正規分布の畳み込み・相関関数を以下の方法で計算してみる．
準備
この記事では，関数$f$のフーリエ変換を\begin{aligned}
\mathcal{F}[f](\xi)
= \int_{-\infty}^{\infty} f(x) e^{-i2\pi\xi x}\,\mathrm{d}x
\end{aligned}
とする．いくつかフーリエ変換の式を導いておく．\mathcal{F}[f](\xi)
= \int_{-\infty}^{\infty} f(x) e^{-i2\pi\xi x}\,\mathrm{d}x
\end{aligned}
1) $f(x)=e^{-\alpha x^{2}}$の場合，指数部が
\begin{aligned}
-\alpha x^{2} -i2\pi\xi x
&= -\alpha \biggl(x + i\frac{\pi\xi}{\alpha} \biggr)^{2}
-\frac{\pi^{2}\xi^{2}}{\alpha}
\end{aligned}
となるので，-\alpha x^{2} -i2\pi\xi x
&= -\alpha \biggl(x + i\frac{\pi\xi}{\alpha} \biggr)^{2}
-\frac{\pi^{2}\xi^{2}}{\alpha}
\end{aligned}
\begin{aligned}
\mathcal{F}[f](\xi)
= \sqrt{\frac{\pi}{\alpha}} e^{-\pi^{2}\xi^{2/}\alpha}
\end{aligned}
である．詳しくは次の記事を参照：ガウス積分と派生公式 - Notes_JP\mathcal{F}[f](\xi)
= \sqrt{\frac{\pi}{\alpha}} e^{-\pi^{2}\xi^{2/}\alpha}
\end{aligned}
2) $f(x) = g(x-\beta)$の場合は
\begin{aligned}
\mathcal{F}[f](\xi)
&= \int_{-\infty}^{\infty} g(x-\beta) e^{-i2\pi\xi (x-\beta)}\,\mathrm{d}(x-\beta)
\times e^{-i2\pi\xi \beta} \\
&= e^{-i2\pi\xi \beta} \mathcal{F}[g](\xi)
\end{aligned}
\mathcal{F}[f](\xi)
&= \int_{-\infty}^{\infty} g(x-\beta) e^{-i2\pi\xi (x-\beta)}\,\mathrm{d}(x-\beta)
\times e^{-i2\pi\xi \beta} \\
&= e^{-i2\pi\xi \beta} \mathcal{F}[g](\xi)
\end{aligned}
3) 1)，2)より$f(x)=e^{-\alpha (x-\beta)^{2}}$の場合，
\begin{aligned}
\mathcal{F}[f](\xi)
= \sqrt{\frac{\pi}{\alpha}} e^{-\pi^{2}\xi^{2/}\alpha} e^{-i2\pi\xi \beta}
\end{aligned}
\mathcal{F}[f](\xi)
= \sqrt{\frac{\pi}{\alpha}} e^{-\pi^{2}\xi^{2/}\alpha} e^{-i2\pi\xi \beta}
\end{aligned}
畳み込み
$f_{i}(x)=e^{-\alpha_{i} (x-\mu_{i})^{2}}$のとき，畳み込み\begin{aligned}
(f_{1}*f_{2})(y)
=\int_{-\infty}^{\infty} f_{1}(x) f_{2}(y - x)\,\mathrm{d}x
\end{aligned}
を求める．(f_{1}*f_{2})(y)
=\int_{-\infty}^{\infty} f_{1}(x) f_{2}(y - x)\,\mathrm{d}x
\end{aligned}
$1/\alpha = 1/\alpha_{1} + 1/\alpha_{2}=(\alpha_{1}+\alpha_{2})/\alpha_{1}\alpha_{2}$，$\mu=\mu_{1}+\mu_{2}$とすると，
\begin{aligned}
(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\mathcal{F}[f_{1}]\cdot \mathcal{F}[f_{2}]\biggr](y) \\
&=\mathcal{F}^{-1} \biggl[
\sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
e^{-i2\pi\xi \mu_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi \mu_{2}} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\mathcal{F}^{-1} \biggl[e^{-i2\pi\xi \mu} e^{-\pi^{2}\xi^{2}/\alpha} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\sqrt{ \frac{\alpha}{\pi}}
e^{-\alpha(y - \mu)^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha(y - \mu)^{2}}
\end{aligned}
となる．(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\mathcal{F}[f_{1}]\cdot \mathcal{F}[f_{2}]\biggr](y) \\
&=\mathcal{F}^{-1} \biggl[
\sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
e^{-i2\pi\xi \mu_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi \mu_{2}} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\mathcal{F}^{-1} \biggl[e^{-i2\pi\xi \mu} e^{-\pi^{2}\xi^{2}/\alpha} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\sqrt{ \frac{\alpha}{\pi}}
