`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:

- The effect of a change in the predictors depends on the level of the response.
- An additive change in the predictors has a multiplicative effect on the response.
- 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*