Neg Binomial und der Prior von Jeffreys

Ich versuche, den Jeffreys-Prior für eine negative Binomialverteilung zu erhalten. Ich kann nicht sehen, wo ich falsch liege. Wenn also jemand darauf hinweisen könnte, wäre das sehr willkommen.

Okay, die Situation ist also folgende: Ich soll die vorherigen Verteilungen vergleichen, die unter Verwendung eines Binomials und eines negativen Binomials erhalten wurden, wobei (in beiden Fällen) Versuche und vorliegen $n$ $m$ Erfolge vorliegen. Ich bekomme die richtige Antwort für den Binomialfall, aber nicht für das negative Binomial.

Nennen wir Jeffreys 'vorheriges . Dann, $\pi_J(\theta)$

π_{J} (θ) \propto [I (θ)]^{1 / 2} .

$\pi_J(\theta)\propto [I(\theta)]^{1/2}.$

Unter den Regelmäßigkeitsbedingungen (erfüllt, wenn wir es mit der exponentiellen Familie zu tun haben),

wobei für das negative Binomim obigen Ausdruckist(die Gesamtzahl der Erfolgeist fest,nicht). Die Verteilung - ich denke - ist

I (θ) = - E (\frac{\partial^{2} \log L (θ | x)}{\partial θ^{2}})

$I(\theta)=-E\left(\frac{\partial^2 \log L(\theta|x)}{\partial \theta^2}\right)$

n

$n$

x

$x$

m

$m$

n

$n$

da als Erfolgswahrscheinlichkeit definiert ist und die Anzahl der Erfolge ist. Dies ist auch die Wahrscheinlichkeit, da ein Skalar und kein Vektor ist. Daher,

p (m | θ) \propto θ^{m} (1 - θ)^{n - m}

$p(m|\theta)\propto\theta^m(1-\theta)^{n-m}$

θ

$\theta$

m

$m$

m

$m$

L (θ | n) \propto θ^{m} (1 - θ)^{n - m} \log L (θ | n) = m \log θ + (n - m) \log (1 - θ) \frac{\partial \log L (θ | n)}{\partial θ} = \frac{m}{θ} - \frac{n - m}{1 - θ} \frac{\partial^{2} \log L (θ | n)}{\partial θ^{2}} = - \frac{m}{θ^{2}} - \frac{n - m}{(1 - θ)^{2}}

$L(\theta|n)\propto\theta^m(1-\theta)^{n-m}\\ \log L(\theta|n)=m\log\theta +(n-m)\log (1-\theta)\\ \frac{\partial\log L(\theta|n)}{\partial \theta}=\frac{m}{\theta}-\frac{n-m}{1-\theta}\\ \frac{\partial^2\log L(\theta|n)}{\partial \theta^2}=-\frac{m}{\theta^2}-\frac{n-m}{(1-\theta)^2}$ so the Fisher information is

I (θ) = - E (\frac{\partial^{2} \log L (θ | n)}{\partial θ^{2}}) = \frac{m}{θ^{2}} + \frac{E (n) - m}{(1 - θ)^{2}} = \frac{m}{θ^{2}} + \frac{\frac{m θ}{1 - θ} - m}{(1 - θ)^{2}} = \frac{m (1 - θ)^{2} + \frac{m θ^{3}}{(1 - θ)} - m θ^{2}}{θ^{2} (1 - θ)^{2}} = \frac{m (1 - 2 θ) + \frac{m θ^{3}}{(1 - θ)}}{θ^{2} (1 - θ)^{2}} = \frac{m (1 - 2 θ) (1 - θ) + m θ^{3}}{θ^{2} (1 - θ)^{3}} = \frac{m (1 - 3 θ + 2 θ^{2} + θ^{3})}{θ^{2} (1 - θ)^{3}} \propto \frac{1 - 3 θ + 2 θ^{2} + θ^{3}}{θ^{2} (1 - θ)^{3}}

$I(\theta)=-E\left(\frac{\partial^2\log L(\theta|n)}{\partial \theta^2}\right)=\frac{m}{\theta^2}+\frac{E(n)-m}{(1-\theta)^2}=\frac{m}{\theta^2}+\frac{\frac{m\theta}{1-\theta}-m}{(1-\theta)^2}\\ =\frac{m(1-\theta)^2+\frac{m\theta^3}{(1-\theta)}-m\theta^2}{\theta^2(1-\theta)^2}=\frac{m(1-2\theta)+\frac{m\theta^3}{(1-\theta)}}{\theta^2(1-\theta)^2}\\ =\frac{m(1-2\theta)(1-\theta)+m\theta^3}{\theta^2(1-\theta)^3}=\frac{m(1-3\theta+2\theta^2+\theta^3)}{\theta^2(1-\theta)^3}\\ \propto\frac{1-3\theta+2\theta^2+\theta^3}{\theta^2(1-\theta)^3}$

This, however, does not give me the correct answer. The correct answer is

π_{J} (θ) \propto \frac{1}{θ (1 - θ)^{1 / 2}}

$\pi_J(\theta)\propto \frac{1}{\theta(1-\theta)^{1/2}}$ which means that the Information I get should be

I (θ) = \frac{1}{θ^{2} (1 - θ)}

$I(\theta)=\frac{1}{\theta^2(1-\theta)}$ since the prior should be proportional to the square root of the information.

Can anyone find any mistakes? I wouldn't be surprised if I screwed something up with the set up of the distribution (successes vs failures with their respective probabilities, etc).

I used the expected value from Wikipedia and I know the correct answer from here (page 3).

— hejseb
quelle

The problem arises because the negative binomial distribution can be formulated differently. As a consequence, the expectation differs for different formulations. The way you have specified the negative binomial distribution, the expectation of $n$ is $E(n) = m/\theta$ (e.g. see here on page 3). With that, the Fisher information simplifies to

I (θ) = m (\frac{1}{θ^{2} (1 - θ)})

$I(\theta) = m\left(\frac{1}{\theta^2(1-\theta)}\right)$

Thus the Jeffreys' prior is

π_{J} (θ) = | I (θ) |^{1 / 2} \propto θ^{- 1} (1 - θ)^{- 1 / 2}

$\pi_{J}(\theta) = |I(\theta)|^{1/2}\propto \theta^{-1}(1-\theta)^{-1/2}$

as you already noted.

— COOLSerdash
quelle

Terrific! That's very helpful and also an excellent reference as it goes through the very problem I was struggling with. Thank you!

— hejseb

I've found a solution that uses another formulation, see here. Glad I could help. You're welcome.

— COOLSerdash