方針
3.19とほとんど同じ問題．
答案
\begin{align}
P[Y\geq \lambda]&= \int_\lambda^\infty\frac{1}{\Gamma(x)}y^{x-1}\exp(-y)dy\\
&= \frac{1}{(x-1)!}\int_\lambda^\infty y^{x-1}\exp(-y)dy\\
&= \frac{1}{(x-1)!}\int_\lambda^\infty y^{x-1}\{-\exp(-y)\}'dy\\
&= \frac{1}{(x-1)!}\{\left[-y^{x-1}\exp(-y)\right]_\lambda^\infty+\int_\lambda^\infty(x-1)y^{x-2}\exp(-y)dy\}\\
&= \frac{\lambda^{x-1}}{(x-1)!}\exp(-\lambda)+\frac{1}{(x-2)!}\int_\lambda^\infty y^{x-2}\exp(-y)dy\tag{1}
\end{align}
ここで，
P_{x-1}=\frac{1}{(x-1)!}\int_\lambda^\infty y^{x-1}\exp(-y)dy
とおくと，
\begin{align}
P_0&= \int_\lambda^\infty\exp(-y)dy\\
&= \left[-\exp(-y)\right]_\lambda^\infty\\
&= \exp(-\lambda)\\
&= \frac{\lambda^0}{0!}\exp(-\lambda)
\end{align}
したがって．(1)より，
\begin{align}
P[Y\geq\lambda]&= P_{x-1}\\
&= \frac{\lambda^{x-1}}{(x-1)!}\exp(-\lambda)+P_{x-2}\\
&= \frac{\lambda^{x-1}}{(x-1)!}\exp(-\lambda)+\frac{\lambda^{x-2}}{(x-2)!}\exp(-\lambda)+P_{x-3}\\
&\vdots\\
&= \frac{\lambda^{x-1}}{(x-1)!}\exp(-\lambda)+\frac{\lambda^{x-2}}{(x-2)!}\exp(-\lambda)+...+\frac{\lambda^0}{0!}\exp(-\lambda)\\
&= \sum_{k=0}^{x-1}\frac{\lambda^x}{k!}\exp(-\lambda)\\
&= P[X\leq x-1]
\end{align}
参考文献
- 『現代数理統計学の基礎』(久保川達也 著)