S1: Chapter 8Discrete Random VariablesDr J Frost (jfrost@tiffin.kingston.sch.uk)

Last modified: 25th 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.

A random variable represents a single experiment/trial, and has 2 ingredients:

x 1 2 3 4 5 6

P(X=x) 0.3 0.2 0.1 0.25 0.05 0.1

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 FUNCTIONe.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:(a) The outputs have to be between 0 and 1, i.e. .(b) The sum of the outputs is 1, i.e. ?

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

No Yes

No Yes

This is a continuous random variable.

It does not vary, so is not a variable!

Notation and Terminology

e.g. P(C = blue)

The are two equivalent ways of writing a probability:

e.g. p(blue)

Full notation: Shorthand:

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.

Example

The random variable represents the number of heads when three coins are tossed.

Underlying Sample Space

{ HHH, HHT, HTT, HTH, THH, THT, TTH, TTT }

Probability Function

Num heads x 0 1 2 3

P(X=x) 1/8 3/8 3/8 1/8

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!

? ?

Exam Question

(Hint: Use your knowledge that )

Edexcel S1 May 2012

p(-1) = 4k, p(0) = k, p(1) = 0, p(2) = kAnd since , 4k + k + 0 + k = 6k = 1Therefore ?

Exercise 8A

The random variable X has a probability functionP(X = x) = kx, x = 1, 2, 3, 4.Show that

The random variable X has a probability function:

where k is a constant.

a) Find the value of k.b) Construct a table giving the probability distribution of X.

7a) k = 0.1257b)

5

7

x 1 2 3 4

P(X = x) 0.125 0.125 0.375 0.375?

Probabilities of ranges of values

x 1 2 3 4 5 6P(X=x) 0.1 0.2 0.3 0.25 0.1 0.05

???

?

Cumulative Distribution Function (CDF)

How could we express “the probability that the age of someone is at most 40”?

F is known as the cumulative distribution function, where

(note the capital F)

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x 0 1 2P(X=x) 0.25 0.5 0.25

x 0 1 2F(x) 0.25 0.75 1

If X is the number of heads thrown in 2 throws...

? ? ?

? ? ?

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.

Example

x 1 2 3F(x) 3/4 7/8 1? ? ?

a

x 1 2 3P(X=x) 3/4 1/8 1/8

b

c

d

? ? ?

?

?

CDF

Shoe Size (x)

p(x)

Shoe Size (x)

F(x)

1

It’s just like how we’d turn a frequency graph into a cumulative frequency graph.

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Exam Questions

Edexcel S1 May 2013 (Retracted)

x 1 2 3

P(X = x) 0.4 0.25 0.35= 0.4?

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Edexcel S1 Jan 2013

F(3) = 1, so (27 + k)/40 = 1, ...

x 1 2 3

P(X = x) 0.35 0.175 0.475?

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Expected Value, E[X]Suppose that we throw a single fair die 60 times, and see the following outcomes:

x 1 2 3 4 5 6

Frequency 9 11 10 8 12 10

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)

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Quickfire E[X]Find the expected value of the following distributions (in your head!).

x 1 2 3

P(X = x) 0.1 0.6 0.3

E[X] = 2.2

x 4 6 8

P(X = x) 0.5 0.25 0.25

E[X] = 5.5

x 10 20 30

P(X = x) 1/3 1/3 1/3

E[X] = 20

? ?

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Harder Example

x 1 2 3 4 5

P(X = x) 0.1 p 0.3 q 0.2

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?

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!

x 1 2 3

P(X = x) 0.1 0.5 0.4

E[X2] = (12 x 0.1) + (22 x 0.5) + (32 x 0.4) = 5.7E[2X] = (2 x 0.1) + (4 x 0.5) + (6 x 0.4) = 4.6E[1 – X] = (0 x 0.1) + (-1 x 0.5) + (-2 x 0.4) = -1.3

??

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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

E[X2] – E[X]2? ? ?

x 1 2 3

P(X = x) 0.1 0.5 0.4(We already worked out that E[X2] = 5.7)

Var[X] = 5.7 – 2.32 = 0.41

Var[X] =

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Exam Questions

Edexcel S1 May 2010

Edexcel S1 Jan 2009

a = 1/4

= 1

E[X2] = 3.1 So Var[X] = 3.1 – 12 = 2.1

= 1

= P(X <= 1.5) = P(X <= 1) = 0.7

E[X2] = 2. So Var[X] = 2 – 12 = 1

???

???

Coding!Oh dear god, not again...

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 :

Recap

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] + 10E[3X] = 3E[X]

Var[3X] = 9Var[X]

???

Adding 10 to all values adds 10 to the expected value.

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Quickfire CodingExpress these in terms of the original E[X] and Var[X].

E[4X + 1] = 4E[X] + 1E[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)/2Var[(X-1)/2] = ¼ Var[X]

???

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Exercise 8E

E[X] = 2, Var[X] = 6Finda) E[3X] = 3E[X] = 6 d) E[4 – 2X] = 4 – 2E[X] = 0f) Var[3X + 1] = 9Var[X] = 54

The random variable Y has mean 2 and variance 9.Find:a) E[3Y+1] = 3E[Y] + 1 = 7c) Var[3Y+1] = 9Var[Y] = 81e) E[Y2] = Var[Y] + E[Y]2 = 13f) E[(Y-1)(Y+1)] = E[Y2 – 1] = E[Y2] – 1 = 12

2

5

???

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Discrete Uniform distribution

x 1 2 3 4 5 6

P(X = x) 1/6 1/6 1/6 1/6 1/6 1/6

If X is the throw of a fair die, this obviously is its distribution...

We call this a discrete uniform distribution.?

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If had say an n-sided fair die, then:

𝐸 [ 𝑋 ]=𝑛+12

𝑉𝑎𝑟 [ 𝑋 ]= 112

(𝑛+1 ) (𝑛−1 )

You won’t have exam questions on these, but they’re useful to know.

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ExampleDigits are selected at random from a table of random numbers.a) Find the mean and standard deviation of a single digit.b) 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 + 1Then E[R] = E[X – 1] = E[X] – 1 = 11/2 – 1 = 4.5

Var[R] = Var[X – 1]= Var[X]= = 8.25So = 2.87 (to 2sf)

b) We want

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