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  • Variance

Variance - Transformation

• #tk theorem 38
• Example:
  • We have a Probability Density Function for Continuous Random Variable.
  • We can determine the effect of the mean ($\mu$) and Variance (${\sigma }^2$) of applying a linea transformation on $X$
  • We have $X$ and $E(x)$ and $var(x)={\sigma }^2$
  • Apply $y=aX+b$, where $a,b\in \mathbb{R}$
  • $E(Y)=E(aX+b)=aE(x)+b$
  • $var(Y)=var(aX+b)=a^{2}var(x)$
    • Since ${\sigma }^2=\sum (x_i-\mu {)}^{2}p(x_i)$
      • ${\sigma }^2=\sum (a{x}_i-a\mu {)}^{2}p(x_i)$
    • ${\sigma }^2=\int (x-\mu {)}^{2}f(x)dx$
      • ${\sigma }^2=\int (ax-a\mu {)}^{2}f(x)dx$
    • This is why $a^2$ is there.
    • $b$ doesn't effect anything because a shift up and down doesn't change the fact that you're just measuring side to side for Variance
• Proof:
  • $\mathrm{\forall }a,b\in \mathbb{R},var(aX+b)=a^{2}var(x)$
  • Mean: $E(aX+b)=E(aX)+E(b)$
  • $E(aX+b)=aE(X)+\underset{\begin{array}{c}\text{The expected value}\\ \text{of a constant}\\ \text{is the constant}\end{array}}{\underbrace{E(b)}}$
  • $=a\underset{\mu }{\underbrace{E(x)}}+b$
  • So $E(aX+b)=a\mu +b$
  • $var(x)=E[(x-\mu {)}^2]$
  • $var(ax+b)=E[(ax+b-(a\mu +b){)}^2]$
  • $var(ax+b)=E[(ax+\cancel{b}-a\mu -\cancel{b}{)}^2]$
  • $var(ax+b)=E[(ax-a\mu {)}^2]$
  • $var(ax+b)=E[(a(x-\mu ){)}^2]$
  • $var(ax+b)=E[a^2(x-\mu {)}^2]$
  • $var(ax+b)=a^2\underset{var(x)}{\underbrace{E[(x-\mu {)}^2]}}$
  • $var(ax+b)=a^{2}var(x)$