東京工業大学 工学院 システム制御系 2022年度 問題3
Author
Description
Kai
問1
\[
\begin{aligned}
\sigma_S^2
&= E \left[ \left( S - E[S] \right)^2 \right]
\\
&= E \left[ S^2 - 2E[S]S + \left( E[S] \right)^2 \right]
\\
&= E \left[ S^2 \right] - 2 \left( E[S] \right)^2 + \left( E[S] \right)^2
\\
&= E \left[ S^2 \right] - \left( E[S] \right)^2
\end{aligned}
\]
問2
\[
\begin{aligned}
\frac{dM_X(t)}{dt}
&= \lambda e^t \exp \left( \lambda (e^t-1) \right)
\\
\frac{d^2 M_X(t)}{dt^2}
&= \lambda e^t \left(\lambda e^t + 1 \right) \exp \left( \lambda (e^t-1) \right)
\end{aligned}
より、
\begin{aligned}
\mu_X
&= \frac{dM_X(0)}{dt}
\\
&= \lambda
\\
E \left[ X^2 \right]
&= \frac{d^2 M_X(0)}{dt^2}
\\
&= \lambda (\lambda + 1)
\\
\sigma_X^2
&= E \left[ X^2 \right] - \mu_X^2
\\
&= \lambda
\end{aligned}
\]
を得る。
問3
\[
\begin{aligned}
M_Y(t)
&= \sum_{y=1}^\infty e^{yt} \cdot (1-q) q^{y-1}
\\
&= \frac{1-q}{q} \sum_{y=1}^\infty \left( q e^t \right)^y
\\
&= \frac{1-q}{q} \cdot q e^t \cdot \frac{1}{1-qe^t}
\\
&= \frac{(1-q)e^t}{1-qe^t}
\end{aligned}
\]
問4
\(X,Y\) が独立なので \(E[XY]=E[X]E[Y]\) が成り立つことを考慮して、
\[
\begin{aligned}
M_Z(t)
&= E \left[ e^{tZ} \right]
\\
&= E \left[ e^{t(X+Y)} \right]
\\
&= E \left[ e^{tX} e^{tY} \right]
\\
&= E \left[ e^{tX} \right] E \left[ e^{tY} \right]
\\
&= M_X(t) M_Y(t)
\end{aligned}
\]
を得る。
問5
(1)
\[
\begin{aligned}
M_Z(t)
&= M_X(t) M_Y(t)
\\
&= e^t \exp \left( e^t-1 \right) \cdot \frac{e^t}{2-e^t}
\\
&= \frac{\exp \left( e^t-1 \right)}{2e^{-t}-1}
\end{aligned}
\]
(2)
\[
\begin{aligned}
\frac{dM_Z(t)}{dt}
&= \frac{\exp \left( e^t-1 \right) \left( 2 - e^t + 2e^{-t} \right)}{\left( 2e^{-t}-1 \right)^2}
\end{aligned}
\]
(3)
\[
\begin{aligned}
\mu_Z
&= \left. \frac{dM_Z(t)}{dt} \right|_{t=0}
\\
&= 3
\end{aligned}
\]