نفرض أن $X$ متغير عشوائي يتبع التوزيع الطبيعي (Normal distribution) بمتوسط $\mu$ وتباين $\sigma^{2}$
\begin{equation*}
X\sim \mathcal{N}(\mu ,\sigma ^{2})
\end{equation*}
بالتالي دالة الكثافة الإحتمالية (The probability density function) يمكن كتابتها كما يلي:
\begin{equation*}
f(x;\mu ,\sigma )=\frac{1}{\sqrt{2\pi \sigma ^{2}}}e^{-\frac{\left( x-\mu
\right) ^{2}}{2\sigma ^{2}}}
\end{equation*}
كذالك يمكن حساب التوقع كما يلي:
\begin{eqnarray}
E(x) &=&\int_{-\infty }^{\infty }xf(x;\mu ,\sigma )dx \notag \\
&=&\int_{-\infty }^{\infty }\frac{x}{\sqrt{2\pi \sigma ^{2}}}e^{-\frac{%
\left( x-\mu \right) ^{2}}{2\sigma ^{2}}}dx \notag \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }xe^{-\frac{%
\left( x-\mu \right) ^{2}}{2\sigma ^{2}}}dx
\end{eqnarray}
نفرض أن:
\begin{equation*}
u=x-\mu
\end{equation*}
بالتالي:
\begin{equation*}
x=u+\mu
\end{equation*}
\begin{equation*}
dx=du
\end{equation*}
\begin{eqnarray*}
E\left( x\right) &=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty
}\left( u+\mu \right) e^{-\frac{u^{2}}{2\sigma ^{2}}}du \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }ue^{-\frac{u^{2}%
}{2\sigma ^{2}}}+\mu e^{-\frac{u^{2}}{2\sigma ^{2}}}du \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }ue^{-\frac{u^{2}%
}{2\sigma ^{2}}}du+\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty
}\mu e^{-\frac{u^{2}}{2\sigma ^{2}}}du \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\underset{0}{\underbrace{\int_{-\infty
}^{\infty }ue^{-\frac{u^{2}}{2\sigma ^{2}}}du}}+\mu \underset{1}{\underbrace{%
\int_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi \sigma ^{2}}}e^{-\frac{u^{2}}{%
2\sigma ^{2}}}du}} \\
&=&\mu
\end{eqnarray*}
ايضاً، يمكن حساب التباين كمايلي:
\begin{equation*}
Var(x)=E\left[ \left( x-\mu \right) ^{2}\right]
\end{equation*}
\begin{eqnarray*}
Var(x) &=&E\left[ \left( x-\mu \right) ^{2}\right] \\
&=&\int_{-\infty }^{\infty }\left( x-\mu \right) ^{2}f(x;\mu ,\sigma )dx \\
&=&\int_{-\infty }^{\infty }\frac{\left( x-\mu \right) ^{2}}{\sqrt{2\pi
\sigma ^{2}}}e^{-\frac{\left( x-\mu \right) ^{2}}{2\sigma ^{2}}}dx
\end{eqnarray*}
نفرض أن:
\begin{equation*}
y=x-\mu
\end{equation*}
بالتالي:
\begin{equation*}
dy=dx
\end{equation*}
\begin{equation*}
Var(x)=\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }y^{2}e^{-%
\frac{y^{2}}{2\sigma ^{2}}}dy
\end{equation*}
بإستخدام التكامل بالتجزئة:
\begin{equation*}
\int udv=uv-\int vdu
\end{equation*}
نفرض أن:
\begin{eqnarray*}
u &=&y \\
dv &=&ye^{-\frac{y^{2}}{2\sigma ^{2}}}dy
\end{eqnarray*}
بالتالي:
\begin{eqnarray*}
du &=&dy \\
v &=&-\sigma ^{2}e^{-\frac{y^{2}}{2\sigma ^{2}}}
\end{eqnarray*}
أخيراً:
\begin{eqnarray*}
Var(x) &=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }y^{2}e^{-%
\frac{y^{2}}{2\sigma ^{2}}}dy \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\int_{-\infty }^{\infty }udv \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\left( \left[ uv\right] _{-\infty
}^{\infty }-\int_{-\infty }^{\infty }vdu\right) \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\left( \left[ -\sigma ^{2}ye^{-\frac{%
y^{2}}{2\sigma ^{2}}}\right] _{-\infty }^{\infty }-\int_{-\infty }^{\infty
}-\sigma ^{2}e^{-\frac{y^{2}}{2\sigma ^{2}}}dy\right) \\
&=&\frac{1}{\sqrt{2\pi \sigma ^{2}}}\left[ -\sigma ^{2}ye^{-\frac{y^{2}}{%
2\sigma ^{2}}}\right] _{-\infty }^{\infty }-\frac{1}{\sqrt{2\pi \sigma ^{2}}}%
\int_{-\infty }^{\infty }-\sigma ^{2}e^{-\frac{y^{2}}{2\sigma ^{2}}}dy \\
&=&\frac{-\sigma ^{2}}{\sqrt{2\pi \sigma ^{2}}}\underset{0}{\underbrace{%
\left[ ye^{-\frac{y^{2}}{2\sigma ^{2}}}\right] _{-\infty }^{\infty }}}%
+\sigma ^{2}\underset{1}{\underbrace{\int_{-\infty }^{\infty }\frac{1}{\sqrt{%
2\pi \sigma ^{2}}}e^{-\frac{y^{2}}{2\sigma ^{2}}}dy}} \\
&=&\sigma ^{2}
\end{eqnarray*}