Timeline for What is the difference between a "link function" and a "canonical link function" for GLM
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S Jan 26, 2022 at 12:37 | history | suggested | IWillDominate | CC BY-SA 4.0 |
Improved the density equation formatting to make it easier to interpret scale parameter phi. Also moved tau before the exponential to emphasise that it doesnt depend on the parameter of interest theta but just the scale.
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| Jan 26, 2022 at 8:38 | review | Suggested edits | |||
| S Jan 26, 2022 at 12:37 | |||||
| Oct 11, 2021 at 14:52 | comment | added | InTheSearchForKnowledge | Hi, could you please explain in detail why "the sum of the residuals is 0" when we use the canonical link function ? Thank you very much for your help! | |
| S Mar 25, 2018 at 9:35 | history | suggested | GZ1995 | CC BY-SA 3.0 |
Correct a typo, canonical link should be $\eta \equiv\theta$ rather than $\mu \equiv \theta$
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| Mar 25, 2018 at 4:48 | review | Suggested edits | |||
| S Mar 25, 2018 at 9:35 | |||||
| May 29, 2015 at 10:16 | comment | added | Ziyuan | Can I have the references for the "desirable statistical properties"? | |
| Jan 28, 2015 at 9:20 | comment | added | Leo Alekseyev | It seems that there's a typo in key sentence of the answer: shouldn't it read "if the function connects $\mu$ and $\theta$ s.t. $\eta \equiv \theta$"? | |
| Mar 14, 2014 at 15:22 | comment | added | Druss2k | Thank you very much. Using the previous example, we've that $\gamma'(\theta) = \pi = \frac{exp(\theta)}{1+exp(\theta)}$. Hence $(\gamma')^{-1}(.) = \text{logit(.)}$. As you said (I just rephrase), we only have $\eta = \theta$ if $g(.)$ is the canonical link, which is the logit. Then we will have $\theta = logit(\pi) = \eta$. So the equality between $\theta$ and the predictor $\eta$ only exists, if we use the canonical link function. | |
| Mar 14, 2014 at 10:10 | comment | added | Momo | I hope I understand your confusion correctly: In the exponential family you talk about, the canonical parameter is $\theta$ and the canonical link is when $\eta=\theta$ which is when $g(\mu)=\theta$. As also $\theta=(\gamma')^{-1}(\mu)$ (if you calculate the expected value of the first derivative with respect to $\theta$ of the likelihood function) the only case when $\theta \equiv \mu$ appears when $g(.)=(\gamma')^{-1}(.)$. | |
| Mar 13, 2014 at 18:04 | comment | added | Druss2k | I still get smth wrong. For instance the binomial distribution. Here, the canonical link is the logit and for one r.v. $Y_i$, we've $\mu=\pi$. Hence $\gamma'(\pi) = \theta(\pi)=\text{log}(\frac{\pi}{1-\pi})$ and $\gamma(\theta) = n\cdot\text{log}(1+\text{exp}(\theta))$ . By the definition that the canonical link is the one which connects $\mu$ and $\theta$ such that $\mu\equiv\theta$, I dont see how this is true if I apply presumably $(\gamma')^{-1}(\theta) = \pi(\theta) = \frac{\text{exp}(\theta)}{1+\text{exp}(\theta)}$? Do I apply $(\gamma')^{-1}$ to the LHS of $\theta(\pi)$? | |
| Feb 21, 2014 at 12:18 | comment | added | Wei | Does a canonical link function always exist in GLM? What are the necessary conditions for it to exist? Thanks. | |
| Oct 21, 2012 at 20:07 | vote | accept | steadyfish | ||
| Oct 21, 2012 at 15:26 | history | edited | Momo | CC BY-SA 3.0 |
added 214 characters in body
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| Oct 21, 2012 at 15:13 | comment | added | gung - Reinstate Monica | +1, this is a really nice answer, @Momo. I did find some of the equations harder to read when they were buried in the paragraphs, so I 'blocked' them out by using double dollar-signs (ie \$$). I hope that's OK (if not, you can rollback, w/ my apologies). | |
| Oct 21, 2012 at 15:11 | history | edited | gung - Reinstate Monica | CC BY-SA 3.0 |
added double dollars ($$) for 'blocked' latex
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| Oct 21, 2012 at 15:09 | history | edited | Momo | CC BY-SA 3.0 |
added 501 characters in body
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| Oct 21, 2012 at 14:58 | history | answered | Momo | CC BY-SA 3.0 |