統計検定 1級過去問解答/解答例と解説
2017年11月26日 (日) 試験

統計数理問4 [1]

設問の要約

$Z$ の確率分布は何か

解答例

M_{X} (t) = M_{Y} (t) = \exp [\frac{1}{2} t^{2}]

$\begin{equation} M_X(t) = M_Y(t) = \exp\left[\frac{1}{2}t^2\right] \end{equation}$

\begin{aligned} M_{k X} (t) & = E [e^{k X t}] \\ = \int_{\infty}^{\infty} e^{k x t} \frac{1}{\sqrt{2 π}} \exp [- \frac{x^{2}}{2}] d x \\ = \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{x^{2}}{2} + k x t] d x \\ = \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{1}{2} (x^{2} - 2 k t x)] d x \\ = \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{1}{2} (x^{2} - 2 k t x + k^{2} t^{2} - k^{2} t^{2})] d x \\ = \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{1}{2} (x^{2} - 2 k t x + k^{2} t^{2}) + \frac{1}{2} k^{2} t^{2}] d x \\ = \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{1}{2} (x - k t)^{2} + \frac{1}{2} k^{2} t^{2}] d x \\ = \exp [\frac{k^{2}}{2} t^{2}] \frac{1}{\sqrt{2 π}} \int_{\infty}^{\infty} \exp [- \frac{(x - k t)^{2}}{2}] d x \\ = \exp [\frac{k^{2}}{2} t^{2}] \end{aligned}

$\begin{align} M_{kX}(t) &= E[e^{kXt}] \\ &= \int_{\infty}^{\infty}e^{kxt} \frac{1}{\sqrt{2\pi}}\exp\left[-\frac{x^2}{2}\right]dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{x^2}{2} + kxt\right]dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{1}{2}(x^2 - 2ktx)\right]dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{1}{2}(x^2 - 2ktx + k^2t^2 - k^2t^2)\right]dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{1}{2}(x^2 - 2ktx + k^2t^2) + \frac{1}{2}k^2t^2\right]dx \\ &= \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{1}{2}(x - kt)^2 + \frac{1}{2}k^2t^2\right]dx \\ &= \exp\left[\frac{k^2}{2}t^2\right] \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}\exp\left[-\frac{(x - kt)^2}{2}\right]dx \\ &= \exp\left[\frac{k^2}{2}t^2\right] \end{align}$

\begin{aligned} M_{Z} (t) & = E [e^{Z t}] \\ = E [e^{(a + k X + Y) t}] \\ = e^{a t} E [e^{k X t} e^{Y t}] \\ = e^{a t} E [e^{k X t}] E [e^{Y t}] \\ = e^{a t} \cdot \exp [\frac{k^{2}}{2} t^{2}] \cdot \exp [\frac{1}{2} t^{2}] \\ = \exp [a t + \frac{k^{2} + 1}{2} t^{2}] \end{aligned}

$\begin{align} M_Z(t) &= E[e^{Zt}] \\ &= E[e^{(a + kX + Y)t}] \\ &= e^{at} E[e^{kXt} e^{Yt}] \\ &= e^{at} E[e^{kXt}]E[e^{Yt}] \\ &= e^{at} \cdot \exp\left[\frac{k^2}{2}t^2\right] \cdot \exp\left[\frac{1}{2}t^2\right] \\ &= \exp\left[at + \frac{k^2 + 1}{2}t^2\right] \end{align}$

よって，

Z \sim N (a, k^{2} + 1)

$\begin{equation} Z \sim N(a,~k^2 + 1) \end{equation}$

統計検定 1級 過去問 解答/解答例と解説 2017年11月26日 (日) 試験

統計数理 問4 [1]

設問の要約

解答例

統計検定 1級過去問解答/解答例と解説
2017年11月26日 (日) 試験

統計数理問4 [1]