UA MATH566 例题 Poisson回归、Overdispersion与负二项回归
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,nYi?～iid?Pois(βxi?),i=1,?,n, β∈R1\beta \in \mathbb{R}^1β∈R1.
Part (a) Find MLE of β\betaβ. Denote as β^\hat\betaβ^?
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_iE[Yi?∣xi?]=βxi?, and let μi\mu_iμi? denote βxi\beta x_iβxi?. 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_iH0?:Var[Yi?∣xi?]=μi?Ha?:Var[Yi?∣xi?]>μi?
Assume Var[Yi∣xi]=μi+αμiVar[Y_i|x_i] = \mu_i + \alpha \mu_iVar[Yi?∣xi?]=μi?+αμi?, α>0\alpha>0α>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_iVar[Yi?∣xi?]=μi?+αμi??E[(Yi??μi?)2?Yi?]=αμi?
Suppose the data generating process is
(Yi?μi)2?Yi=αμi+?i(Y_i-\mu_i)^2 - Y_i = \alpha \mu_i + \epsilon_i(Yi??μi?)2?Yi?=αμi?+?i?
Use weighted least square to estimate α\alphaα, assuming weights are wi=1/xi,i=1,?,nw_i=1/x_i,i=1,\cdots,nwi?=1/xi?,i=1,?,n.
Answer.
Part (a)
Joint likelihood of model Yi～iidPois(βxi),i=1,?,nY_i \sim_{iid} Pois(\beta x_i),i=1,\cdots,nYi?～iid?Pois(βxi?),i=1,?,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}}L(β)=i=1∏n?Yi?!(βxi?)Yi?e?βxi??=β∑i=1n?Yi?e?β∑i=1n?xi?i=1∏n?Yi?!xiYi???lnL(β)=nYˉlnβ?βnxˉ?lni=1∏n?Yi?!xiYi????β?lnL(β)?=βnYˉ??nxˉ=0?β^?=xˉYˉ?
Part (b)
Residual is
?i=(Yi?μi)2?Yi?αμi\epsilon_i = (Y_i-\mu_i)^2 - Y_i - \alpha \mu_i?i?=(Yi??μi?)2?Yi??αμi?
Replace μi\mu_iμi? 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}}μ^?i?=EYi?=β^?xi?=xˉYˉxi??
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)^2wi??i2?=wi?((Yi??xˉYˉxi??)2?Yi??αxˉYˉxi??)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)^2αmin?Q=i=1∑n?wi?((Yi??xˉYˉxi??)2?Yi??αxˉYˉxi??)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}}?α?Q?=?2i=1∑n?wi?xˉYˉxi??((Yi??xˉYˉxi??)2?Yi??αxˉYˉxi??)=0?α^=xˉYˉ?∑i=1n?wi?xi2?∑i=1n?wi?xi?[(Yi??xˉYˉxi??)2?Yi?]?=nYˉ∑i=1n?(Yi??xˉYˉxi??)2?nYˉ?
總結
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