# Fisher Information for Cox Model

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.

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.