STA256 Lecture 10

• Review for STA256 Test 1 Prep
• 1
  • Let $X$ be a Continuous Random Variable with Probability Density Function and Cumulative Distribution Function
  • $f(x)=\frac{1}{25}x+\frac{1}{10},0<x<5$
  • $F(x)=\left\{\begin{array}{cc}0& x\le 0\\ \frac{1}{50}{x}^2+\frac{1}{10}x& 0\le x\le 5\\ 1& x\ge 1\end{array}\right\}$
  • a
    • verify $F(x)$ is the Cumulative Distribution Function of $f(x)$
    • Since $F(x)$ is the integral of $f(x)$ from $0<x<5$. Just find the derivative.
    • $F(x)=P(X\le x)={\int }_{-\mathrm{\infty }}^{x}f(t)dt$
    • ${\int }_{t=0}^{t=x}\frac{1}{25}t+\frac{1}{10}dt$
    • Antiderive
      • $\frac{1}{25}\frac{1}{2}{t}^2+\frac{1}{10}t$
      • $(\frac{1}{50}{t}^2+\frac{1}{10}t){}_{t=0}^{t=x}$
      • $\frac{1}{50}{x}^2+\frac{1}{10}x$
      • This is $F(x)$ from $0\le x\le 5$
  • b
    • Find $P(X\le 2)$
    • To use the Probability Density Function, we need to integrate from $0\le 2$
    • To use the Cumulative Distribution Function, we evaluate $F(2)$
    • $F(2)=\frac{1}{50}(2{)}^2+\frac{1}{10}(2)=0.08+0.2=0.28$
    • $P(X\le 2)={\int }_0^2\frac{1}{25}x+\frac{1}{10}dx=0.28$
  • c
    • Find $P(X>2)$
    • Probability Density Function
    • Cumulative Distribution Function
      • Because $P(X\le 2{)}^c=P(X>2)=1-\underset{F(2)}{\underbrace{P(X\le 2)}}$
      • $1-F(2)=1-0.28=0.72$
  • d
    • Find the ${50}^{\text{th}}$ Percentile $x_{50}$
    • Probability Density Function
      • ${\int }_{-\mathrm{\infty }}^{x_{50}}f(x)dx=0.5$
      • $(\frac{1}{50}{x}^2+\frac{1}{10}x)=0.5$
      • $\frac{1}{50}{x}_{50}^2+\frac{1}{10}{x}_{50}=0.5$
      • $\dots$
    • Cumulative Distribution Function
      • $F(x_{50})=0.5$
      • $\frac{1}{50}{x}_{50}^2+\frac{1}{10}{x}_{50}=0.5$
      • $x_{50}^2+5x_{50}=25$
      • $x_{50}^2+5x_{50}-25=0$
      • $x_{50}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
      • $x_{50}=\frac{-5\pm \sqrt{5^2-4(-25)}}{2}$
      • $x_{50}=\frac{-5\pm \sqrt{125}}{2}$
        • $x_{50}=3.0901699438$
        • $x_{50}=-8.0901699438$
      • It doesn't make sense for negative values to be in the Support of a Random Variable.
      • So only $x_{50}\approx 3.09$
      • Verify that $f(x)$ is a valid Probability Density Function, $\int f(x)dx=1$ #tk

        	left=-0.25; right=5.25;
        	top=2; bottom=-0.25;
        	---
        	y=\frac{1}{50}x^{2}+ \frac{1}{10}x|0<=x<=5
        	x=3.09
        

