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战巡
Posted at 2024-10-22 12:41:37
Last edited by 战巡 at 2024-10-22 21:29:00\[P(X=k)=C_{\alpha+k-1}^k\left(\frac{\beta}{1+\beta}\right)^\alpha\left(\frac{1}{1+\beta}\right)^k\]
\[L(\boldsymbol{X}|\alpha,\beta)=\prod_{i=1}^nC_{\alpha+X_i-1}^{X_i}\left(\frac{\beta}{1+\beta}\right)^\alpha\left(\frac{1}{1+\beta}\right)^{X_i}=\left(\frac{\beta}{1+\beta}\right)^{n\alpha}\left(\frac{1}{1+\beta}\right)^{n\bar{X}}\prod_{i=1}^nC_{\alpha+X_i-1}^{X_i}\]
\[=\left(\frac{\beta}{1+\beta}\right)^{n\alpha}\left(\frac{1}{1+\beta}\right)^{n\bar{X}}\prod_{k=0}^{\infty}(C_{\alpha+k-1}^{k})^{n_k}\]
\[l(\boldsymbol{X}|\alpha,\beta)=\ln(L(\boldsymbol{X}|\alpha,\beta))=n\alpha\ln(\frac{\beta}{1+\beta})+n\bar{X}\ln(\frac{1}{1+\beta})+\sum_{k=0}^\infty n_k\ln(C_{\alpha+k-1}^k)\]
这里面
\[\ln(C_{\alpha+k-1}^k)=\ln(\frac{(\alpha+k-1)(\alpha+k-2)...(\alpha+1)\alpha}{k!})=\sum_{j=1}^k\ln(\alpha+j-1)-\ln(k!)\]
\[\frac{d}{d\alpha}\ln(C_{\alpha+k-1}^k)=\frac{d}{d\alpha}[\sum_{j=1}^k\ln(\alpha+j-1)-\ln(k!)]=\sum_{j=1}^k\frac{1}{\alpha+j-1}\]
故此
\[
\begin{cases}\frac{d}{d\alpha}l(\boldsymbol{X}|\alpha,\beta)=n\ln(\frac{\beta}{1+\beta})+\sum_{k=0}^\infty n_k\sum_{j=1}^k\frac{1}{\alpha+j-1}=0\\\frac{d}{d\beta}l(\boldsymbol{X}|\alpha,\beta)=(\alpha-\beta\bar{X})\cdot\frac{n}{\beta(1+\beta)}=0\end{cases}\]
然后
\[\hat{\beta}=\frac{\hat{\alpha}}{\bar{X}}\]
\[n\ln(\frac{\hat{\alpha}}{\hat{\alpha}+\bar{X}})+\sum_{k=0}^\infty n_k\sum_{j=1}^k\frac{1}{\hat{\alpha}+j-1}=0\]
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