Normal Distribution

This is a short discussion of testing normally distributed variables. This will give an introduction to the use of maximum likelihood methods for a continuous distribution and provide results which will be use to provide approximate tests for binomial models. The normal distribution will be covered in more detail in Linear Normal Models .

1 Maximum Likelihood

X is a normally distributed random variable, with mean μ and variance σ², ...X ∼ N(μ, σ²) . ...Suppose σ² is known, we want to estimate μ. For an observed value x, define the likelihood function:

L ( μ ) ...:= ...L ( μ | x ) ...:= ...f ( x | μ ) ...= ...

	1
	√2πσ²

exp

– (x – μ)2 2σ²

Where f ( x ) is the density function of the normal distribution. L ( μ ) reaches its maximum when ...μ = x, ...so ...

∧ μ . ...= ...x

2 Likelihood Ratio

Consider the hypothesis ...H: μ = μ₀ . ...The Likelihood Ratio Test is

LR (x) ...= ...

	L(μ0)
	L( ∧ μ . )

...= ...

	exp { – (x – μ0)2 / 2σ²}
	exp { – (x – x)2 / 2σ²}

...........................= ...exp { – (x – μ₀)² / 2σ²}

3 Significance Probability

Recall that for a given hypothesis H: μ = μ₀, the significance probability for the result x is:

SP( x ) ...= ...P{ y | LR (y) ≤ LR (x) }

Now .........LR (y) ≤ LR (x) ......⇔ ......(y – μ₀)² ≥ (x – μ₀)²

case 1) ...x > μ₀, ...then

LR (y) ≤ LR (x) ......⇔ ......y ≥ x ......or ......y ≤ 2μ₀ – x

SP ...= ...P(LR (y) ≤ LR (x)) ...= ...P(y ≥ x) ......+ ......P(y ≤ 2μ₀ – x)

.........= ...

1 – Φ x – μ0 σ

...+ ...Φ

μ0 – x σ

......= ......2Φ

μ0 – x σ

case 2) ...x < μ₀, ...then

LR (y) ≤ LR (x) ......⇔ ......y – μ₀ ...≥ ...μ₀ – x ......or ......μ₀ – y ...≥ ...x – μ₀

..............................⇔ ......y ≥ 2μ₀ – x ......or ......y ≤ x

SP ...= ...P(LR (y) ≤ LR (x)) ...= ...P(y ≤ x) ......+ ......P(y ≥ 2μ₀ – x)

.........= ...Φ

x – μ0 σ

...+ ...

1 – Φ μ0 – x σ

......= ......2Φ

x – μ0 σ

In general

SP( x ) ...= ...2Φ – | x – μ0 | σ

4 Confidence Interval

For which values of μ₀ will we accept the hypothesis at level α? I.e. SP(x) ≥ α, given μ₀?

Let ε_α be the number such that Φ(ε_α) = α. ...ε_α is the α × 100 percentile of the N(0,1) distribution.

case 1) ...x > μ₀, ...and ...2Φ((μ₀ – x) / σ) ≥ α

Φ

μ0 – x σ

......≥ ......α/2

	μ0 – x
	σ

......≥ ......ε_α/2

μ₀ ......≥ ......x + σε_α/2

case 2) ...x < μ₀, ...and ...2Φ((x – μ₀) / σ) ≥ α

Φ

x – μ0 σ

......≥ ......α/2

	x – μ0
	σ

......≥ ......ε_α/2

μ₀ ......≤ ......x – σε_α/2

Combining the two results, we can define the confidence interval as:

x + σεα/2 ......≤ ......μ0 ......≤ ......x – σεα/2

[Remember that ε_α/2 ≤ 0 ...if ...α/2 ≤ 0.5 ...– ...which it is as α < 1 always.]

 

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