+2(0+2)(0+1)²: By CLT, Sn (T, - E(x)N (0, Vancxi)Vas (Xi)):-()(0+2) CO+1)We will. use delta method,N(0, (8'0+2) (o+1→ 9'a) = (1-x)-x(-1)(1-x)²Hele, g o) = xg'cx)3=13D22.1111-0XuCo ti)CO+2) (0+1)²|© (©+1)=N(0, 0(0+)*)Cot1)%D Let X1...,X, be independent random variables where each X; has a pdf f(x; 8) = 01°-1 for 0 <r< 1, where 0 > 0 is an unknown parameter.(a) Show that the method-of-moment of 6 is 6n =(b) Prove that yn(en - 0) converges in distribution to a normal distribution with mean zero andvariance v(e) and find v(0) as a function in terms of 0

In such type of problems, we first attempt to find the asymptotic distribution of sample mean, X¯…

Answered: Let X1...,Xn be independent random…

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Please see attached mathematical statistics question attached along with attempt of proof.

part(b)

How to prove the convergence in distribution to a normal distribution with mean zero and variance v(theta) and nd v(theta) as a function in terms of theta? Is my solution correct?

Page-1
0-1
) ECx)
%3D
%3D
1.
0+2
%3D
O+2
0+2
.'. Vas (x) = E (x³) - E²(x) = _©
2.
0+2
(0+2)(0+1)²
: By CLT, Sn (T, - E(x)N (0, Vancxi)
Vas (Xi))
:-
()
(0+2) CO+1)
We will. use delta method,
N(0, (8'
0+2) (o+1
→ 9'a) = (1-x)-x(-1)
(1-x)²
Hele, g o) = x
g'cx)3=
13D
2
2.
1
11
1-0
Xu
Co ti)
CO+2) (0+1)²
|
© (©+1)
=N(0, 0(0+)*)
Cot1)
%D

Let X1...,X, be independent random variables where each X; has a pdf f(x; 8) = 01°-1 for 0 <
r< 1, where 0 > 0 is an unknown parameter.
(a) Show that the method-of-moment of 6 is 6n =
(b) Prove that yn(en - 0) converges in distribution to a normal distribution with mean zero and
variance v(e) and find v(0) as a function in terms of 0

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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