Linear Functions of a Discrete Random Variable

Linear Functions of a Discrete Random Variable
“Teach A Level Maths” Statistics 1 Linear Functions of a Discrete Random Variable © Christine Crisp

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We have already looked at the effect of scaling ( or coding ) data.
If x gives the values of a variable and y is a linear function of that variable, we found the following results: If , the mean of y changes in the same way, so However, adding a constant has no effect on the standard deviation, so If we want variance instead of standard deviation, we square the result giving:

We get similar results if we are dealing with a discrete random variable ( which is used as a model for the data ). e.g. The probability table for a random variable X is shown below: 0·1 0·5 0·4 P(X = x) 3 2 1 x Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) Solution: (a) E(X) is the alternative notation for the mean, m So,

4x = 4 is the same as the probability that x = 1,
2 3 P(X = x) 0·4 0·5 0·1 Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (b) Find E(4X). We start with a table of values of 4x. 12 8 4 4x 3 2 1 x The probability that 4x = 4 is the same as the probability that x = 1, and so on, so we get

x 1 2 3 P(X = x) 0·4 0·5 0·1 x 1 2 3 4x 4 8 12 0·1 0·5 0·4 P(X = x)
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (b) Find E(4X). We start with a table of values of 4x. x 1 2 3 The probability that 4x = 4 is the same as the probability that x = 1, and so on, so we get 4x 4 8 12 0·1 0·5 0·4 P(X = x) So,

x 1 2 3 P(X = x) 0·4 0·5 0·1 x 1 2 3 4x 4 8 12 0·1 0·5 0·4 P(X = x)
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (b) Find E(4X). We start with a table of values of 4x. x 1 2 3 The probability that 4x = 4 is the same as the probability that x = 1, and so on, so we get 4x 4 8 12 0·1 0·5 0·4 P(X = x) So, So,

x 1 2 3 P(X = x) 0·4 0·5 0·1 4x + 5 17 13 9 P(X = x) 0·1 0·5 0·4
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (c) E(5) simply means the “expected value of 5” or the mean value of 5. So, E(5) = 5 (d) For E(4X + 5) we need 4x + 5 17 13 9 P(X = x) 0·1 0·5 0·4 So,

x 1 2 3 P(X = x) 0·4 0·5 0·1 0·1 0·5 0·4 P(X = x) 17 13 9 4x + 5
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (c) E(5) simply means the “expected value of 5” or the mean value of 5. So, E(5) = 5 (d) For E(4X + 5) we need 0·1 0·5 0·4 P(X = x) 17 13 9 4x + 5 So, =

x 1 2 3 P(X = x) 0·4 0·5 0·1 0·1 0·5 0·4 P(X = x) 17 13 9 4x + 5
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5) (c) E(5) simply means the “expected value of 5” or the mean value of 5. So, E(5) = 5 (d) For E(4X + 5) we need 0·1 0·5 0·4 P(X = x) 17 13 9 4x + 5 So, = 5

10 8 6 4 2 x The results we have found can be generalised to give
e.g. The probability distribution for the r.v. X is given by 10 8 6 4 2 x Find (a) E(X), (b) Hence find E(2X - 3) Solution: (a) “Hence” in part (b) of the question means that we must use the answer to part (a) rather than using the values and probabilities of 2X - 3. (b)

Before we can find an expression for it’s useful to define a new notation for .
We know that Using an extension of the notation for ( i.e. E(X) ), we write So, This, in turn, can be written as and, because the last term is clumsy, we write

We must be very careful if we use
This, is the mean of and this is the mean of X which has been squared.

0·1 0·5 0·4 P(X = x) 3 2 1 x We’ll now find a formula for e.g.
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5) We already have the following results: (a) (b) N.B. Variance measures variability or spread. There is no spread if we just have 5.

x 1 2 3 P(X = x) 0·4 0·5 0·1 We’ll now find a formula for e.g.
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5) (c)

4x 4 8 12 P(X = x) 0·4 0·5 0·1 We’ll now find a formula for e.g.

Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5) (c) (d)

4x + 5 9 13 17 P(X = x) 0·4 0·5 0·1 We’ll now find a formula for e.g.

So, The general result is This is just like the rule for coding data: Multiplying by a constant results in the variance being multiplied by the square of the constant. Adding a constant has no effect on the variance since it does not change the spread of the values.

SUMMARY The mean of a linear function of a discrete random variable, X, is given by The variance of X, is given by which can be written as The variance of a linear function of X is given by

e.g. The probability distribution for the r.v. X is given by
10 8 6 4 2 x Find (a) Var(X), (b) Hence find Var(2X - 3) Solution: (a) We already have the result So, (b)

Exercise 1. The probability distribution for the r.v. X is given by x 1 2 3 4 5 Find (a) E(X) (b) E(3X + 1) (c) Var(X) (d) Var(3X + 1) Solution: (a) E(X) = 3. ( We can get this directly from the symmetry of the table. ) (b) E(3X + 1) = 3E(X) + 1 = 10 (c) (d)

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

SUMMARY The variance of X, is given by which can be written as The variance of a linear function of X is given by The mean of a linear function of a discrete random variable, X, is given by

e.g. The probability distribution for the r.v. X is given by
10 8 6 4 2 x Find (a) E(X) (b) E(2X - 3) (c) Var(X) (d) Var(2X - 3) Solution: (a) (b)

So, (d) (c)

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