15
$\begingroup$
Call:
glm(formula = darters ~ river + pH + temp, family = poisson, data = darterData)

Deviance Residuals:
    Min      1Q   Median     3Q    Max
-3.7422 -1.0257   0.0027 0.7169 3.5347

Coefficients:
              Estimate Std.Error z value Pr(>|z|)
(Intercept)   3.144257  0.218646  14.381  < 2e-16 ***
riverWatauga -0.049016  0.051548  -0.951  0.34166
pH            0.086460  0.029821   2.899  0.00374 **
temp         -0.059667  0.009149  -6.522  6.95e-11 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for poisson family taken to be 1)
Null deviance: 233.68 on 99 degrees of freedom
Residual deviance: 187.74 on 96 degrees of freedom
AIC: 648.21

I want to know how to interpret each parameter estimate in the table above.

$\endgroup$
5
  • $\begingroup$ The interpretation is identical: stats.stackexchange.com/a/126225/7071 $\endgroup$
    – dimitriy
    Commented Dec 13, 2014 at 0:31
  • 8
    $\begingroup$ This question appears to be off-topic because it is about explaining an R output without any form of intelligent question behind. This is the category "I dump my computer output there and you run the stat analysis for me"... $\endgroup$
    – Xi'an
    Commented Dec 13, 2014 at 7:19
  • 2
    $\begingroup$ Your dispersion parameter seems to indicate that there are some problems with your model. Perhaps you should consider using a quasipoisson distribution instead. I bet your parameter estimates will change drastically and so will the interpretation. If you run "plot(model)" you will get some plots of your residuals, have a look at these plots for unwanted patterns before you start interpreting your actual model. For quickly plotting the fit of your model you can also use "visreg(modelfit)" from the visreg package $\endgroup$
    – Robbie
    Commented Dec 13, 2014 at 8:56
  • 4
    $\begingroup$ @Xi'an, although the question is sparse & required editing, I don't think it is off-topic. Consider these questions that are not considered off-topic: Interpretation of R's lm() output, & Interpretation of R's output for binomial regression. It does appear to be a duplicate, however. $\endgroup$ Commented Dec 13, 2014 at 16:03
  • 2
    $\begingroup$ This is a duplicate of How to interpret coefficients in a Poisson regression? Please read the linked thread. If you still have a question after reading that, come back here & edit your question to state what you have learned & what you still need to know, then we can provide the information you need without simply duplicating material elsewhere that already didn't help you. $\endgroup$ Commented Dec 13, 2014 at 16:04

2 Answers 2

34
$\begingroup$

I don't think the title of your question accurately captures what you're asking for.

The question of how to interpret the parameters in a GLM is very broad because the GLM is a very broad class of models. Recall that a GLM models a response variable $y$ that is assumed to follow a known distribution from the exponential family, and that we have chosen an invertible function $g$ such that $$ \mathrm{E}\left[y\,|\,x\right] = g^{-1}{\left(x_0 + x_1\beta_1 + \dots + x_J\beta_J\right)} $$ for $J$ predictor variables $x$. In this model, the interpretation of any particular parameter $\beta_j$ is the rate of change of $g(y)$ with respect to $x_j$. Define $\mu \equiv \mathrm{E}{\left[y\,|\,x\right]} = g^{-1}{\left(x\right)}$ and $\eta \equiv x \cdot \beta$ to keep the notation clean. Then, for any $j \in \{1,\dots,J\}$, $$ \beta_j = \frac{\partial\,\eta}{\partial\,x_j} = \frac{\partial\,g(\mu)}{\partial\,x_j} \text{.} $$ Now define $\mathfrak{e}_j$ to be a vector of $J-1$ zeroes and a single $1$ in the $j$th position, so that for example if $J=5$ then $\mathfrak{e}_3 = \left(0,0,1,0,0\right)$. Then $$ \beta_j = g{\left(\mathrm{E}{\left[y\,|\,x + \mathfrak{e}_j \right]}\right)} - g{\left(\mathrm{E}{\left[y\,|\,x\right]}\right)} $$

Which just means that $\beta_j$ is the effect on $\eta$ of a unit increase in $x_j$.

You can also state the relationship in this way: $$ \frac{\operatorname{\partial}\mathrm{E}{\left[y\,|\,x\right]}}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{d}\mu}{\operatorname{d}\eta}\frac{\operatorname{\partial}\eta}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mu}{\operatorname{\partial}\eta} \beta_j = \frac{\operatorname{d}g^{-1}}{\operatorname{d}\eta} \beta_j $$ and $$ \mathrm{E}{\left[y\,|\,x + \mathfrak{e}_j \right]} - \mathrm{E}{\left[y\,|\,x\right]} \equiv \operatorname{\Delta_j} \hat y = g^{-1}{\left( \left(x + \mathfrak{e}_j\right)\beta \right)} - g^{-1}{\left( x\,\beta \right)} $$

Without knowing anything about $g$, that's as far as we can get. $\beta_j$ is the effect on $\eta$, on the transformed conditional mean of $y$, of a unit increase in $x_j$, and the effect on the conditional mean of $y$ of a unit increase in $x_j$ is $g^{-1}{\left(\beta\right)}$.


