Write

v = σ^{2}

. For one observation, the score is

s_{μ} (x; μ, v) = \frac{x - μ}{v}, s_{v} (x; μ, v) = - \frac{1}{2 v} + \frac{( x - μ ) ^{2}}{2 v ^{2}},

and

I_{1} (μ, v) = (v^{- 1} 0 0 (2 v^{2})^{- 1}) .

Under the restriction

v = 1

μ = \overset{ˉ}{X}

. At

(\overset{ˉ}{X}, 1)

the first score component is zero, while the second is

\frac{1}{2} \sum_{i = 1}^{n} ((X_{i} - \overset{ˉ}{X})^{2} - 1) .

Since

I_{1} (\overset{ˉ}{X}, 1)^{- 1} = diag (1, 2)

, substitution into the score statistic gives

T_{n} = [\frac{1}{2 n} i = 1 \sum n ((X_{i} - \overset{ˉ}{X})^{2} - 1)]^{2} .

ii

Solution

Under

H_{0}

, regardless of the unknown value of

μ

Q_{n} = \sum_{i = 1}^{n} (X_{i} - \overset{ˉ}{X})^{2} \sim χ_{n - 1}^{2} .

Write

Q_{n}

as a sum of

n - 1

independent squared standard normals. Since such a square has mean one and variance two, the central limit theorem gives

\frac{Q _{n} - ( n - 1 )}{2 ( n - 1 )} d N (0, 1) .

Now

\frac{Q _{n} - n}{2 n} = \frac{n - 1}{n} \frac{Q _{n} - ( n - 1 )}{2 ( n - 1 )} - \frac{1}{2 n},

so Slutsky's theorem makes this converge to

N (0, 1)

. Squaring and using the continuous mapping theorem yields

T_{n} d χ_{1}^{2} .

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