&& f(x|a,b) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}\\ && \textrm{Find mode:} \\ && f'(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} [ (a-1)x^{a-2} (1-x)^{b-1} - x^{a-1} (b-1)(1-x)^{b-2} ] = 0 \\ && \Rightarrow (a-1) x^{a-2} (1-x)^{b-1} - x^{a-1} (b-1) (1-x)^{b-2} = 0 \\ && \Rightarrow (a-1) (1-x) - x (b-1) = 0 \\ && \Rightarrow a - ax - 1 + x - xb + x = 0 \\ && \Rightarrow x(2 - a - b) = 1 - a \\ && \Rightarrow x = \frac{1-a}{2-a-b} = \frac{a-1}{a+b-2}

ベータ分布のモードを求める！

aka.tkf+formula@gmail.com - about 1 year ago
Embed the image to your blog in HTML, Markdown, Textile, BBCode.
+Name +Password