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  • STA256 Lecture 09

Variance and Standard Deviation

• The mean on its own, doesn't give the complete story about a distribution.
• Like Average True Range is used to evaluate the volatility of a security.
• Example:
  • The mean doesn't tell us the exact story.
  • Around the mean at st.george, the spread is larger, at utm the spread is smaller.
• Variance
• Standard Deviation
• Example: #tk slide 100
  • Discrete Random Variable
    • Either 8 or 12 emails a day
    • $\begin{array}{ccc}X=x& 8& 12\\ P(X=x)& \frac{1}{2}& \frac{1}{2}\end{array}$
    • Mean: $\mu =E(x)=\sum x_{i}p(x_i)$
      • $8\frac{1}{2}+12\frac{1}{2}=10$
    • Variance: ${\sigma }^2=E[(x-\mu {)}^2]$
      • First way
        • $=\sum (x_i-\mu {)}^{2}p(x_i)$
        • $=(8-10{)}^2\frac{1}{2}+(12-10{)}^2\frac{1}{2}$
        • $=4$
      • Second way (recommended)
        • First step
          • $E(x^2)=\sum x_i^{2}p(x_i)$
          • $8^2\frac{1}{2}+{12}^2\frac{1}{2}$
          • $104$
        • Second Step
          • ${\sigma }^2=E[x^2]-{\mu }^2=104-{10}^2=4$
    • Standard Deviation: $SD(x)=\sqrt{{\sigma }^2}$
      • $SD(x)=\sqrt{4}=2$
    • On average they recieve $10$ emails a day with a Standard Deviation of $2$ emails
  • Continuous Random Variable
    • Let $X$ be a Continuous Random Variable with a Probability Density Function given by $f(x)=\left\{\begin{array}{cc}\frac{x}{4}-\frac{1}{2}& 3<x<5\\ 0& \text{otherwise}\end{array}\right\}$
    • Find the mean, Variance and Standard Deviation.
    • $\mu$
      • $\mu =E(x)={\int }_3^{5}xf(x)dx$
      • $\mu =E(x)={\int }_3^{5}x(\frac{x}{4}-\frac{1}{4})dx$
      • $\mu =E(x)={\int }_3^5\frac{x^2}{4}-\frac{x}{4}dx$
      • $\frac{25}{6}$
    • ${\sigma }^2$
      • $E(x^2)={\int }_3^5{x}^{2}f(x)dx$
      • $E(x^2)={\int }_3^5{x}^2(\frac{x}{4}-\frac{1}{4})dx$
      • $E(x^2)={\int }_3^5\frac{1}{4}{x}^3-\frac{1}{4}{x}^{2}dx$
      • Antiderive
        • $\frac{1}{4}\frac{1}{4}{x}^4-\frac{1}{4}\frac{1}{3}{x}^3$
        • $(\frac{1}{16}{x}^4-\frac{1}{12}{x}^3){}_3^5$
        • $(\frac{1}{16}(5{)}^4-\frac{1}{12}(5{)}^3)-(\frac{1}{16}(3{)}^4-\frac{1}{12}(3{)}^3)$
        • $\frac{53}{3}$
      • ${\sigma }^2=E(x^2)-{\mu }^2$
      • ${\sigma }^2=\frac{53}{3}-(\frac{25}{6}{)}^2$
      • ${\sigma }^2=\frac{11}{36}$
    • $\sigma =\sqrt{\frac{11}{36}}=\frac{\sqrt{11}}{6}$

      	left=-0.5; right=6;
      	top=1; bottom=-0.1;
      	---
      	y=\frac{x}{4}- \frac{1}{2}|3<=x<=5
      	x= 25/6
      	(25/6,0)|label:This is the Mean