STAT 211 Topic 4

Lecture 8

Lecture 8 Notes

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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

$P(a\leq X\leq b)=\int _{a}^{b}f(x)\ dx$

Uniform distribution

Simplest PDF $f$ as a straight horizontal line: $f(x)={\begin{cases}{\frac {1}{B-A}}&A\leq x\leq B\\0&{\mbox{otherwise}}\end{cases}}$

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):

$F(x)=P(X\leq x)=\int _{-\infty }^{x}f(y)\ dy$

PDF is derivative of CDF:

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

Example

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

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

$\int _{0.5}^{1.3}0.5x\ dx=0.25x^{2}{\big |}_{0.5}^{1.3}=0.25\left(1.3^{2}-0.5^{2}\right)$

Lecture 9

Lecture 9 Notes

Tuesday, February 15, 2011

Expected Value

Expected value of Continuous Random Variable is:

$E(X)=\int _{-\infty }^{\infty }xf(x)\ dx$

Variance

Variance of a Continuous random variable is:

$V(X)=\int _{-\infty }^{\infty }(x-\mu )^{2}f(x)\ dx\quad \mu =E(X)$

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

$f(x)={\begin{cases}k\left[1-(x-3)^{2}\right]&2\leq x\leq 4\\0&{\mbox{otherwise}}\end{cases}}$

${\begin{aligned}E(X)&=\int _{2}^{4}x\cdot k\left[1-(x-3)^{2}\right]\ dx\\E(X^{2})&=\int _{x}^{4}x^{2}\cdot k\left[1-(x-3)^{2}\right]\ dx\end{aligned}}$

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"

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

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 $\Phi (z)=P(Z\leq z)$

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:

• $P(Z\leq -1.5)=\Phi (-1.5)=0.067$
• $P(Z\leq 2.5)=\Phi (2.5)=0.994$

We can find:

• $P(-1.5\leq Z\leq 2.5)=\Phi (2.5)-\Phi (-1.5)$
• $P(-2.5\leq Z)=\Phi (2.5)$ (symmetric around 0)
• $P(-2.5\leq Z\leq 1.5)=\Phi (2.5)-\Phi (-1.5)$

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

Percentile of Standard Normal Distribution

$\Phi (2.33)=0.99$ 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 $Y=aX+b\quad (a\neq 0)$, $Y$ is also normal rv.
2. $E(Y)=a\times \mu +b$ and $V(Y)=a^{2}\times \sigma ^{2}$. Therefore $Y\sim N(a\mu +b,a^{2}\sigma ^{2})\,\!$

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

Example 1

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

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

Example 2

$X\sim N(10,4)$

$X=2Z+10$

5^th Percentile = ?

1. Find 95^th percentile of Z: what $z$ satisfies $\Phi (z)=0.95$? (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 $\Phi (1)=1-0.16=0.84$

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 ${\tfrac {1}{\beta }}$

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

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

$\Gamma (\alpha )\approx \alpha !$

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?

  $P(-1.5\sigma +\mu <X<1.5\sigma +\mu )=P(-1.5<Z<1.5)=\Phi (1.5)-\Phi (-1.5)=\Phi (1.5)-(1-\Phi (1.5))\,\!$

• farther than 2.5 standard deviations of its mean value?

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