解答
1
$$\varphi_{X}(0) = \mathbb{E}\left(e^{i\cdot0\cdot X}\right) = \mathbb{E}(1) = 1$$
2
$$
\begin{aligned}
\frac{d^m\varphi_{X}}{dt^m}
&= \underset{\text{m個}}{0+\cdots+0} + \frac{m!\cdot i^m}{m!}\mathbb{E}\left(X^m\right) \\&+\frac{\left\{(m+1)\cdot m\cdots 3\cdot2\right\} i^m\cdot(it)}{(m+1)!}\mathbb{E}\left(X^{m+1}\right) + \frac{\left\{(m+2)\cdot (m+1)\cdots 4\cdot3\right\}i^m\cdot(it)^2}{(m+2)!}\mathbb{E}\left(X^{m+2}\right)+\cdots\\
&= \frac{m!\cdot i^m}{m!}\mathbb{E}\left(X^m\right) + \sum_{n}^{\infty}\frac{i^m(it)^n}{n!}\mathbb{E}\left(m+n\right)\\
\therefore\left.\frac{d^{m} \varphi_{x}}{d t^{m}}\right|_{t=0}
&= i^m \mathbb{E}\left(X^m\right) + 0\qquad \\
\therefore \frac{1}{i^m}\left.\frac{d^{m} \varphi_{x}}{d t^{m}}\right|_{t=0}
&=\mathbb{E}\left(X^m\right)\\
\end{aligned}
$$
3
$$\begin{aligned}
\varphi_X(t) &= \sum_{n=0}^{\infty}\frac{\lambda^k}{k!}e^{-\lambda}e^{itk}\\
&=\sum_{n=0}^{\infty}\frac{\left(\lambda e^{it}\right)^k}{k!}e^{-\lambda}\\
&= e^{\lambda e^{it}}e^{-\lambda}\\
&= e^{\lambda\left(e^{it}-1\right)}
\end{aligned}$$
4
オイラーの公式より、
$$e^{it} = \cos t+i\sin t$$
が成り立つので、
$$
\begin{aligned}
e^{it\left(aX+bY\right)} =& e^{itaX}e^{itbY}\\
=& \left(\cos \left(taX\right) + i\sin\left(taX\right)\right)\left(\cos \left(tbY\right) + i\sin\left(tbY\right)\right)\\
=& \left(\cos\left(taX\right)\cos\left(tbY\right) - \sin\left(taX\right)\sin\left(tbY\right)\right) \\
&+ i\left(\cos \left(taX\right)\sin\left(tbY\right) + \sin\left(taX\right)\cos \left(tbY\right)\right)
\end{aligned}
$$
と分解できる。したがって、\(X,Y\) が独立なので、
$$
\begin{aligned}
\varphi_{aX+bY}(t)
=&\mathbb{E}\left(e^{it\left(aX+bY\right)}\right)\\
=& \mathbb{E}\left(\cos\left(taX\right)\right)\mathbb{E}\left(\cos\left(tbY\right)\right) - \mathbb{E}\left(\sin\left(taX\right)\right)\mathbb{E}\left(\sin\left(tbY\right)\right) \\
&+ i\left(\mathbb{E}\left(\cos \left(taX\right)\right)\mathbb{E}\left(\sin\left(tbY\right)\right) + \mathbb{E}\left(\sin\left(taX\right)\right)\mathbb{E}\left(\cos \left(tbY\right)\right)\right)\\
=& \left(\mathbb{E}\left(\cos taX\right) + i\mathbb{E}\left(\sin taX\right)\right)\cdot\left(\mathbb{E}\left(\cos tbY\right) + i\mathbb{E}\left(\sin tbY\right)\right)\\
=&\mathbb{E}\left(e^{itaX}\right)\cdot\mathbb{E}\left(e^{itbY}\right)\\
=&\varphi_{X}(at)\varphi_{Y}(bt)
\end{aligned}
$$