Fraga pa tenta 140603

Asymptotiska fordelningen ges i facit av

\(\hat \gamma \approx N(\gamma, \theta(1 - \theta)^2/(n(1 - \theta)^2))\)

Hur fas uttrycket \(\theta(1 - \theta)^2/(n(1 - \theta)^2))\)? Min tanke var att anvanda \(I_x(\hat \gamma)^{-1}\). Men detta blir, givet att \(I(\theta) = \frac{nk}{\theta(1-\theta)^2}\),

1. \(I_x(\hat \gamma)^{-1} =\)

2. \(= \frac{\hat\gamma(1 - \hat\gamma)^2}{nk}\)

3. \(\frac{\frac{\hat\theta}{1 - \hat\theta}(1 - \frac{\hat\theta}{1 - \hat\theta})^2}{nk}\)

4. \(\frac{\frac{\hat\theta}{1 - \hat\theta}(\frac{1- \hat\theta}{1 - \hat\theta} - \frac{\hat\theta}{1 - \hat\theta})^2}{nk}\)

5. \(\frac{\frac{\hat\theta}{(1 - \hat\theta)}(\frac{1 - 2\hat\theta}{1 - \hat\theta})^2}{nk}\)

6. \(\frac{\hat\theta(1-2\hat\theta)^2}{(1 - \hat\theta)^3nk}\)

Vilket ger ett annat resultat an i facit.