STA256 Lecture 17

• Covariance
  • $cov(x_1,x_2)=0\text{⧸}⟹\text{Independence}$
  • But $\text{Independence}⟹cov(x_1,x_2)=0$
• Example:
  • $\begin{array}{cccc}& & & y\\ & & -1& 0& 1\\ & -1& 0& \frac{1}{4}& 0\\ x& 0& \frac{1}{4}& 0& \frac{1}{4}\\ & 1& 0& \frac{1}{4}& 0\end{array}$
  • Find the marginal of $x$
  • $\begin{array}{cccc}X=x& -1& 0& 1\\ P(X=x)\end{array}$
  • $\begin{array}{cccc}Y=y& -1& 0& 1\\ P(Y=y)\end{array}$
  • Show that $cov(x,y)=E(XY)-E(X)E(Y)=0$
    • #tk
  • $P(X=-1,Y=-1)=0$
  • $P(X=-1)\cdot P(Y=-1)$
    • $=\frac{1}{4}\frac{1}{4}$
    • $=\frac{1}{16}$
  • See that $0\ne \frac{1}{16}$
  • So $P(X=x,Y=y)\ne P(X=x)P(Y=y)$
  • So they're not independent
• Example:
  • Suppose we see moderate of strong correlation between two variables
  • Does this mean that selling more ice cream, results in more crime?
  • The conditions under which ice cream sales increase also result in an increase in crime rate, but not directly does it influence it.
  • There is a hidden relationship that's not captured.
  • "York university is closed, brampton mans need to find a way to pass the time" - Nishan Mudalige
• Linear Combinations of Random Variables
• Chapter 3:
• Special Distributions