• 2
  • Let $X$ be a Discrete Random Variable, with a Probability Mass Function
  • $\begin{array}{ccccccc}X=x& 0& 1& 2& 3& 4& 5\\ P(X=x)& 0.1& 0.25& 0.2& 0.2& 0.15& 0.1\end{array}$
  • a
    • Find the $P(X<4)$ gives $x>0$
    • $P(X<4|X\ge 1)$
    • $P(X<4|X\ge 1)=\frac{P(1\le X\le 3)}{P(X\ge 1)}$
    • $P(1\le X\le 3)=P(X=1)+P(X=2)+P(X=3)$
    • $P(1\le X\le 3)=0.25+0.2+0.2=0.65$
    • $P(X\ge 1)=1-P(X\le 0)$
    • $P(X\ge 1)=1-0.1=0.9$
    • $P(X<4|X\ge 1)=\frac{0.65}{0.9}=0.7\overline{22}=\frac{13}{18}$
  • b
    • Consider the Transformations - Univariate given by $Y=\left\{\begin{array}{cc}x& \text{if }x\text{ is even}\\ -1& \text{if }x\text{ is odd}\end{array}\right\}$
    • Find the Probability Mass Function of $Y$
    • $Y=\left\{\begin{array}{cc}x& x=0,2,4\\ -1& x=1,3,5\end{array}\right\}$
    • By the Transformations - Univariate theorem
    • $f_y(y)=f_x(g^{-1}(y))$
    • $f_y(y=0)=f_x(g^{-1}(y=0))=f_x(x=0)=0.1$
    • $f_y(y=2)=f_x(g^{-1}(y=2))=f_x(x=2)=0.2$
    • $f_y(y=4)=f_x(g^{-1}(y=4))=f_x(x=4)=0.15$
    • $f_y(y=-1)=f_x(g^{-1}(y=-1))=f_x(x=1,3,5)=0.25+0.2+0.1=0.55$
    • Probability Mass Function $Y$
    • $\begin{array}{ccccc}Y=y& -1& 0& 2& 4\\ P(Y=y)& 0.55& 0.1& 0.2& 0.15\end{array}$
    • #tk Find the Cumulative Distribution Function of $Y$.
      • $F_y(y)=\left\{\begin{array}{cc}0& y<-1\\ 0.55& -1\le y<0\\ 0.65& 0\le y<2\\ 0.85& 2\le y<4\\ 1& y\ge 4\end{array}\right\}$
      • Sketch the Probability Distribution Function and Cumulative Distribution Function of $Y$
• 3
  • An organization reviews electronic threats
  • For 2024, it recorded:
  • $\text{Threats}=\begin{array}{cc}\text{Phishing}& 15\mathrm{\%}\\ \text{Malware}& 35\mathrm{\%}\\ \text{Unauthorized Transfer}& 30\mathrm{\%}\\ \text{Unauthorized Access}& 20\mathrm{\%}\end{array}$
  • Their system was able to detect threats at:
  • $\text{Detected}=\begin{array}{cc}\text{Phishing}& 92\mathrm{\%}\\ \text{Malware}& 94\mathrm{\%}\\ \text{Unauthorized Transfer}& 92\mathrm{\%}\\ \text{Unauthorized Access}& 89\mathrm{\%}\end{array}$
  • Suppose a threat is selected at random, what is the probability it was successfully detected.
  • a) Probability of being detected $P(D)$
    • $P(Ph)\cdot P(D|Ph)+P(Ma)\cdot P(D|Ma)+P(Ut)\cdot P(D|Ut)+P(Ua)\cdot P(D|Ua)$
    • $0.15(0.96)+\cdots +0.2(0.89)$
    • $=0.927$
  • b) Give a threat was detected, what's the probability it was an unauthorized access attempt?
    • Conditional Probability
    • $P(Ua|D)=\frac{P(Ua\cap D)}{P(D)}=\frac{0.2(0.89)}{0.927}=\frac{0.178}{0.927}=0.19201726\approx 19.20\mathrm{\%}$
• 4
  • Let $A,B$ be two events in sample space $S$.
  • Using Axioms of Probability, prove that the probability of $P(A\cup B)\le P(A)+P(B)$
  • If they're pairwise disjoint, or they partition the space, then we have $P(A\cup B)=P(A)+P(B)$
  • If they're not disjoint, then $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
  • Or $P(A\cup B)=P(A)+P(B\setminus A)$
  • $A$ and $B\setminus A$ are disjoint
    So we don't need to subtract the intersection
  • $P(A\cup B)=P(A)+P(B\setminus A)$
  • Notice $B=(A\cap B)\cup B$
  • $P(B)=P(A\cap B)+P(B\setminus A)$
  • $P(B\setminus A)=P(B)-P(A\cap B)$
  • By Axiom 1, we know that $P(A\cap B)\ge 0$.
  • $P(B\setminus A)\le P(B)$
  • By transitivity we have that $P(A\cup B)=P(A)+\underset{\le P(B)}{\underbrace{P(B\setminus A)}}$
  • $P(A\cup B)\le P(A)+P(B)$