Probability and Statistics Lecture 6

Continued from Probability and Statistics Lecture 5

Notes on Probability and Random Variables

Example 1: Probability Density Function (PDF)

A continuous random variable $X$ has the following probability law:

$$f(x)=\{\begin{array}{ll}k{x}^2,& 0\le x\le 2\\ 0,& \text{otherwise}\end{array}$$

To determine $k$, we use the property of PDFs: ${\int }_{-\infty }^{\infty }f(x)dx=1$ ${\int }_0^{2}k{x}^{2}dx=1$ $k{\left[\frac{x^3}{3}\right]}_0^2=1$ $k\frac{8}{3}=1⟹k=\frac{3}{8}$

Probability Calculations:

1. $P(0.1\le X\le 0.4)$: ${\int }_{0.1}^{0.4}\frac{3}{8}{x}^{2}dx=\frac{1}{8}\left[(0.4{)}^3-(0.1{)}^3\right]=0.007875$

2. $P(0.2\le X\le 3)$: ${\int }_{0.2}^2\frac{3}{8}{x}^{2}dx=\frac{1}{8}\left[2^3-(0.2{)}^3\right]=0.999$

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Example 2: Exponential PDF

Suppose a continuous random variable $X$ has the PDF:

$$f(x)=\{\begin{array}{ll}\frac{k}{4}{e}^{-x/4},& x>0\\ 0,& \text{otherwise}\end{array}$$

To determine $k$, ensure the total probability equals 1: ${\int }_0^{\infty }\frac{k}{4}{e}^{-x/4}dx=1$ Using the integral $\int e^{ax}dx=\frac{e^{ax}}{a}$: $\frac{k}{4}{\left[-4e^{-x/4}\right]}_0^{\infty }=1$ $k=1$

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Example 3: Rainfall Distribution

The daily rainfall $X$ (in inches) is a continuous random variable with PDF:

$$f(x)=\{\begin{array}{ll}\frac{3}{4}(2x-x^2),& 0<x<2\\ 0,& \text{otherwise}\end{array}$$

Probability Calculations:

1. $P(X\le 1)$: ${\int }_0^1\frac{3}{4}(2x-x^2)dx=\frac{1}{2}$

2. $P(X>1.5)$: ${\int }_{1.5}^2\frac{3}{4}(2x-x^2)dx=0.15625$

3. $P(1<X\le 1.5)$: ${\int }_1^{1.5}\frac{3}{4}(2x-x^2)dx=0.34375$

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Example 4: PMF of $Y$

Let $X$ be a random variable taking values $1,2,3,\dots$ with $P(X=n)={\left(\frac{1}{2}\right)}^n$.

Define $Y$:

• $Y=1$ if $X$ is even.
• $Y=-1$ if $X$ is odd.

The PMF of $Y$:

• $P(Y=1)=P(X=2)+P(X=4)+\dots$ $={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^4+\dots$ $=\frac{{\left(\frac{1}{2}\right)}^2}{1-{\left(\frac{1}{2}\right)}^2}=\frac{1}{3}$

• $P(Y=-1)=1-P(Y=1)=\frac{2}{3}$

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Example 5: Symmetric PDF

Given PDF:

$$f(x)=\{\begin{array}{ll}kx(1-x),& 0<x<1\\ 0,& \text{otherwise}\end{array}$$

1. Find $k$: ${\int }_0^{1}kx(1-x)dx=1$ $k{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=1⟹k=6$

2. Find $b$ such that $P(X\le b)=P(X>b)$: ${\int }_0^{b}6x(1-x)dx={\int }_b^{1}6x(1-x)dx$ Solving the cubic equation yields $b=\frac{1}{2}$.

References

• Continued to Probability and Statistics Lecture 7