STAT 211 Topic 4

Lecture 8

Lecture 8 Notes

« previous | Thursday, February 10, 2011 | next »

World series: 2 teams Ends when winning team wins 4 games

Evenly matched (λ = .5)

Continuous Random Variables

Probability density function (PDF)

Distribution (PDF) is the smooth curve that fits over a histogram.

Comparable to PMF for discrete random variables:

• Area under PDF histogram should equal 1
• PDF is always positive (≥ 0)

Note: $P(X=c)=0$ for any constant $c$, so inclusive/exclusive ranges do not matter

$P(a\le X\le b)=\underset{a}{\overset{b}{\int }}f(x)dx$

Uniform distribution

Simplest PDF $f$ as a straight horizontal line: $f(x)=\{\begin{array}{ll}\frac{1}{B-A}& A\le x\le B\\ 0& \text{otherwise}\end{array}$

Area under line should be 1.

Cumulative distribution function (CDF)

"Sum" (integral) of all PDFs for random variable X, except continuous (no jumps between probabilities):

$F(x)=P(X\le x)=\underset{-\mathrm{\infty }}{\overset{x}{\int }}f(y)dy$

PDF is derivative of CDF:

$f(x)=\frac{dF}{dx}$

Example

Suppose a random variable $X$ has PDF $f(x)=\{\begin{array}{ll}0.5x& 0\le x\le 2\\ 0& otherwise\end{array}$

What is $P(0.5\le X\le 1.3)$?

$\underset{0.5}{\overset{1.3}{\int }}0.5xdx=0.25x^2{|}_{0.5}^{1.3}=0.25(1.3^2-0.5^2)$

Lecture 9

Lecture 9 Notes

Tuesday, February 15, 2011

Expected Value

Expected value of Continuous Random Variable is:

$E(X)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf(x)dx$

Variance

Variance of a Continuous random variable is:

$V(X)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(x-\mu {)}^{2}f(x)dx\mu =E(X)$

Also equivalent to:

$V(X)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2$

Note: when plugging $X^2$ into Expected Value function, do not put it into the PDF function

Example

$f(x)=\{\begin{array}{ll}k[1-(x-3{)}^2]& 2\le x\le 4\\ 0& \text{otherwise}\end{array}$

$\begin{array}{rl}E(X)& =\underset{2}{\overset{4}{\int }}x\cdot k[1-(x-3{)}^2]dx\\ E(X^2)& =\underset{x}{\overset{4}{\int }}{x}^2\cdot k[1-(x-3{)}^2]dx\end{array}$

Percentiles

• Medain: What is $a$ such that $F(a)=0.5$?
• Q₁ (25^th percentile): What is $a$ such that $F(a)=0.25$?
• Q₃ (75^th percentile): What is $a$ such that $F(a)=0.75$?
• in general: ($100p$^th percentile): What is $a$ such that $F(a)=p$?

Normal Distribution

For normal distributions,

• μ will be used for mean and median on symmetric unimodal curve
• σ² will be used for variance

"Gaussian Distribution"

$f(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}}-\mathrm{\infty }\le x\le \mathrm{\infty }$

Written: $X\sim N(\mu ,{\sigma }^2)$

Special Case: Normal distribution with μ = 0 and σ = 1 is called Standard Normal distribution: $Z\sim N(0,1)$

Curve of this function is called a Z-curve

CDF of standard normal random variable Z is $\mathrm{\Phi }(z)=P(Z\le z)$

Lecture 10

Lecture 10 Notes

Thursday, February 17, 2011

Topic: How to you get $P(X<3)$ when $X\sim N(-1,10)$?

Probabilities using the Z-Curve

Given:

• $P(Z\le -1.5)=\mathrm{\Phi }(-1.5)=0.067$
• $P(Z\le 2.5)=\mathrm{\Phi }(2.5)=0.994$

We can find:

• $P(-1.5\le Z\le 2.5)=\mathrm{\Phi }(2.5)-\mathrm{\Phi }(-1.5)$
• $P(-2.5\le Z)=\mathrm{\Phi }(2.5)$ (symmetric around 0)
• $P(-2.5\le Z\le 1.5)=\mathrm{\Phi }(2.5)-\mathrm{\Phi }(-1.5)$

Generalization: for any $z$, $\mathrm{\Phi }(z)=1-\mathrm{\Phi }(-z)$

Percentile of Standard Normal Distribution

$\mathrm{\Phi }(2.33)=0.99$ implies 2.33 is 99^th percentile.

