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S1: Chapter 8 Discrete Random Variables

S1: Chapter 8 Discrete Random Variables. Dr J Frost (jfrost@tiffin.kingston.sch.uk) . Last modified: 25 th October 2013. Variables and Random Variables. In Chapter 2, we saw that just like in algebra, we can use a variable to represent some quantity, such as height.

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S1: Chapter 8 Discrete Random Variables

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  1. S1: Chapter 8Discrete Random Variables Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 25th October 2013

  2. Variables and Random Variables In Chapter 2, we saw that just like in algebra, we can use a variable to represent some quantity, such as height. A random variablerepresents a single experiment/trial, and has 2 ingredients: A random variable representing the throw of an unfair die: 1. THE OUTCOMES /SUPPORT VECTOR We tend to use a lowercase variable letter to represent the value of the outcome. 2. PROBABILITY FUNCTION e.g. P(X = 3) = 0.1. It is just a function that maps an outcome to the probability. The probability function for a discrete variable has two obvious constraints: The outputs have to be between 0 and 1, i.e. . The sum of the outputs is 1, i.e. ?

  3. Is it a discrete random variable? The height of a person randomly chosen. The number of cars that pass in the next hour. The number of countries in the world.  No  Yes This is a continuous random variable.  No  Yes  No  Yes It does not vary, so is not a variable!

  4. Notation and Terminology The are two equivalent ways of writing a probability: Full notation: Shorthand: e.g. P(C = blue) e.g. p(blue) This says “the probability the outcome of an experiment, represented by the random variable , is the value . Note the lowercase instead of uppercase. Because a probability function is ultimately just a function, on the rare occasion it’s written as , where x is a particular outcome. Don’t be upset by this.

  5. Example The random variable represents the number of heads when three coins are tossed. Underlying Sample Space Probability Function ? ? { HHH, HHT, HTT, HTH, THH, THT, TTH, TTT } The second way of writing it allows us to conflate outcomes with the same probability. In the Edexcel syllabus, we call the table a probability distribution and the latter form a probability function. The true distinction is slightly abstract/subtle: don’t worry about it for now!

  6. Exam Question Edexcel S1 May 2012 (Hint: Use your knowledge that ) ? p(-1) = 4k, p(0) = k, p(1) = 0, p(2) = k And since , 4k + k + 0 + k = 6k = 1 Therefore

  7. Exercise 8A The random variable X has a probability function P(X = x) = kx, x = 1, 2, 3, 4. Show that The random variable X has a probability function: where k is a constant. Find the value of k. Construct a table giving the probability distribution of X. 7a) k = 0.125 7b) 5 7 ?

  8. Probabilities of ranges of values ? ? ? ?

  9. Cumulative Distribution Function (CDF) How could we express “the probability that the age of someone is at most 40”? If X is the number of heads thrown in 2 throws... ? ? ? ? ? F is known as the cumulative distribution function, where (note the capital F) ? ? ?

  10. Example The discrete random variable X has a cumulative distribution function defined by: ; x = 1, 2 and 3 Find the value of k. F(3) = 1. Thus k = 5. Draw the distribution table for the cumulative distribution function. Write down F(2.6) F(2.6) = F(2) = 7/8 Find the probability distribution of X. a ? b ? ? ? c ? d ? ? ?

  11. CDF ? 1 p(x) F(x) Shoe Size (x) Shoe Size (x) It’s just like how we’d turn a frequency graph into a cumulative frequency graph.

  12. Exam Questions Edexcel S1 May 2013 (Retracted) ? = 0.4 ? Edexcel S1 Jan 2013 ? F(3) = 1, so (27 + k)/40 = 1, ... ?

  13. Expected Value, E[X] Suppose that we throw a single fair die 60 times, and see the following outcomes: What is the mean outcome based on our sample? ? But using the actual probabilities of each outcome (i.e. 1/6 for each), what would we expect the average outcome to be? 3.5 ? If X is the random variable, is known as the expected value of . You could think of it as the weighted sum of the outcomes (where the weights are the probabilities)

  14. Quickfire E[X] Find the expected value of the following distributions (in your head!). E[X] = 2.2 E[X] = 5.5 ? ? E[X] = 20 ?

