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UA MATH566 例题 Poisson回归、Overdispersion与负二项回归

發布時間：2025/4/14 编程问答 24 豆豆

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UA MATH566 例題 Poisson回歸、Overdispersion

Poisson regression is widely used in modelling count data. Model assumption is $Yi～iidPois(βxi),i=1,?,nY_i \sim_{iid} Pois(\beta x_i),i=1,\cdots,n$ , $β∈R1\beta \in \mathbb{R}^1$ .

Part (a) Find MLE of $β\beta$ . Denote as $β^\hat\beta$
Part (b) A drawback of Poisson regression is that mean and variance of reponse are assumed to be equal. If variance of response is greater than mean of reponse, we say the count data is overdispersion. Cameron and Trivedi (1990) developed regression-based method to test overdispersion. Mean response of Poisson regression is $E[Yi∣xi]=βxiE[Y_i|x_i] = \beta x_i$ , and let $μi\mu_i$ denote $βxi\beta x_i$ . Hypothesis for overdispersion test is
$H0:Var[Yi∣xi]=μiHa:Var[Yi∣xi]>μiH_0:Var[Y_i|x_i] = \mu_i \\ H_a:Var[Y_i|x_i]>\mu_i$

Assume $Var[Yi∣xi]=μi+αμiVar[Y_i|x_i] = \mu_i + \alpha \mu_i$ , $α>0\alpha>0$ indicating overdispersion,
$Var[Yi∣xi]=μi+αμi?E[(Yi?μi)2?Yi]=αμiVar[Y_i|x_i] = \mu_i + \alpha \mu_i \Rightarrow E[(Y_i-\mu_i)^2-Y_i] = \alpha \mu_i$

Suppose the data generating process is
$(Yi?μi)2?Yi=αμi+?i(Y_i-\mu_i)^2 - Y_i = \alpha \mu_i + \epsilon_i$

Use weighted least square to estimate $α\alpha$ , assuming weights are $wi=1/xi,i=1,?,nw_i=1/x_i,i=1,\cdots,n$ .

Answer.
Part (a)
Joint likelihood of model $Yi～iidPois(βxi),i=1,?,nY_i \sim_{iid} Pois(\beta x_i),i=1,\cdots,n$ :
$L(β)=∏i=1n(βxi)Yie?βxiYi!=β∑i=1nYie?β∑i=1nxi∏i=1nxiYiYi!ln?L(β)=nYˉln?β?βnxˉ?ln?∏i=1nxiYiYi!?ln?L(β)?β=nYˉβ?nxˉ=0?β^=YˉxˉL(\beta) = \prod_{i=1}^n \frac{(\beta x_i)^{Y_i}e^{-\beta x_i}}{Y_i!} = \beta^{\sum_{i=1}^n Y_i}e^{-\beta \sum_{i=1}^n x_i} \prod_{i=1}^n \frac{x_i^{Y_i}}{Y_i!} \\ \ln L(\beta) = n\bar{Y}\ln \beta - \beta n\bar{x} - \ln \prod_{i=1}^n \frac{x_i^{Y_i}}{Y_i!} \\ \frac{\partial \ln L(\beta)}{\partial \beta} = \frac{n\bar{Y}}{\beta} - n\bar{x} = 0 \Rightarrow \hat\beta = \frac{\bar{Y}}{\bar{x}}$

Part (b)
Residual is
$?i=(Yi?μi)2?Yi?αμi\epsilon_i = (Y_i-\mu_i)^2 - Y_i - \alpha \mu_i$

Replace $μi\mu_i$ with fitted value from Poisson regression,
$μ^i=EYi=β^xi=Yˉxixˉ\hat\mu_i = EY_i = \hat \beta x_i = \frac{\bar{Y} x_i}{\bar{x}}$

So weighted residual square is
$wi?i2=wi((Yi?Yˉxixˉ)2?Yi?αYˉxixˉ)2w_i\epsilon_i^2 = w_i \left( (Y_i-\frac{\bar{Y} x_i}{\bar{x}})^2 - Y_i - \alpha \frac{\bar{Y} x_i}{\bar{x}} \right)^2$

Optimization for WLS is
$min?αQ=∑i=1nwi((Yi?Yˉxixˉ)2?Yi?αYˉxixˉ)2\min_{\alpha} Q = \sum_{i=1}^n w_i \left( (Y_i-\frac{\bar{Y} x_i}{\bar{x}})^2 - Y_i - \alpha \frac{\bar{Y} x_i}{\bar{x}} \right)^2$

Calculate
$?Q?α=?2∑i=1nwiYˉxixˉ((Yi?Yˉxixˉ)2?Yi?αYˉxixˉ)=0?α^=∑i=1nwixi[(Yi?Yˉxixˉ)2?Yi]Yˉxˉ∑i=1nwixi2=∑i=1n(Yi?Yˉxixˉ)2?nYˉnYˉ\frac{\partial Q}{\partial \alpha} =-2\sum_{i=1}^n w_i \frac{\bar{Y} x_i}{\bar{x}} \left( (Y_i-\frac{\bar{Y} x_i}{\bar{x}})^2 - Y_i - \alpha \frac{\bar{Y} x_i}{\bar{x}} \right) = 0 \\ \Rightarrow \hat{\alpha} = \frac{\sum_{i=1}^n w_ix_i[(Y_i-\frac{\bar{Y} x_i}{\bar{x}})^2 - Y_i]}{\frac{\bar{Y}}{\bar{x}}\sum_{i=1}^n w_i x_i^2} = \frac{\sum_{i=1}^n (Y_i-\frac{\bar{Y} x_i}{\bar{x}})^2 - n\bar{Y}}{n\bar{Y}}$

總結

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