PRML 第1章 1.10(基本)

解答

$${x}$$，$${z}$$が独立であることから，$${p(x,z) = p(x)p(z)}$$となる．

連続変数の場合：

$$
\begin{array}{lll}
\mathbb{E}[x+z]
&=& {\displaystyle \iint} (x+z) p(x,z) dxdz\\[1em]
&=& {\displaystyle \iint} (x+z) p(x)p(z) dxdz\\[1em]
&=& {\displaystyle \iint} x p(x)p(z) dxdz + {\displaystyle \iint} z p(x)p(z) dxdz\\[1em]
&=& {\displaystyle \int} xp(x)dx {\displaystyle \int} p(z)dz + {\displaystyle \int} p(x)dx {\displaystyle \int} zp(z)dz\\[1em]
&=& {\displaystyle \int} xp(x)dx + {\displaystyle \int} zp(z)dz\\[1em]
&=& \mathbb{E}[x] + \mathbb{E}[z]
\end{array}
$$

$$
\begin{array}{lll}
\mathrm{var}[x+z]
&=& \mathbb{E}[\{(x+z) - \mathbb{E}[x+z]\}^2]\\[1em]
&=& \mathbb{E}[(x - \mathbb{E}[x] + z - \mathbb{E}[z])^2]\\[1em]
&=& {\displaystyle \iint} (x - \mathbb{E}[x] + z - \mathbb{E}[z])^2 p(x)p(z) dxdz\\[1em]
&=& {\displaystyle \iint} (x - \mathbb{E}[x])^2 p(x)p(z) dxdz\\[1em]
&& + {\displaystyle \iint} (z - \mathbb{E}[z])^2 p(x)p(z) dxdz\\[1em]
&& + 2 {\displaystyle \iint} (x - \mathbb{E}[x])(z - \mathbb{E}[z]) p(x)p(z) dxdz\\[1em]
&=& {\displaystyle \int} (x - \mathbb{E}[x])^2 p(x) dx + {\displaystyle \int} (z - \mathbb{E}[z])^2 p(z) dz\\[1em]
&& + 2 {\displaystyle \iint} (x - \mathbb{E}[x])(z - \mathbb{E}[z]) p(x)p(z) dxdz\\[1em]
&=& \mathrm{var}[x] + \mathrm{var}[z]\\[1em]
&& + 2 {\displaystyle \int} (x - \mathbb{E}[x])p(x) dx {\displaystyle \int} (z - \mathbb{E}[z]) p(z) dz\\[1em]
&=& \mathrm{var}[x] + \mathrm{var}[z] + 2 (\mathbb{E}[x] - \mathbb{E}[x])(\mathbb{E}[z] - \mathbb{E}[z])\\[1em]
&=& \mathrm{var}[x] + \mathrm{var}[z]
\end{array}
$$

離散変数の場合：

$$
\begin{array}{lll}
\mathbb{E}[x+z]
&=& {\displaystyle \sum_x \sum_z} (x+z) p(x,z)\\[1.5em]
&=& {\displaystyle \sum_x \sum_z} (x+z) p(x)p(z)\\[1.5em]
&=& {\displaystyle \sum_x \sum_z} x p(x)p(z) + {\displaystyle \sum_x \sum_z} z p(x)p(z)\\[1.5em]
&=& {\displaystyle \sum_x} xp(x) + {\displaystyle \sum_z} zp(z)\\[1.5em]
&=& \mathbb{E}[x] + \mathbb{E}[z]
\end{array}
$$

$$
\begin{array}{lll}
\mathrm{var}[x+z]
&=& \mathbb{E}[\{(x+z) - \mathbb{E}[x+z]\}^2]\\[1em]
&=& \mathbb{E}[(x - \mathbb{E}[x] + z - \mathbb{E}[z])^2]\\[1em]
&=& {\displaystyle \sum_x \sum_z} (x - \mathbb{E}[x] + z - \mathbb{E}[z])^2 p(x)p(z)\\[1.5em]
&=& {\displaystyle \sum_x \sum_z} (x - \mathbb{E}[x])^2 p(x)p(z)\\[1.5em]
&& + {\displaystyle \sum_x \sum_z} (z - \mathbb{E}[z])^2 p(x)p(z)\\[1.5em]
&& + 2 {\displaystyle \sum_x \sum_z} (x - \mathbb{E}[x])(z - \mathbb{E}[z]) p(x)p(z)\\[1.5em]
&=& {\displaystyle \sum_x} (x - \mathbb{E}[x])^2 p(x) + {\displaystyle \sum_z} (z - \mathbb{E}[z])^2 p(z)\\[1.5em]
&& + 2 {\displaystyle \sum_x \sum_z} (x - \mathbb{E}[x])(z - \mathbb{E}[z]) p(x)p(z)\\[1.5em]
&=& \mathrm{var}[x] + \mathrm{var}[z]\\[1em]
&& + 2 {\displaystyle \sum_x} (x - \mathbb{E}[x])p(x) {\displaystyle \sum_z} (z - \mathbb{E}[z]) p(z)\\[1.5em]
&=& \mathrm{var}[x] + \mathrm{var}[z] + 2 (\mathbb{E}[x] - \mathbb{E}[x])(\mathbb{E}[z] - \mathbb{E}[z])\\[1em]
&=& \mathrm{var}[x] + \mathrm{var}[z]
\end{array}
$$