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I understand the deviance residuals $r_D$ for a Poisson GLM with log link function are given by $r_D = \mu_{ij} \log(\mu_{ij}/\hat{\mu}_{ij}) + (\hat{\mu}_{ij} - \mu_{ij})$

I was wondering though what the formula would be for a Poisson GLM with identity link function? Is it the same or not?

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  • $\begingroup$ I don't think your formula is correct. (For example, note that the $i$th deviance residual is a function of the $i$th observed, $y_i$, which your fomula lacks). Where did your formula come from? $\endgroup$
    – Glen_b
    Commented Mar 18, 2019 at 11:12
  • $\begingroup$ I saw this formula mentioned in an answer posted here stats.stackexchange.com/questions/99065/… - what would be the correct formula then? $\endgroup$
    – Tom Wenseleers
    Commented Mar 18, 2019 at 12:04
  • $\begingroup$ I think that answer conflates two different things. I've put what I believe is the correct formula into my answer. $\endgroup$
    – Glen_b
    Commented Mar 18, 2019 at 12:07

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The deviance of the $i$th observation - and hence the corresponding deviance residual - is determined by the distribution family; it is not affected by the link function (except in that the link function affects the estimate of $\mu_i$).

I think the $i$th deviance residual for the Poisson model is

$\text{sign}(y_i-\hat{\mu}_i)\sqrt{2\{y_i\log(y_i/\hat{\mu}_i)-(y_i-\hat{\mu}_i)\}}$

(and as Gordon points out, $0$ in the case that $y_i$ and $\hat{\mu}_i$ are both zero)

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  • $\begingroup$ Many thanks for this - much appreciated - makes sense! $\endgroup$
    – Tom Wenseleers
    Commented Mar 18, 2019 at 12:04
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    $\begingroup$ It might be worthwhile to add that the deviance residual is defined to be zero when $y_i$ and $\hat\mu_i$ are both zero. $\endgroup$
    – Gordon Smyth
    Commented Mar 19, 2019 at 23:59

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