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  • STA256 Lecture 05

Cumulative Distribution Function

• Right continuous, non-decreasing function approaching $1$.
• Theorem 28 ^[1]
• Properties of the CDF
  • For the semi open interval $(a,b]$
    • $P(a<x\le b)=F(b)-F(a)$
  • $F(x)=P(X\le x)$
    • $P(X>x)=1-F(x)$
  • For a Discrete Random Variable
    • $P(X=x)=F(x)-F(x^{-})$
      • $F(x)$ is the cdf at value $x$
      • $F(x^{-})$ is at the jump before
  • Where $x_1<x_2<x_3<\cdots <x_k$
    • $\begin{array}{cccccc}X=x& x_1& x_2& x_3& \dots & x_k\\ P(X=x)\end{array}$
  • Sketch:
    • Looks like a staircase.
• To find the Cumulative Distribution Function of a Continuous Random Variable, we use a change of variable.
• $F(x)=P(X\le x)$
  • ${\int }_{-\mathrm{\infty }}^{x}f(?)d?$
  • ${\int }_{-\mathrm{\infty }}^{x}f(t)dt$
  • We need to make $f(x)$ in terms of $f(t)$
  • You can use the Cumulative Distribution Function of a Continuous Random Variable to recover the Probability Density Function.
  • $f_X(x)=\frac{d}{dx}{F}_X(x)$
  • The derivative of an integral is the original integrand
  • $f(x)=\begin{array}{cc}-x^2+\frac{8x}{3}& 0\le x\le 1\\ 0& \text{otherwise}\end{array}$
  • $F(x)=P(X\le x)$
  • ${\int }_{t=-\mathrm{\infty }}^{t=x}f(t)dt$
    • $f(x)=-x^2+\frac{8x}{3}$
    • $f(1)=-(1{)}^2+\frac{8(1)}{3}$
    • $f(1)=-(1{)}^2+\frac{8(1)}{3}$
    • $f(t)=-t^2+\frac{8t}{3}$
  • ${\int }_{t=0}^{t=x}-t^2+\frac{8t}{3}dt$
  • $(\frac{1}{3}(-t^3)+\frac{1}{6}(8t^2)){}_{t=0}^{t=x}$
  • $(-\frac{x^3}{3}+\frac{8x^2}{6})$
  • Cumulative Distribution Function
    • $F(x)=\begin{array}{cc}0& x<0\\ -\frac{x^3}{3}+\frac{8x^2}{6}& 0\le x\le 1\\ 1& x>1\end{array}$

      	left=-1; right=3;
      	top=3; bottom=-1;
      	---
      	y=0|x<0
      	y=\frac{x^{3}}{3}+\frac{8x^{2}}{6}|0<=x<=1
      	y=1|x>1