MLE von
Sei eine Zufallsstichprobe aus einer Verteilung mit pdf X1,X2,X3,...,XnX1,X2,X3,...,XnX_{1},X_{2},X_{3},...,X_{n}f(x;α,θ)=e−x/θθαΓ(α)xα−1I(0,∞)(x),α,θ>0f(x;α,θ)=e−x/θθαΓ(α)xα−1I(0,∞)(x),α,θ>0f(x;\alpha,\theta)=\frac{e^{-x/\theta}}{\theta^{\alpha}\Gamma(\alpha)}x^{\alpha-1}I_{(0,\infty)}(x ),\alpha,\theta>0 Finden Sie den Maximum-Likelihood-Schätzer für und . Seiαα\alphaθθ\thetaΨ(α)=dΓ(α)dαΨ(α)=dΓ(α)dα\Psi(\alpha)=\frac{d\Gamma(\alpha)}{d\alpha} Mein Versuch, L(α,θ)===∏i=1nf(xi)∏i=1ne−xi/θθαΓ(α)xα−1i1Γn(α)⋅θnα(∏i=1nxi)α−1exp(−∑i=1nxiθ)L(α,θ)=∏i=1nf(xi)=∏i=1ne−xi/θθαΓ(α)xiα−1=1Γn(α)⋅θnα(∏i=1nxi)α−1exp(−∑i=1nxiθ)\begin{eqnarray*} \mathcal{L}(\alpha,\theta)&=&\prod_{i=1}^{n}f(x_i)\\ &=&\prod_{i=1}^{n}\frac{e^{-x_i/\theta}}{\theta^{\alpha}\Gamma(\alpha)}x_i^{\alpha-1}\\ &=&\frac{1}{\Gamma^{n}(\alpha)\cdot \theta^{n \alpha}}(\prod_{i=1}^{n}x_i)^{\alpha-1}\exp(-\sum_{i=1}^{n}\frac{x_i}{\theta}) \end{eqnarray*} ℓ(α,θ)δℓ(α,θ)δθ1θ2∑i=1nxiθ^=====−nlog(Γ(α))−nαlog(θ)+(α−1)∑i=1nlog(xi)−1θ∑i=1nxi−nαθ+1θ2∑i=1nxi=0nαθ∑ni=1xinα1αx¯ℓ(α,θ)=−nlog(Γ(α))−nαlog(θ)+(α−1)∑i=1nlog(xi)−1θ∑i=1nxiδℓ(α,θ)δθ=−nαθ+1θ2∑i=1nxi=01θ2∑i=1nxi=nαθθ^=∑i=1nxinα=1αx¯\begin{eqnarray*} \ell(\alpha,\theta)&=&-n\log(\Gamma(\alpha))-n\alpha\log(\theta)+(\alpha-1)\sum_{i=1}^{n}\log(x_i)-\frac{1}{\theta}\sum_{i=1}^{n}x_i\\ \frac{\delta \ell(\alpha,\theta)}{\delta \theta}&=&-\frac{n\alpha}{\theta}+\frac{1}{\theta^2}\sum_{i=1}^{n}x_i=0\\ \frac{1}{\theta^2}\sum_{i=1}^{n}x_i&=&\frac{n\alpha}{\theta}\\ \hat{\theta}&=&\frac{\sum_{i=1}^{n}x_i}{n\alpha}\\ &=&\frac{1}{\alpha}\bar{x}\\ \end{eqnarray*} dℓ(α,θ^)dαlog(α)−Γ′(α)Γ(α)===−n⋅Γ′(α)Γ(α)−nlog(1αx¯)+∑i=1nlog(xi)=0−n⋅Γ′(α)Γ(α)+nlog(α)−nlog(x¯)+∑i=1nlog(xi)=0log(x¯)−∑ni=1log(xi)ndℓ(α,θ^)dα=−n⋅Γ′(α)Γ(α)−nlog(1αx¯)+∑i=1nlog(xi)=0=−n⋅Γ′(α)Γ(α)+nlog(α)−nlog(x¯)+∑i=1nlog(xi)=0log(α)−Γ′(α)Γ(α)=log(x¯)−∑i=1nlog(xi)n\begin{eqnarray*} \frac{d \ell(\alpha,\hat{\theta})}{d\alpha}&=&\frac{-n \cdot \Gamma'(\alpha)}{\Gamma(\alpha)}-n\log(\frac{1}{\alpha}\bar{x})+\sum_{i=1}^{n}\log(x_i)=0\\ &=&\frac{-n \cdot \Gamma'(\alpha)}{\Gamma(\alpha)}+n\log(\alpha)-n\log(\bar{x})+\sum_{i=1}^{n}\log(x_i)=0\\ \log(\alpha)-\frac{\Gamma'(\alpha)}{\Gamma(\alpha)}&=&\log(\bar{x})-\frac{\sum_{i=1}^{n}\log(x_i)}{n} \end{eqnarray*} Ich konnte nicht mehr …