e^{-\alpha(y - \mu)^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha(y - \mu)^{2}}
\end{aligned}
特に，$f_{i}$が正規分布（正規分布の覚え方 - Notes_JP）
\begin{aligned}
f_{i}(x)
=\frac{1}{\sqrt{2\pi}\sigma_{i}} e^{-(x-\mu_{i})^{2}/2\sigma_{i}^2}
\end{aligned}
のとき，$\alpha_{i} = 1/2\sigma_{i}^2$とすると$1/\alpha=2(\sigma_{1}^{2} + \sigma_{2}^{2})$だから，f_{i}(x)
=\frac{1}{\sqrt{2\pi}\sigma_{i}} e^{-(x-\mu_{i})^{2}/2\sigma_{i}^2}
\end{aligned}
\begin{aligned}
(f_{1}*f_{2})(y)
&= \frac{1}{2\pi\sigma_{1}\sigma_{2}}
\sqrt{\frac{2\pi}{1/\sigma_{1}^{2}+1/\sigma_{2}^{2}}}
e^{-(y-(\mu_{1}+\mu_{2}))^{2}/2(\sigma_{1}^{2} + \sigma_{2}^{2})} \\
&= \frac{1}{\sqrt{2\pi(\sigma_{1}^{2}+\sigma_{2}^{2}})}
e^{-(y-(\mu_{1}+\mu_{2}))^{2}/2(\sigma_{1}^{2} + \sigma_{2}^{2})}
\end{aligned}
というよく知られた結果が得られる（$X_{i}\sim N(\mu_{i}, \sigma_{i}^{2}) \Rightarrow X_{1}+X_{2} \sim N(\mu_{1}+\mu_{2}, \sigma_{1}^{2}+\sigma_{2}^{2})$）．(f_{1}*f_{2})(y)
&= \frac{1}{2\pi\sigma_{1}\sigma_{2}}
\sqrt{\frac{2\pi}{1/\sigma_{1}^{2}+1/\sigma_{2}^{2}}}
e^{-(y-(\mu_{1}+\mu_{2}))^{2}/2(\sigma_{1}^{2} + \sigma_{2}^{2})} \\
&= \frac{1}{\sqrt{2\pi(\sigma_{1}^{2}+\sigma_{2}^{2}})}
e^{-(y-(\mu_{1}+\mu_{2}))^{2}/2(\sigma_{1}^{2} + \sigma_{2}^{2})}
\end{aligned}
相関関数
$f_{i}(x)=e^{-\alpha_{i} (x-\mu_{i})^{2}}$のとき，相関関数\begin{aligned}
(f_{1} \star f_{2})(y)
=\int_{-\infty}^{\infty} \bar{f}_{1}(x) f_{2}(x+y)\,\mathrm{d}x
\end{aligned}
を求める．(f_{1} \star f_{2})(y)
=\int_{-\infty}^{\infty} \bar{f}_{1}(x) f_{2}(x+y)\,\mathrm{d}x
\end{aligned}
$1/\alpha = 1/\alpha_{1} + 1/\alpha_{2}=(\alpha_{1}+\alpha_{2})/\alpha_{1}\alpha_{2}$，$\tilde{\mu}=\mu_{2} - \mu_{1}$とすると，
\begin{aligned}
(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\overline{\mathcal{F}[f_{1}]} \cdot \mathcal{F}[f_{2}]\biggr](y) \\
&=\mathcal{F}^{-1} \biggl[
\sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
e^{\textcolor{red}{+}i2\pi\xi \mu_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi \mu_{2}} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\mathcal{F}^{-1} \biggl[e^{-i2\pi\xi \tilde{\mu}} e^{-\pi^{2}\xi^{2}/\alpha} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\sqrt{ \frac{\alpha}{\pi}}
e^{-\alpha(y - \tilde{\mu})^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha(y - \tilde{\mu})^{2}}
\end{aligned}
となる．(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\overline{\mathcal{F}[f_{1}]} \cdot \mathcal{F}[f_{2}]\biggr](y) \\
&=\mathcal{F}^{-1} \biggl[
\sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
e^{\textcolor{red}{+}i2\pi\xi \mu_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi \mu_{2}} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\mathcal{F}^{-1} \biggl[e^{-i2\pi\xi \tilde{\mu}} e^{-\pi^{2}\xi^{2}/\alpha} \biggr](y)\\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
\sqrt{ \frac{\alpha}{\pi}}
e^{-\alpha(y - \tilde{\mu})^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha(y - \tilde{\mu})^{2}}
\end{aligned}
確率変数で言えば，
\begin{aligned}
X_{i}\sim N(\mu_{i}, \sigma_{i}^{2})
