Poisson deviance

For independent Poisson observations with fitted means

μ_{i}

, the deviance from the saturated model is

D = 2 \sum_{i} {y_{i} lo g \frac{y _{i}}{μ _{i}} - (y_{i} - μ_{i})},

with

0 lo g 0 = 0

.

Table of contents
- Quadratic Pearson approximation to the Poisson deviance Poisson deviance
- Poisson deviance simplifies when an intercept is fitted Poisson deviance

Quadratic Pearson approximation to the Poisson deviance ( $D ≃ \sum_{i} (y_{i} - e_{i})^{2} / e_{i}$ )

Write

y_{i} = e_{i} + r_{i}

. For

∣ r_{i} ∣/ e_{i}

small, the cell contribution to the Poisson deviance expands as

2 [(e_{i} + r_{i}) lo g (1 + r_{i} / e_{i}) - r_{i}] = \frac{r _{i}^{2}}{e _{i}} - \frac{r _{i}^{3}}{3 e _{i}^{2}} + O (r_{i}^{4} / e_{i}^{3}) .

Summing gives the Pearson chi-squared statistic as the quadratic approximation. The approximation can be inaccurate for sparse expected counts or cells with relative residuals near minus one.

Poisson deviance simplifies when an intercept is fitted ( $D = 2 \sum_{i} y_{i} lo g (y_{i} / e_{i})$ )

The full Poisson deviance is

2 \sum_{i} [y_{i} lo g (y_{i} / e_{i}) - y_{i} + e_{i}]

. An unpenalized log-mean intercept has likelihood score

\sum_{i} (y_{i} - e_{i})

, which vanishes at an interior fit. The linear terms then cancel, giving the displayed expression. Without that intercept score equation, discarding the linear terms is not justified. Use the limiting convention

0 lo g (0/ e_{i}) = 0

.

 Ancestors (8)

 Incoming links (28)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version