Comprehensive Solution: Stat 6202 Midterm Exam on Probability

This document provides solutions to a Stat 6202 midterm exam, focusing on key concepts in probability and statistics. The solution covers topics such as the asymptotic normal distribution of a chi-square distributed random variable, maximum likelihood estimation (MLE) for Bernoulli distributions, and the derivation of asymptotic distributions for estimators. Specifically, it addresses the convergence of the square root of a chi-square distribution to a normal distribution, the properties of Bernoulli distributions, and the application of the central limit theorem. The solution also includes analysis of the Rayleigh distribution. Desklib offers a platform to explore more solved assignments and past papers for students.

Online Statistics Test
By
Name ()
Course Name ()
Instructor Name ()
Name of University ()
Submission Date ()

Question one
The mean of a chi-square distribution is given by the number of its degrees of freedom and
the variance is twice the degrees of freedom. The square root of a chi-square distribution,
therefore is expected to to have variance and mean in the same ratio as is parent distribution.
Degrees of freedom = n = mean. Variance = 2n
Zn = square roo(Yn-root(n))
n-√n, 2n-√n
N(0,1/2), represents a normal distribution with mean zero and variance half a unit., the new
distribution Yn can be viewed to be half the standard normal distribution. The chi-square
distribution follows the normal distribution with increased number of observations. Applying the
central limit theorem and applying the limits, we have
With limits as n tends to infinity, the chi-square with mean n and var 2n,
n-√n, 2n-√n ∞-sqrt( ∞)=0
and variance 2n-(n)^1/2 with limits a n tends to infitnity we get ½
N(mean =0,var =1/2)
2 | P a g e

Question two
From statistical concepts and probability distribution, the sum of Bernoulli random variables
gives a probability distribution with a binomial probability density function. When n is large
enough, the Bernoulli distribution converges to normal distribution.
If xi is a Bernoulli random variables, the sum of xis follows binomial distribution
^∅ is is the ratios of the two Bernoulli distributions.
p
1− p
q
1−q
Where p is the mean of the first
distribution and q the mean of the second distribution. Using the MLE, the estimate of p is x bar
which is unbiased and the estimate of q is y bar which is also unbiased.
Il( ^∅ ) = Il(
^p
1− ^p
^q
1− ^q
) =
Il ( p
1.− p )
Il( q
1−q )
, from this expression we can deliver the distribution of Il( ^∅ )
which id Bernoulli with mean Il(θ), where θ is the ratio of p/q of the two given distributions. The
Bernoulli distribution can be viewed as a binomial distribution where the value of n =1.
Given as ( n
x ) px qn−x
, putting the value of n as, we have a Bernoulli distribution.
( n
x ) px qn−x . px qn−x
lim
n → ∞ ( n
x ) px qn− x px qn−x
, this translate to n/p as the mean and var p n2
np applying the limits we
get mean as 0 and var 1….which a standard distribution with N(0,1)
3 | P a g e

Question four
The mean of a chi-square distribution is given by the number of its degrees of freedom and
the variance is twice the degrees of freedom. The square root of a chi-square distribution,
therefore is expected to to have variance and mean in the same ratio as is parent distribution.
Degrees of freedom = n = mean. Variance = 2n
Zn = square roo(Yn-root(n))
n-√n, 2n-√n
N(0,1/2), represents a normal distribution with mean zero and variance half a unit., the new
distribution Yn can be viewed to be half the standard normal distribution. The chi-square
distribution follows the normal distribution with increased number of observations. Applying the
central limit theorem and applying the limits, we have
With limits as n tends to infinity, the chi-square with mean n and var 2n,
n-√n, 2n-√n ∞-sqrt( ∞)=0
and variance 2n-(n)^1/2 with limits a n tends to infitnity we get ½
N(mean =0,var =1/2)
4 | P a g e

5 | P a g e