How to interpret parameter estimates in Poisson GLM results [closed]

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.

Answer

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.

Attribution
Source : Link , Question Author : tomjerry001 , Answer Author : shadowtalker

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