STAT 305 Introduction to Statistical Inference

Transcription

1 STAT 305 Introduction to Statistical Inference Pavel Krupskiy (Instructor) January April 2018

2 Course information Time and place: Mon, Wed, Fri, 13:00 14:00, LSK 201 Instructor: Pavel Krupskiy, ESB 3144, Teaching Assistants: TBA Clicker: You will need a clicker for this course (available in the bookstore) Course textbook: STAT 305, Introduction to Statistical Inference, by Welch, W. J. (available in the bookstore) Course website: ugrad.stat.ubc.ca/ stat305 Canvas: Course materials will be available at canvas.ubc.ca Piazza: piazza.com/ubc.ca/winterterm22017/stat305

3 Course information Course prerequisites: 1. One of STAT 200, BIOL 300, STAT 241, STAT 251, COMM 291, ECON 325, FRST 231, PSYC 218, PSYC 366 (STAT 200 or BIOL 300 is recommended) 2. One of MATH 302, STAT 302 Course outline: 1. Probability distributions and their properties 2. Parameter estimation, maximum likelihood estimators 3. Bayes rule, prior and posterior distributions 4. Hypothesis testing and confidence intervals 5. Categorical data and multinomial distribution

4 Course information Grading: Two midterm exams (35% total): February 7th, March 14th Final exam 50% Weekly labs (5% total) Clicker questions (5% total) WeBWorK assignments (5% total) Exercises from the textbook (not marked) To pass the course, you need to get at least 50%, based on the five marked components listed above

5 Course information Exams: Please bring your student ID to the midterm exams and final exam A two-sided A4 formula sheet can be used at the midterm exams and final exam (with formulas but not explanations) The midterm exams and final exam will be based on the lectures, textbook exercises, lab problems and WeBWorK questions If you cannot attend a midterm exam for a valid documented reason, the weights for the remaining midterm exam and final exam will be readjusted

6 Course information Labs and WeBWorK assignments: You will work in teams at the labs. A joint report for the team will be handed in at the end and marked We will use the R statistical software: Familiarity with R will also be tested on the quizzes and final examination WeBWorK assignments will be available online and graded automatically: STAT W2

7 Let s get started!

8 The cumulative distribution function (cdf) for a continuous random variable Y is F Y (y) = Pr(Y < y). The probability density function (pdf) is f Y (y) := F Y (y) y = lim y 0 F Y (y + y) F Y (y). y For small y, Pr(y < Y < y + y) = F Y (y + y) F Y (y) y f Y (y).

9 Some properties: F Y (y) is an increasing function of y; F Y ( ) = 0, F Y ( ) = 1; f Y (y) 0 and f Y (y)dy = 1. In particular, Pr(y 1 < Y < y 2 ) = Pr(Y < y 1 ) Pr(Y < y 2 ) = F Y (y 2 ) F Y (y 1 ) Pr(Y > y) = Pr(y < Y < ) = F Y ( ) F Y (y) = 1 F Y (y)

10 The mean is E(Y ) = The k-th moment is E(Y k ) = The variance is yf Y (y)dy. y k f Y (y)dy, k = 1, 2,... Var(Y ) = E{(Y EY ) 2 } = E(Y 2 ) {E(Y )} 2. Standard deviation is σ(y ) = Var(Y ).

11 A discrete random variable Y can take values y 1, y 2,.... The probability mass function (pmf) for a discrete random variable Y is f Y (y i ) = Pr(Y = y i ), i = 1, 2,... The cdf is F Y (y) = Pr(Y y) = i:y i y Pr(Y = y i ) = All values of the pmf must sum up to one: Pr(Y = y i ) = f Y (y i ) = 1. i i i:y i y f Y (y i ).

12 The mean is E(Y ) = i y i f Y (y i ). The k-th moment is E(Y k ) = i y k i f Y (y i ). The variance and standard deviation are defined in the same way as for a continuous random variable.

13 Example 1: Y has an exponential distribution with the rate parameter λ > 0 : F Y (y) = 1 e λy, f Y (y) = λe λy, y > 0. We find E(Y ) = 0 y λe λy dy = 1/λ. Example 2: Y has a Poisson distribution with parameter λ: f Y (k) = e We find E(Y ) = λk k=0 ke λ k! = λ. λ λk, k = 0, 1, 2,... k!

14 Example 1 (Exponential distribution): In R, With λ = 2, pexp(y, λ) computes F Y (y), dexp(y, λ) computes f Y (y). F Y (1) = 1 e 2 = , and f Y (1) = 2e 2 = The function exp(y) computes e y. You can check in R, that pexp(1, 2) and 1-exp(-2) give you the same answer.

15 Example 2 (Poisson distribution): In R, With λ = 1, F Y (2) = 2 k=0 ppois(y, λ) computes F Y (y), dpois(y, λ) computes f Y (y). 1 1k e k! = e 1 + e e 1 = 5 2 e 1 = , f Y (2) = 1 2 e 1 = You can check in R, that ppois(2, 1) and 2.5*exp(-1) give you the same answer.

16 Example 2 (Poisson distribution): The pmf values should sum up to one for this distribution: or e λ = k=0 k=0 λ λk e k! = 1, λ k k! = 1 + λ 1! + λ2 2! + λ3 3! +... This is a Taylor series expansion (at zero) of the exponential function.

17 Example 3: Let the cdf F Y (y) = Pr(Y < y) = 1 y 1/2, y > 1. The pdf f Y (y) = F Y (y)/ y = (1/2) y 3/2. The mean does not exist because E(Y ) = 1 y f Y (y)dy = (1/2) 1 y 1/2 dy =. Example 4: Let the pmf f Y (y k ) = Pr(Y = y k ) = 1/2 k for y k = 2 k, k = 1, 2,.... The mean does not exist because E(Y ) = y k f Y (y k ) = k=1 2 k (1/2 k ) = k=1 1 =. k=1