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Sampling from the Normal Distribution Confidence Intervals

Interval Estimation

Maura Mezzetti

Department of Economics and Finance Universit`a Tor Vergata

Sampling from the Normal Distribution Confidence Intervals

Outline

1 Sampling from the Normal Distribution

2 Confidence Intervals The Pivotal Approach Example

Sampling from the Normal Distribution Confidence Intervals

Chi Squared random Variable

The notation X ∼ χ^2 p indicates a Chi Squared Random Variable with p degrees of freedom: If Z ∼ N(0, 1), then Z 2 ∼ χ^12 , that is, the square of a standard normal random variable is a Chi Squared random variable with one degree of freedom. If X 1 , X 2 ,... , Xn are independent and Xi ∼ χ^2 pi then

X 1 + X 2 +... + Xn ∼ χ^2 p 1 +p 2 +...+pn ,

that is, independent chi squared variables add to a chi squared variable, and the degrees of freedom also adds.

Sampling from the Normal Distribution Confidence Intervals

The Derived distributions: Students’s t and Snedecor’s F

Let X 1 , X 2 ,... , Xn be a random sample from N(μ, σ^2 ) distribution, then the quantity:

T =

X¯ − μ S/

n has Student’s t distribution with n degrees of freedom and we write T ∼ tn− 1. Let X 1 , X 2 ,... , Xn be a random sample from N(μx , σ^2 x ) and let Y 1 , Y 2 ,... , Ym be a random sample from N(μy , σ y^2 ) distribution, then the quantity:

F =

S X^2 /σ X^2 S Y^2 /σ Y^2

has Snedecor’s F distribution with n − 1 and m − 1 degrees of freedom and we write F ∼ Fn− 1 ,m− 1.

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Interval Estimation

The purpose of using an interval estimator rather than a point estimator is to have some guarantee of capturing the parameter of interest. The certainty of this guarantee is quantified in the following definitions.

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Standard Error and Precision

Let X 1 ,... , Xn a random sample with pdf f (x; θ). Let T = T (X) a point estimator for θ.

Definition The standard deviation of an estimator is called the standard error of the estimator (SE ). In symbol

SE (T ) =

p Var (T )

The precision of an estimator is measured by the SE of the estimator A precise estimator has a small standard error, but exactly how are the precision and the standard error related? (The answer will be given by the confidence intervals)

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

From point to interval estimators

Example Let X 1 ,... , Xn i.i.d X ∼ N(μ, 1). X^ ¯ is a point estimator for μ [ ¯X − 1; ¯X + 1] is an interval estimator for μ

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

From point to interval estimators

Let X 1 ,... , X 9 i.i.d X ∼ N(μ, 1)

[ ¯X − 1; ¯X + 1] is an interval estimator for μ

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

From point to interval estimators

Let X 1 ,... , X 9 i.i.d X ∼ N(μ, 1)

[ ¯X − 1; ¯X + 1] is an interval estimator for μ

What is the covarage probability of [ X¯ − 1; X¯ + 1]?

1. 9973

P( ¯X − 1 ≤ μ ≤ X¯ + 1) = 0. 9973

Necessary Distribution of X¯ X^ ¯ − μ 1 3

∼ N(0, 1)

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

From point to interval estimators

Let X 1 ,... , X 9 i.i.d X ∼ N(μ, 1)

[ ¯X − 1; ¯X + 1] is an interval estimator for μ

What is the covarage probability of [ X¯ − 1; X¯ + 1]?

1. 9973

P( ¯X − 1 ≤ μ ≤ X¯ + 1) = 0. 9973

Necessary Distribution of X¯ X^ ¯ − μ 1 3

∼ N(0, 1)

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Example: Uniform distribution

Example Let (X 1 , X 2 ,... , Xn) i.i.d. Uniform random variables on [0, θ]. The MLE of θ is Y = max(X 1 , X 2 ,... , Xn), the sample maximum. Find the coverage probability for the interval estimators: (^1) [aY , bY ], 1 ≤ a < b (^2) [Y + c, Y + d], 0 ≤ c < d

Distribution of Y?

P(Y ≤ y ) =

 (^) y

θ

n

fY (y ) = n

 (^) y

θ

n− (^1 )

θ

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Example: Uniform distribution

Example Let (X 1 , X 2 ,... , Xn) i.i.d. Uniform random variables on [0, θ]. The MLE of θ is Y = max(X 1 , X 2 ,... , Xn), the sample maximum. Find the coverage probability for the interval estimators: (^1) [aY , bY ], 1 ≤ a < b (^2) [Y + c, Y + d], 0 ≤ c < d

Distribution of Y?

P(Y ≤ y ) =

 (^) y

θ

n

fY (y ) = n

 (^) y

θ

n− (^1 )

θ

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Example: Uniform distribution

Example Let (X 1 , X 2 ,... , Xn) i.i.d. Uniform random variables on [0, θ]. The MLE of θ is Y = max(X 1 , X 2 ,... , Xn), the sample maximum. Find the coverage probability for the interval estimators: (^1) [aY , bY ], 1 ≤ a < b (^2) [Y + c, Y + d], 0 ≤ c < d

Prove these results (^1) Pθ(θ ∈ [aY , bY ]) = Pθ

b ≤^

Y θ ≤^

1 a

a

n −

b

n

(^2) Pθ(θ ∈ [Y + c, Y + d]) =

1 − c θ

n −

1 − d θ

n

Sampling from the Normal Distribution Confidence Intervals

The Pivotal Approach Example

Example: Uniform distribution

Example Let (X 1 , X 2 ,... , Xn) i.i.d. Uniform random variables on [0, θ]. The MLE of θ is Y = max(X 1 , X 2 ,... , Xn), the sample maximum. Find the coverage probability for the interval estimators: (^1) [aY , bY ], 1 ≤ a < b (^2) [Y + c, Y + d], 0 ≤ c < d

In the first case the coverage probability does not depend on θ and the confident coefficient is

a

n −

b

n . In the second case the coverage probability depends on θ and the confident coefficient is zero: limθ→∞

h 1 − c θ

n −

1 − d θ

ni = 0.