\begin{align} E \left( X_n \right) &= \sum_{k=1}^{2^n} \frac{k}{2^n} \frac{1}{2^n} \\ &= \frac{1}{2^{2n}} \sum_{k=1}^{2^n} k \\ &= \frac{1}{2^{2n}} \cdot \frac{1}{2} 2^n (2^n+1) \\ &= \frac{1}{2} + \frac{1}{2^{n+1}} \\ &\to \frac{1}{2} \ \ (n \to \infty) \\ E(X) &= \int_0^1 x \ dx = \frac{1}{2} \end{align}
\begin{align} E \left( X_n^2 \right) &= \sum_{k=1}^{2^n} \left( \frac{k}{2^n} \right)^2 \frac{1}{2^n} \\ &= \frac{1}{2^{3n}} \sum_{k=1}^{2^n} k^2 \\ &= \frac{1}{2^{3n}} \cdot \frac{1}{6} 2^n (2^n+1) (2^{n+1} + 1) \\ &= \frac{(2^n+1) (2^{n+1} + 1)}{6 \cdot 2^{2n}} \\ E \left( X^2 \right) &= \int_0^1 x^2 \ dx = \frac{1}{3} \\ V \left( X_n \right) &= E \left( X_n^2 \right) - E \left( X_n \right)^2 \\ &= \frac{2^{2n} - 1}{12 \cdot 2^{2n}} \\ &\to \frac{1}{12} \ \ (n \to \infty) \\ V \left( X \right) &= E \left( X^2 \right) - E \left( X \right)^2 \\ &= \frac{1}{12} \end{align}
\begin{align} m_{X_n} (t) &= \frac{1}{2^n} \sum_{k=1}^{2^n} e^{\frac{k}{2^n} t} \\ &= \frac{e^{\frac{t}{2^n}}}{2^n} \frac{1-e^{t}}{1-e^{\frac{t}{2^n}}} \\ &\to \frac{e^t - 1}{t} \ \ (n \to \infty) \tag{1} \\ M_X (t) &= \int_0^1 e^{tx} \ dx \\ &= \frac{e^t - 1}{t} \end{align}
ただし、式 (1) は、 $\varepsilon = 1/2^n$ として、次のようにわかる： \begin{align} (1-e^t) \frac{\varepsilon e^{\varepsilon t}}{1 - e^{\varepsilon t}} &= (1-e^t) \frac{\varepsilon + \varepsilon^2 t + \cdots} {- \varepsilon t - \frac{1}{2} \varepsilon^2 t^2 - \cdots} \\ &= (1-e^t) \frac{1 + \varepsilon t + \cdots} {- t - \frac{1}{2} \varepsilon t^2 - \cdots} \\ &\to \frac{e^t - 1}{t} \ \ (\varepsilon \to 0) \end{align}