📈パレート分布の期待値・分散2024/08/21に公開統計統計学機械学習数理統計statstech 確率密度関数 fX(x)=βαx−(β+1), x∈[α,∞) f_{X}(x) = \beta \alpha x^{-(\beta+1)}, \; x \in[\alpha, \infty) fX(x)=βαx−(β+1),x∈[α,∞) 期待値 1<β1 < \beta1<β に注意. E(X)=∫α∞xfX(x)dx=∫α∞xβαβx−(β+1)dx=∫βαβx−(β+1)+1dx=∫βαβx−(β−1)−1dx=∫β(β−1)−1(β−1)αβαβ−1α−{β−1}x−{(β−1)+1}=αβ(β−1)−1 \begin{aligned} \mathbb{E}(X) & = \int_{\alpha}^{\infty} x f_{X}(x) d x \\ & = \int_{\alpha}^{\infty} x \beta \alpha^{\beta} x^{-(\beta+1)} d x \\ & = \int \beta \alpha^{\beta} x^{-(\beta+1)+1} d x \\ & = \int \beta \alpha^{\beta} x^{-(\beta-1)-1} d x \\ & = \int \beta(\beta-1)^{-1}(\beta-1) \alpha^{\beta} \alpha^{\beta-1} \alpha^{-\{\beta-1\}} x^{-\{(\beta-1)+1\}} \\ & = \alpha \beta(\beta-1)^{-1} \\ \end{aligned} E(X)=∫α∞xfX(x)dx=∫α∞xβαβx−(β+1)dx=∫βαβx−(β+1)+1dx=∫βαβx−(β−1)−1dx=∫β(β−1)−1(β−1)αβαβ−1α−{β−1}x−{(β−1)+1}=αβ(β−1)−1 分散 2<β2 < \beta2<β に注意. E(X2)=∫α∞x2fX(x)dx=∫α∞x2βαβx−(β+1)dx=∫βαβx−(β+1)+2dx=∫βαβx−{(β−2)+1}dx=∫β(β−2)−1(β−2)αβαβ−2α−{β−2}x−{(β−2)+1}=α2β(β−2)−1 \begin{aligned} \mathbb{E}(X^{2}) & = \int_{\alpha}^{\infty} x^{2} f_{X}(x) d x \\ & = \int_{\alpha}^{\infty} x^{2} \beta \alpha^{\beta} x^{-(\beta+1)} d x \\ & =\int \beta \alpha^{\beta} x^{-(\beta+1)+2} d x \\ & =\int \beta \alpha^{\beta} x^{-\{(\beta-2)+1\}} d x \\ & =\int \beta(\beta-2)^{-1}(\beta-2) \alpha^{\beta} \alpha^{\beta-2} \alpha^{-\{\beta-2\}} x^{-\{(\beta-2)+1\}} \\ & =\alpha^{2} \beta(\beta-2)^{-1} \\ \end{aligned} E(X2)=∫α∞x2fX(x)dx=∫α∞x2βαβx−(β+1)dx=∫βαβx−(β+1)+2dx=∫βαβx−{(β−2)+1}dx=∫β(β−2)−1(β−2)αβαβ−2α−{β−2}x−{(β−2)+1}=α2β(β−2)−1 V(X)=E(X2)−E(X)2=α2β(β−2)−1−α2β2(β−1)−2=α2β(β−2)−1(β−1)−2(β−1)2−α2β(β−1)−2(β−2)−1(β−2)β=α2β(β−2)−1(β−1)−2{(β−1)2−(β−2)β}=α2β(β−2)−1(β−1)−2{β2−2β+1−β2+2β}=α2β(β−2)−1(β−1)−2 \begin{aligned} \mathbb{V}(X) & = \mathbb{E}(X^{2})-\mathbb{E}(X)^{2} \\ & = \alpha^{2} \beta(\beta-2)^{-1}-\alpha^{2} \beta^{2}(\beta-1)^{-2} \\ & = \alpha^{2} \beta(\beta-2)^{-1}(\beta-1)^{-2}(\beta-1)^{2}-\alpha^{2} \beta(\beta-1)^{-2}(\beta-2)^{-1}(\beta-2) \beta \\ & = \alpha^{2} \beta(\beta-2)^{-1}(\beta-1)^{-2}\left\{(\beta-1)^{2}-(\beta-2) \beta\right\} \\ & = \alpha^{2} \beta(\beta-2)^{-1}(\beta-1)^{-2}\left\{\beta^{2}-2 \beta+1-\beta^{2}+2 \beta\right\} \\ & = \alpha^{2} \beta(\beta-2)^{-1}(\beta-1)^{-2} \end{aligned} V(X)=E(X2)−E(X)2=α2β(β−2)−1−α2β2(β−1)−2=α2β(β−2)−1(β−1)−2(β−1)2−α2β(β−1)−2(β−2)−1(β−2)β=α2β(β−2)−1(β−1)−2{(β−1)2−(β−2)β}=α2β(β−2)−1(β−1)−2{β2−2β+1−β2+2β}=α2β(β−2)−1(β−1)−2 DiscussionログインするとコメントできますLogin
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