For all problems on this page, suppose you have data
X_1, ..., X_n ~ N(0, 1)
that is a random sample of identically and independently distributed standard normal random variables.
Useful facts:
For a standard normal random variable X_1, we have:
E[X] = 0, E[X^2] = 1, E[|X|] = sqrt(2/pi) approx 0.8
The i.i.d. assumption
Suppose X_1 is an observation for Bob, X_5 is an observation for Alice, X_7 is an observation for Charlie.
Using the following facts about Bob:
P(-2 < X_1 < 2) approx 0.95 and P(-sqrt(2) <= X_1 <= sqrt(2)) approx 0.75,
compute the probability
P(-2 < X_5 < 2, -2 < (X_7)^5 < 2)
(Enter the probability P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) or if the probability is not determined uniquely, then enter -1.)
P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) = ________
Sample mean
Consider the sample mean:
bar{X_n} = (1/n)(X_1 + X_2 + ... + X_n).
What are the mean E[bar{X_n}] and variance Var[bar{X_n}] of bar{X_n}?
E[bar{X_n}] = ________
Var[bar{X_n}] = ________
What kind of distribution does bar{X_n} follow?
- Gaussian
- Student t
- Chi square
- Gamma
- nonparametric
Quantiles
Consider the quantile Q_{n, alpha} of order 1 - alpha for the random variable bar{X_n}, that is, the number Q_{n, alpha} such that
P(bar{X_n} <= Q_{n, alpha}) = 1 - alpha, 0 < alpha < 1.
For alpha < 0.5, as the sample size n increases, does the quantile Q_{n, alpha}:
- Increase
- Decrease
- Stay the same
- Oscillate
Answer (1)
Let's break down the problem step by step:
Given Information :
We have a series of observations X 1 , X 2 , ... , X n , which are independent and identically distributed standard normal random variables, i.e., X i ∼ N ( 0 , 1 ) .
The probability P ( − 2 < X 1 < 2 ) ≈ 0.95 is given, and since these observations are i.i.d., for any X i this remains true: P ( − 2 < X i < 2 ) ≈ 0.95 .
Finding the Probability for the Given Condition :
We are asked to calculate P ( − 2 < X 5 < 2 , − 2 < ( X 7 ) 5 < 2 ) .
Since X i are i.i.d. standard normal variables, X 5 is also a standard normal variable, hence P ( − 2 < X 5 < 2 ) ≈ 0.95 .
The probability P ( − 2 < ( X 7 ) 5 < 2 ) involves calculating the fifth power of X 7 , which is nonlinearly transformed, making it complex to directly determine without specific distribution information.
As a result, the probability P ( − 2 < X 5 < 2 , − 2 < ( X 7 ) 5 < 2 ) is not uniquely determined with the given data because it's challenging to calculate ( X 7 ) 5 analytically without extra distribution data. Hence, it remains indeterminate, so we input -1 .
Sample Mean Information :
The sample mean is defined as X n ˉ = n 1 ( X 1 + X 2 + ... + X n ) .
Mean of the Sample Mean : E [ X n ˉ ] is the mean of all observations divided by n . Since each X i has an expected value of 0, E [ X n ˉ ] = 0 .
Variance of the Sample Mean : Va r ( X n ˉ ) = n σ 2 , where σ 2 is the variance of X i . Since each X i has variance 1, Va r ( X n ˉ ) = n 1 .
Distribution Type :
By the Central Limit Theorem, as n becomes large, X n ˉ follows a normal (Gaussian) distribution.
Quantiles as Sample Size Increases :
With increasing n and α < 0.5 , the quantile Q n , α decreases. As the sample size increases, the distribution's standard error decreases, concentrating the distribution around its mean more tightly.
Therefore, the answers are:
P ( − 2 < X 5 < 2 , − 2 < ( X 7 ) 5 < 2 ) = − 1
E [ X n ˉ ] = 0
Va r [ X n ˉ ] = n 1
The distribution of X n ˉ is Gaussian
As n increases, Q n , α decreases.