\Rightarrow X_{2} - X_{1} \sim N(\mu_{2} - \mu_{1}, \sigma_{1}^{2}+\sigma_{2}^{2})
\end{aligned}
を意味する．X_{i}\sim N(\mu_{i}, \sigma_{i}^{2})
\Rightarrow X_{2} - X_{1} \sim N(\mu_{2} - \mu_{1}, \sigma_{1}^{2}+\sigma_{2}^{2})
\end{aligned}
別解
畳み込み
wikiの方法（Sum of normally distributed random variables - Wikipedia）を見つける前にやった計算のメモ．途中で平方完成している分，余計な手間がかかっている．まず
\begin{aligned}
(f_{1}*f_{2})(y)
=\int_{-\infty}^{\infty} f_{1}(x+\mu_{1}) f_{2}(y - (x+\mu_{1}))\,\mathrm{d}x
\end{aligned}
であるから，$f_{1}(x)=e^{-\alpha_{1} x^{2}}$，$f_{2}(x)=e^{-\alpha_{2} (x-(\mu_{1}+\mu_{2}))^{2}}$としてよい．(f_{1}*f_{2})(y)
=\int_{-\infty}^{\infty} f_{1}(x+\mu_{1}) f_{2}(y - (x+\mu_{1}))\,\mathrm{d}x
\end{aligned}
上の結果から，
\begin{aligned}
& \mathcal{F}[f_{1}](\xi)\cdot \mathcal{F}[f_{2}](\xi) \\
&= \sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi (\mu_{1}+\mu_{2})}
\end{aligned}
である．この指数部は，$1/\alpha = 1/\alpha_{1} + 1/\alpha_{2}=(\alpha_{1}+\alpha_{2})/\alpha_{1}\alpha_{2}$，$\mu=\mu_{1}+\mu_{2}$とすると& \mathcal{F}[f_{1}](\xi)\cdot \mathcal{F}[f_{2}](\xi) \\
&= \sqrt{\frac{\pi}{\alpha_{1}}} e^{-\pi^{2}\xi^{2}/\alpha_{1}}
\cdot
\sqrt{\frac{\pi}{\alpha_{2}}} e^{-\pi^{2}\xi^{2}/\alpha_{2}}
e^{-i2\pi\xi (\mu_{1}+\mu_{2})}
\end{aligned}
\begin{aligned}
-\frac{\pi^{2}}{\alpha} \xi^{2} - i2\pi\xi \mu
&= -\frac{\pi^{2}}{\alpha}\biggl(\xi + i\frac{\alpha \mu}{\pi}\biggr)^{2}
-\alpha \mu^{2}
\end{aligned}
となる．-\frac{\pi^{2}}{\alpha} \xi^{2} - i2\pi\xi \mu
&= -\frac{\pi^{2}}{\alpha}\biggl(\xi + i\frac{\alpha \mu}{\pi}\biggr)^{2}
-\alpha \mu^{2}
\end{aligned}
よって，
\begin{aligned}
(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\mathcal{F}[f_{1}]\cdot \mathcal{F}[f_{2}]\biggr](y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}} e^{-\alpha \mu^{2}}
\cdot
\mathcal{F}^{-1}
\Biggl\{\exp
\biggl[-\frac{\pi^{2}}{\alpha}\biggl(\xi + i\frac{\alpha \mu}{\pi}\biggr)^{2}
\biggr]\Biggr\}(y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
e^{-\alpha \mu^{2}}
\cdot
e^{-i2\pi y \cdot i\alpha\mu/\pi}
\mathcal{F}^{-1}
\biggl[e^{-\pi^{2}\xi^{2}/\alpha}\biggr](y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
e^{-\alpha \mu^{2}}
\cdot
e^{2 y \alpha\mu}
\sqrt{\frac{\alpha}{\pi}}
e^{-\alpha y^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha (y-\mu)^{2}}
\end{aligned}
となる．ただし，逆フーリエ変換は，上記のフーリエ変換の結果を（逆に）用いた．
(f_{1}*f_{2})(y)
&= \mathcal{F}^{-1}\biggl[\mathcal{F}[f_{1}]\cdot \mathcal{F}[f_{2}]\biggr](y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}} e^{-\alpha \mu^{2}}
\cdot
\mathcal{F}^{-1}
\Biggl\{\exp
\biggl[-\frac{\pi^{2}}{\alpha}\biggl(\xi + i\frac{\alpha \mu}{\pi}\biggr)^{2}
\biggr]\Biggr\}(y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
e^{-\alpha \mu^{2}}
\cdot
e^{-i2\pi y \cdot i\alpha\mu/\pi}
\mathcal{F}^{-1}
\biggl[e^{-\pi^{2}\xi^{2}/\alpha}\biggr](y) \\
&= \frac{\pi}{\sqrt{\alpha_{1}\alpha_{2}}}
e^{-\alpha \mu^{2}}
\cdot
e^{2 y \alpha\mu}
\sqrt{\frac{\alpha}{\pi}}
e^{-\alpha y^{2}} \\
&=\sqrt{\frac{\pi}{\alpha_{1}+\alpha_{2}}} e^{-\alpha (y-\mu)^{2}}
\end{aligned}