But you seem to be asking specifically about Poisson regression using R's default link function, which in this case is the natural logarithm. If that's the case, you're asking about a specific kind of GLM in which $y \sim \mathrm{Poisson}{\left(\lambda\right)}$ and $g = \ln$. Then we can get some traction with regard to a specific interpretation.

From what I said above, we know that $\frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{d}g^{-1}}{\operatorname{d}\eta} \beta_j$. And since we know $g(\mu) = \ln(\mu)$, we also know that $g^{-1}(\eta) = e^\eta$. We also happen to know that $\frac{\operatorname{d}e^\eta}{\operatorname{d}\eta} = e^\eta$, so we can say that $$ \frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mathrm{E}{\left[y\,|\,x\right]}}{\operatorname{\partial}x_j} = e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J}\beta_j $$

which finally means something tangible:

Given a very small change in $x_j$, the fitted $\hat y$ changes by $\hat y\,\beta_j$.

Note: this approximation can actually work for changes as large as 0.2, depending on how much precision you need.

And using the more familiar unit change interpretation, we have: \begin{align} \operatorname{\Delta_j} \hat y &= e^{ x_0 + x_1\beta_1 + \dots + \left(x_j + 1\right)\,\beta_j + \dots + x_J\beta_J } - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J + \beta_j} - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J}e^{\beta_j} - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J} \left( e^{\beta_j} - 1 \right) \end{align} which means

Given a unit change in $x_j$, the fitted $\hat y$ changes by $\hat y \left( e^{\beta_j} - 1 \right)$.

There are three important pieces to note here:

  1. The effect of a change in the predictors depends on the level of the response.
  2. An additive change in the predictors has a multiplicative effect on the response.
  3. You can't interpret the coefficients just by reading them (unless you can compute arbitrary exponentials in your head).

So in your example, the effect of increasing pH by 1 is to increase $\ln \hat y$ by $\hat y \left( e^{0.09} - 1 \right)$; that is, to multiply $\hat y$ by $e^{0.09} \approx 1.09$. It looks like your outcome is the number of darters you observe in some fixed unit of time (say, a week). So if you're observing 100 darters a week at a pH of 6.7, raising the pH of the river to 7.7 means you can now expect to see 109 darters a week.

$\endgroup$
15
  • $\begingroup$ I made a couple tweaks here, @ssdecontrol. I think they'll make your post a little easier to follow, but if you don't like them, roll them back with my apologies. $\endgroup$ Commented Dec 13, 2014 at 15:50
  • $\begingroup$ I you can't figure that out from my answer then clearly I need to revise the answer. What are you still confused about? $\endgroup$ Commented Dec 14, 2014 at 4:58
  • 1
    $\begingroup$ @skan no, I mean $E[y|x]$. $x$ and $y$ are random variables representing to a single observation. $x$ is a vector indexed by $j$; $x_j$ is the random variable representing a specific feature/regressor/input/predictor for that observation. $\endgroup$ Commented Aug 24, 2016 at 5:52
  • 3
    $\begingroup$ And don't overthink it. Once you understand all the pieces in a GLM, the manipulations here are just a direct application of calculus principles. It really is as simple as taking the derivative with respect to the variable you're interested in. $\endgroup$ Commented Aug 24, 2016 at 5:55
  • 1
    $\begingroup$ @jocateme You are correct, but it can never happen. In this model, $\hat y$ can never be 0, no matter what the values of $x$ and $\beta$ are. An exponential function can never produce a value of exactly 0, nor can a logarithm (being the inverse of the exponential) accept a value of exactly 0. If 0 is a plausible outcome in your problem, you should not use the logarithmic link function. Note also that the "multiplicative" interpretation of the model specifically depends on using the logarithmic link function. $\endgroup$ Commented Sep 14, 2020 at 19:48
5
$\begingroup$

My suggestion would be to create a small grid consisting of combinations of the two rivers and two or three values of each of the covariates, then use the predict function with your grid as newdata. Then graph the results. It is much clearer to look at the values that the model actually predicts. You may or may not want to back-transform the predictions to the original scale of measurement (type = "response").

$\endgroup$
3
  • 1
    $\begingroup$ As much as I like this approach (I do it all the time) I think it's counterproductive for building understanding. $\endgroup$ Commented Dec 13, 2014 at 3:45
  • 1
    $\begingroup$ I found it disconcerting to see advice to look at predictions being criticized as counterproductive for understanding. $\endgroup$
    – DWin
    Commented Oct 10, 2020 at 20:10
  • 1
    $\begingroup$ Yes, I guess this was 6 years ago, but how can his be counterproductive; and why would a person who thinks so do it all the time? Maybe @shadowtalker would like to explain, even after all this time? $\endgroup$
    – Russ Lenth
    Commented Oct 10, 2020 at 20:16

Not the answer you're looking for? Browse other questions tagged or ask your own question.