[1]
\(X\sim Exp(\lambda)\)(指数部分)の時、
\(E[X]=\displaystyle \int^\infty _0xf(x)dx\)
\(\displaystyle =\int^\infty _0x\lambda e^{-\lambda x}dx\)
\(\displaystyle =\frac{1}{\lambda}\)
よって、
\(E[U]=E[X_1+X_2]\)\(=\displaystyle \frac{2}{\lambda}\)
[2]
\(U=X_1+X_2,V=X_2\)と変数変換すると、\(X_1=U-V,X_2=V\)で、ヤコビアンは\(||J||=1\)なので、\(x_1=u-v >0 ,x_2=v>0\)すなわち\(u>v>0\)の時
\(f(u,v)=\lambda e^{-\lambda X_1}\lambda e^{-\lambda X_2}\)
\(=\lambda e^{-\lambda (u-v)}\lambda e^{-\lambda v}\)
\(=\lambda^2 e^{-\lambda u} \)
\(U\)の周辺確率密度関数を求めて、
\(g(u)\)\(\displaystyle = \int^u_0 \lambda^2 e^{-\lambda u}dv\)
\(=\lambda^2ue^{-\lambda u} \)
以上より、
\(\displaystyle g(u)=\left\{ \begin{array}{} \lambda^2ue^{-\lambda u} &(u > 0)\\ \displaystyle0&( u \leq 0) \end{array}\right. \)
[3]
\(\displaystyle E\left[\frac{1}{U}\right]= \int^\infty_0 \frac{1}{u}g(u)du\)
\(\displaystyle =\int^\infty_0 \frac{1}{u}\lambda^2ue^{-\lambda u}du\)
\(=\lambda\)
[4]
\(\displaystyle E\left[\frac{\alpha \bar{X}}{\theta}+\frac{\theta}{\alpha \bar{X}}-2\right]\)
\(\displaystyle =E\left[\frac{\alpha \lambda U}{2}+\frac{2}{\alpha \lambda U}-2\right]\)
\(\displaystyle =\frac{\alpha \lambda}{2}E[U]\)\(+\displaystyle \frac{2}{\alpha \lambda}E\left[\frac{1}{ U}\right]\)\(-2\)
\(\displaystyle =\frac{\alpha \lambda}{2}\frac{2}{\lambda}\)\(+\displaystyle \frac{2}{\alpha \lambda}\lambda\)\(-2\)
\(\displaystyle =\alpha + \frac{2}{\alpha}-2\)
\(\geq 2\sqrt{2}-2\)(相加相乗平均の大小関係)
等号成立は、\(\alpha=\displaystyle \frac{2}{\alpha}\)すなわち、\(\alpha=\sqrt{2}\)の時。
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