التوقع والتباين (Expectation and Variance)

نفرض أن $X$ متغير عشوائي يتبع التوزيع الطبيعي (Normal distribution) بمتوسط $\mu$ وتباين ${\sigma }^2$

$$X\sim \mathcal{N}(\mu ,{\sigma }^2)$$

بالتالي دالة الكثافة الإحتمالية (The probability density function) يمكن كتابتها كما يلي:

$$f(x;\mu ,\sigma )=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}$$

كذالك يمكن حساب التوقع كما يلي:

$$\begin{array}{rcl}E(x)& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf(x;\mu ,\sigma )dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}dx\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}dx\end{array}$$

نفرض أن:

$$u=x-\mu$$

بالتالي:

$$x=u+\mu$$

$$dx=du$$

$$\begin{array}{rcl}E\left(x\right)& =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(u+\mu )e^{-\frac{u^2}{2{\sigma }^2}}du\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u{e}^{-\frac{u^2}{2{\sigma }^2}}+\mu e^{-\frac{u^2}{2{\sigma }^2}}du\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u{e}^{-\frac{u^2}{2{\sigma }^2}}du+\frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mu e^{-\frac{u^2}{2{\sigma }^2}}du\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}\underset{0}{\underbrace{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}u{e}^{-\frac{u^2}{2{\sigma }^2}}du}}+\mu \underset{1}{\underbrace{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{u^2}{2{\sigma }^2}}du}}\\ & =& \mu \end{array}$$

ايضاً، يمكن حساب التباين كمايلي:

$$Var(x)=E\left[{(x-\mu )}^2\right]$$

$$\begin{array}{rcl}Var(x)& =& E\left[{(x-\mu )}^2\right]\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{(x-\mu )}^{2}f(x;\mu ,\sigma )dx\\ & =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{(x-\mu )}^2}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}dx\end{array}$$

نفرض أن:

$$y=x-\mu$$

بالتالي:

$$dy=dx$$

$$Var(x)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^2{e}^{-\frac{y^2}{2{\sigma }^2}}dy$$

بإستخدام التكامل بالتجزئة:

$$\int udv=uv-\int vdu$$

نفرض أن:

$$\begin{array}{rcl}u& =& y\\ dv& =& y{e}^{-\frac{y^2}{2{\sigma }^2}}dy\end{array}$$

بالتالي:

$$\begin{array}{rcl}du& =& dy\\ v& =& -{\sigma }^2{e}^{-\frac{y^2}{2{\sigma }^2}}\end{array}$$

أخيراً:

$$\begin{array}{rcl}Var(x)& =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^2{e}^{-\frac{y^2}{2{\sigma }^2}}dy\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}udv\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}({\left[uv\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}vdu)\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}({[-{\sigma }^{2}y{e}^{-\frac{y^2}{2{\sigma }^2}}]}_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\sigma }^2{e}^{-\frac{y^2}{2{\sigma }^2}}dy)\\ & =& \frac{1}{\sqrt{2\pi {\sigma }^2}}{[-{\sigma }^{2}y{e}^{-\frac{y^2}{2{\sigma }^2}}]}_{-\mathrm{\infty }}^{\mathrm{\infty }}-\frac{1}{\sqrt{2\pi {\sigma }^2}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\sigma }^2{e}^{-\frac{y^2}{2{\sigma }^2}}dy\\ & =& \frac{-{\sigma }^2}{\sqrt{2\pi {\sigma }^2}}\underset{0}{\underbrace{{\left[y{e}^{-\frac{y^2}{2{\sigma }^2}}\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}}}+{\sigma }^2\underset{1}{\underbrace{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{y^2}{2{\sigma }^2}}dy}}\\ & =& {\sigma }^2\end{array}$$