Therefore, -2.33 is 1^st percentile.

α Notation

From now on, let $z_{\alpha }$ represent a number such that $\alpha$ is the area under the Z-curve to the right of $z_{\alpha }$.

In other words, $z_{\alpha }$ is the ((1 − α) × 100)^th percentile.

From previous example, $z_{0.01}=2.33$

Standardization

Caclculating Standard Normal Distribution from Standard Distribution:

1. if $X$ is normal rv and $Y=aX+b(a\ne 0)$, $Y$ is also normal rv.
2. $E(Y)=a\times \mu +b$ and $V(Y)=a^2\times {\sigma }^2$. Therefore $Y\sim N(a\mu +b,a^2{\sigma }^2)$

$\begin{array}{rl}X& \sim N(\mu ,{\sigma }^2)\\ X& =\sigma Z+\mu \\ Z& =\frac{X-\mu }{\sigma }\end{array}$

Example 1

$X\sim N(60,12^2)$

$P(72<X<90)=P(\frac{72-60}{12}<\frac{X-60}{12}<\frac{90-60}{12})=P(1<Z<2.5)=\mathrm{\Phi }(2.5)-\mathrm{\Phi }(1)$

Example 2

$X\sim N(10,4)$

$X=2Z+10$

5^th Percentile = ?

1. Find 95^th percentile of Z: what $z$ satisfies $\mathrm{\Phi }(z)=0.95$? (suppose 1.96)
2. Find 5^th percentile of Z: -1.96
3. Plug in 5^th percentile of Z into formula for X

Empirical Rule

1. Roughly 68% of values are within 1 standard deviation (σ) of the mean (μ)
2. Roughly 95% of values are within 2 standard deviations (2σ) of the mean (μ)

• (1) states that $P(\mu -\sigma <X<\mu +\sigma )=P(-1<Z<1)=0.68$
• This implies that the tails on either side are $\frac{1-0.68}{2}=0.16$
• Therefore $\mathrm{\Phi }(1)=1-0.16=0.84$

Gamma Distribution

How long do you expect to wait to have α many events happening, if the events are happening with a Poisson distribution with a rate of $\frac{1}{\beta }$

given $\alpha ,\beta >0$, PDF is:

$f(x)=\{\begin{array}{ll}\frac{1}{\mathrm{\Gamma }(\alpha ){\beta }^{\alpha }}{x}^{\alpha -1}{e}^{-x/\beta }& x>0\\ 0& otherwise\end{array}$

$\mathrm{\Gamma }(\alpha )\approx \alpha !$

Expected Value

$E(X)=\alpha \beta$

Variance

$V(X)=\alpha {\beta }^2$

Bonus Continuous Distributions

Lognormal Distribution

If $\log (X)$ is a normal distribution.

Summary of Distributions

We've learned:

• Bernoulli (0,1)
• Binomial (X ~ Bin(n, k))
• Poisson
• Uniform
• Normal
• Lognormal

Most situations in nature can be approximated using one of these distributions

Exercises

If a bolt of thread length is normally distributed, what is the probability that the length of a randomly selected bolt is:

• within 1.5 standard deviations of its mean value?

  $P(-1.5\sigma +\mu <X<1.5\sigma +\mu )=P(-1.5<Z<1.5)=\mathrm{\Phi }(1.5)-\mathrm{\Phi }(-1.5)=\mathrm{\Phi }(1.5)-(1-\mathrm{\Phi }(1.5))$

• farther than 2.5 standard deviations of its mean value?

  $P(X<\mu -2.5\sigma \vee X>\mu +2.5\sigma )=P(Z<-2.5\vee Z>2.5)=2(1-\mathrm{\Phi }(2.5))$