Bagaimana menafsirkan estimasi parameter dalam hasil Poisson GLM [ditutup]

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

Saya ingin tahu cara menafsirkan setiap estimasi parameter dalam tabel di atas.

Saya rasa judul pertanyaan Anda tidak secara akurat menangkap apa yang Anda minta.

Pertanyaan tentang bagaimana menafsirkan parameter dalam GLM sangat luas karena GLM adalah kelas model yang sangat luas. Ingatlah bahwa GLM memodelkan variabel respons yang diasumsikan mengikuti distribusi yang diketahui dari keluarga eksponensial, dan bahwa kami telah memilih fungsi yang tidak dapat dibalik g sedemikian sehingga E [ $y$$g$ untukvariabel prediktor J x . Dalam model ini, interpretasi setiap parameter tertentu β j adalah laju perubahan g ( y ) terhadap x j . Tentukan μ ≡ E [

$$\mathrm{E}\left[y\,|\,x\right] = g^{-1}{\left(x_0 + x_1\beta_1 + \dots + x_J\beta_J\right)}$$ $J$$x$$\beta_j$$g(y)$$x_j$ danη≡x⋅βuntuk menjaga notasi tetap bersih. Kemudian, untuk setiapj∈{1,...,J}, β j$\mu \equiv \mathrm{E}{\left[y\,|\,x\right]} = g^{-1}{\left(x\right)}$$\eta \equiv x \cdot \beta$$j \in \{1,\dots,J\}$ Sekarang tentukanejmenjadi vektor dariJ-1nol dan satu1di posisij, sehingga misalnya jikaJ=5makae3=(0,0,1,0,0). Kemudian βj=g(E $$\beta_j = \frac{\partial\,\eta}{\partial\,x_j} = \frac{\partial\,g(\mu)}{\partial\,x_j} \text{.}$$ $\mathfrak{e}_j$$J-1$$1$$j$$J=5$$\mathfrak{e}_3 = \left(0,0,1,0,0\right)$ $$\beta_j = g{\left(\mathrm{E}{\left[y\,|\,x + \mathfrak{e}_j \right]}\right)} - g{\left(\mathrm{E}{\left[y\,|\,x\right]}\right)}$$

$\beta_j$$\eta$$x_j$

$$\frac{\operatorname{\partial}\mathrm{E}{\left[y\,|\,x\right]}}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{d}\mu}{\operatorname{d}\eta}\frac{\operatorname{\partial}\eta}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mu}{\operatorname{\partial}\eta} \beta_j = \frac{\operatorname{d}g^{-1}}{\operatorname{d}\eta} \beta_j$$ $$\mathrm{E}{\left[y\,|\,x + \mathfrak{e}_j \right]} - \mathrm{E}{\left[y\,|\,x\right]} \equiv \operatorname{\Delta_j} \hat y = g^{-1}{\left( \left(x + \mathfrak{e}_j\right)\beta \right)} - g^{-1}{\left( x\,\beta \right)}$$

$g$$\beta_j$$\eta$$y$$x_j$$y$$x_j$$g^{-1}{\left(\beta\right)}$

---

$y \sim \mathrm{Poisson}{\left(\lambda\right)}$$g = \ln$ . Maka kita bisa mendapatkan daya tarik sehubungan dengan interpretasi tertentu.

$\frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{d}g^{-1}}{\operatorname{d}\eta} \beta_j$$g(\mu) = \ln(\mu)$$g^{-1}(\eta) = e^\eta$$\frac{\operatorname{d}e^\eta}{\operatorname{d}\eta} = e^\eta$

$$\frac{\operatorname{\partial}\mu}{\operatorname{\partial}x_j} = \frac{\operatorname{\partial}\mathrm{E}{\left[y\,|\,x\right]}}{\operatorname{\partial}x_j} = e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J}\beta_j$$

yang akhirnya berarti sesuatu yang nyata:

$x_j$$\hat y$$\hat y\,\beta_j$

Catatan: perkiraan ini sebenarnya dapat bekerja untuk perubahan sebesar 0,2, tergantung pada seberapa banyak presisi yang Anda butuhkan.

$$\begin{align} \operatorname{\Delta_j} \hat y &= e^{ x_0 + x_1\beta_1 + \dots + \left(x_j + 1\right)\,\beta_j + \dots + x_J\beta_J } - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J + \beta_j} - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J}e^\beta_j - e^{x_0 + x_1\beta_1 + \dots + x_J\beta_J} \\ &= e^{ x_0 + x_1\beta_1 + \dots + x_J\beta_J} \left( e^\beta_j - 1 \right) \end{align}$$ 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.

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").