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Actually, I'm working on a Statistical Genetics Article (Schaid and al,2010) in a retrospective likelihood context. In the article, authors present some result about conditional likelihood but I can't find its, specifically the Fisher Information result.

I have the model of this form:
$\frac{\exp^{\beta*xd}*P(G)}{\sum_{G*} \exp^{\beta*xd^*}*P(G^*)}$

I calculate the score function from the log-likelihood which is equal to:

$loglik = \log(P(G)) +\beta*xd - log(\sum_{G^*} \exp^{\beta*xd^*}*P(G^*))$

$U(\beta) = xd - \frac{\sum_{G*} xd^*\exp^{\beta*xd^*}*P(G^*)}{\sum_{G*}\exp^{\beta*xd^*}*P(G^*)}$

I find for the second derivative this following expression:

$ \frac{\delta^2}{\delta²\beta}loglik = \frac{\sum_{G^*}xd*^2*\exp^{\beta*xd^*}*P(G^*)\sum_{G^*}\exp^{\beta*xd^*}*P(G^*) - \sum_{G^*}xd^*\exp^{\beta*xd^*}*P(G^*)\sum_{G^*}xd^*\exp^{\beta*xd^*}*P(G^*) }{(\sum_{G^*}\exp^{\beta*xd^*}*P(G^*))^2}$

$= \frac{\sum_{G^*}xd*^2*\exp^{\beta*xd^*}*P(G^*)}{\sum_{G^*}\exp^{\beta*xd^*}*P(G^*)}- (\frac{\sum_{G*} xd^*\exp^{\beta*xd^*}*P(G^*)}{\sum_{G*}\exp^{\beta*xd^*}*P(G^*)})^2$

However authors find $\sum_{G^*}Q(G_{G^*},\beta)(xd^* - \mu(\beta))^2$ with $Q(G_{G^*},\beta) = \frac{\exp^{\beta*xd^*}*P(G^*)}{\sum_{G^*}\exp^{\beta*xd^*}*P(G^*)}$ and $\mu(\beta) = \frac{\sum_{G*} xd^*\exp^{\beta*xd^*}*P(G^*)}{\sum_{G*}\exp^{\beta*xd^*}*P(G^*)}$

I'm pretty sure they do some algreba but I don't see what precisely.

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1 Answer 1

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Finally the result is pretty simple. All you need to do is to consider the second derivative of log-likelihood as a weighted variance.

Let define the expectation as $\mu(\beta)$ and the weigth as $Q(G^*,\beta)$

We can rewrite the fisher Information as:

$xd*^2Q(G^*,\beta) - \mu(\beta)^2$

This result can be considered as a weighted variance of the form:

$$\sum_{G^*}Q(G^*,\beta)(xd*-\mu(\beta))^2$$
$$ = \sum_{G^*} Q(G^*,\beta)xd*^2 - 2 \sum_{G^*} xd*Q(G^*,\beta)\mu(\beta) + \sum_{G^*}Q(G^*,\beta)\mu(\beta)^2 $$
$$= \sum_{G^*} Q(G^*,\beta)xd*^2 - 2\mu(\beta)^2 + \mu(\beta)^2 $$
$$= \sum_{G^*}xd*^2Q(G^*,\beta) - \mu(\beta)^2 $$

The trick is not pretty obvious but it's really intuitive as notation.

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