  15. Harder Example Given that E[X] = 3, find the values of p and q. p + q + 0.1 + 0.3 + 0.2 = 1 (1 x 0.1) + (2 x q) + (3 x 0.3) + (4 x q) + (5 x 0.2) = 3 Thus q = 0.1, p = 0.3 ?

  16. To E[X2] and beyond Remember with the mean for a sample, we could find the “mean of the squares” when finding variance, e.g. ? We just replaced each value with its square. Unsurprisingly the same applies for the expected value of a random variable. Just replace with whatever is in the square brackets. Sorted! E[X2] = (12 x 0.1) + (22x 0.5) + (32 x 0.4) = 5.7 E[2X] = (2 x 0.1) + (4 x 0.5) + (6 x 0.4) = 4.6 E[1 – X] = (0 x 0.1) + (-1 x 0.5) + (-2 x 0.4) = -1.3 ? ? ?

  17. Variance We know how to find it for experimental data. How about for a random variable? Mean of the Squares Minus Square of the Mean ? ? ? Var[X] = E[X2] – E[X]2 (We already worked out that E[X2] = 5.7) Var[X] = 5.7 – 2.32 = 0.41 ?

  18. Exam Questions Edexcel S1 May 2010 a = 1/4 ? ? = 1 E[X2] = 3.1 So Var[X] = 3.1 – 12 = 2.1 ? Edexcel S1 Jan 2009 ? = 1 = P(X <= 1.5) = P(X <= 1) = 0.7 ? E[X2] = 2. So Var[X] = 2 – 12 = 1 ?

  19. Coding! Oh dear god, not again... Recap Suppose that we have a list of peoples heights x. The mean height is 1.5m and the variance 0.2m. We use the coding : ? ? It’s no different with expected values. What do we expect these to be in terms of the original expected value E[X] and the original variance Var[X]? E[X + 10] = E[X] + 10 E[3X] = 3E[X] Var[3X] = 9Var[X] Adding 10 to all values adds 10 to the expected value. ? ? ?

  20. Quickfire Coding Express these in terms of the original E[X] and Var[X]. E[4X + 1] = 4E[X] + 1 E[1 – X] = 1 – E[X] Var[4X] = 16Var[X] Var[X + 1] = Var[X] Var[3X + 2] = 9Var[X] E[(X-1)/2] = (E[X]-1)/2 Var[(X-1)/2] = ¼ Var[X] ? ? ? ? ? ? ?

  21. Exercise 8E E[X] = 2, Var[X] = 6 Find a) E[3X] = 3E[X] = 6 d) E[4 – 2X] = 4 – 2E[X] = 0 f) Var[3X + 1] = 9Var[X] = 54 The random variable Y has mean 2 and variance 9. Find: a) E[3Y+1] = 3E[Y] + 1 = 7 c) Var[3Y+1] = 9Var[Y] = 81 e) E[Y2] = Var[Y] + E[Y]2 = 13 f) E[(Y-1)(Y+1)] = E[Y2 – 1] = E[Y2] – 1 = 12 2 ? ? ? 5 ? ? ? ?

  22. Discrete Uniform distribution ? If X is the throw of a fair die, this obviously is its distribution... We call this a discrete uniform distribution. ? If had say an n-sided fair die, then: ? ? You won’t have exam questions on these, but they’re useful to know.

  23. Example Digits are selected at random from a table of random numbers. Find the mean and standard deviation of a single digit. Find the probability that a particular digit lies within one standard deviation of the mean. a) Our digits are 0 to 9. We have useful formulae when the numbers start from 1 rather than 0. If the digit is R, let X = R + 1 Then E[R] = E[X – 1] = E[X] – 1 = 11/2 – 1 = 4.5 Var[R] = Var[X – 1] = Var[X] = = 8.25 So = 2.87 (to 2sf) b) We want ? ?

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