Random variables - Guo EXAM P

A random variable is a function that assigns a number to each element of the sample space. We typically write a random variable in a capital letter such as X .

Example 1. If we flip a coin and observe which side is up, the sample space is {H, T}. If we assign H=1 and T=0, our random variable is:

1 with probability of 0.5 0 with probability of 0.5 X =

There’s more than one way to assign a value to each element in the sample space. In the flip of a coin, we can also assign H=0 and T=1. In this case, the random variable is:

0 with probability of 0.5 1 with probability of 0.5 X =

Of course, you can assign H=2 and T=3, etc.

Example 2. If we flip a coin three times, the sample space is:

{

HHH HHT HTH HTT THH THT TTH TTT^, ^, ^, ^, ^, ^, ^,

}

Each element in the sample space has 0.125 chance of occurring. If we assign X =# of heads, Y=# of tails, Z=# of successive flips that have the same outcome, then each element corresponds to the following 3 sets of numerical values:

Element X Y Z Probability

HHH 3 0 3 0.125

HHT 2 1 2 0.125

HTH 2 1 1 0.125

HTT 1 2 2 0.125

THH 2 1 2 0.125

THT 1 2 1 0.125

TTH 1 2 2 0.125

TTT 0 3 3 0.125

Please note that in HHH, we have 3 consecutive heads; so Z=3. In HTH, no two consecutive outcomes are the same; so Z=1.

Here X Y Z, , are three random variables.

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Example 3. If we roll a die and record the side that’s face up, the sample space is

{

1, 2,3, 4, 5, 6

}

= . We can assign a value to each element in the sample space as follows:

Element of the sample space X Probability

1 1 1/6

2 2 1/6

3 3 1/6

4 4 1/6

5 5 1/6

6 6 1/6

Here X is a random variable.

Of course, you can assign values differently as follows:

Element of the sample space X Probability

1 6 1/6

2 5 1/6

3 4 1/6

4 3 1/6

5 2 1/6

6 1 1/6

You see that a random variable is really an arbitrary translation of each element in the sample space into a number. Of course, some translation schemes are more useful than others.

What do we gain by such translation? By mapping the entire sample space into a series of numbers, we can extract relevant information from the sample space to solve the problem at hand, while ignoring other details of the sample space.

Example 4. We flip a coin and are interested in finding the # of times heads are up. If we assign 1 and 0 to H and T respectively, then the information about the # of heads up can be conveniently summarized as follows:

1 with probability of 0.5 0 with probability of 0.5 X =

You see that we have reduced the coin flipping process to a simple, elegant math equation. More importantly, this equation answers our question at hand.

Example 5. If we flip a coin three times. We are concerned about the # of times heads show up. Let X represent the # of times heads show up, then we:

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©Yufeng Guo, Deeper Understanding: Exam P

X 3 2 2 1 2 1 1 0

Probability 0.125 0.125 0.125 0.125 0.125 0.125 0.125 0.125

Example 6. We roll a die and record the side that’s face up. We are interested in finding the probability of getting 1,2,3,4,5, and 6 respectively. If we let random variable X represent the number that’s face up, then we have:

X 1 2 3 4 5 6

Probability 1/6 1/6 1/6 1/6 1/6 1/6

Expressed more succinctly:

( )

P X =n =6, where n=1, 2, 3, 4, 5, 6

Discrete random variable vs. continuous random variable

If a random variable can take on discrete values, then it’s a discrete random variable. If a random variable can take on any value in a range, then it’s a continuous random variable.

Example 7. Let the random variable X represent the # of heads we get from flipping a coin n times. Then X can take on integer values ranging from 0 to n . X is a discrete random variable.

Example 8. Let random variable Yrepresent the number randomly chosen from the range [0,1]. Then Ycan take on any value in [0,1]. Y is a continuous random variable.

PMF and CDF for discrete random variables Probability mass function

The most important way to describe a discrete random variable is through the probability mass function (PMF). If x is a possible value of the random variable X , the probability mass of x , denoted as pX

( )

x , is the probability that X =x:

( ) ( )

pX x =P X =x

Example 9. We flip a coin twice and record the # of times we get heads. Let X represent the # of heads in 2 flips of a coin. The probability mass function of X is:

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A probability mass function must satisfy the 3 axioms:

• pX

( )

x ⁰

So a valid PMF needs to satisfy the following two conditions:

( )

⁰

pX x , X

( )

p x =

Example 10. You are given the following PMF:

( )

pN n e

= n , where n=0,1, 2,...,+ and is a positive constant Verify that this is a legitimate PMF.

Solution

©Yufeng Guo, Deeper Understanding: Exam P So

( )

pN n e

= n is a valid PMF.

Example 11. A special die has 3 sides painted 1, 2, and 3 respectively. If the die is thrown, each side has an equal chance of landing face up on the ground. Two dies are thrown together and let X represent the sum of the two sides facing up.

Find the probability mass function of X . Solution

In the above table, the blue cells represent the values of X . Because each side has 1 3 chance of landing face up, each cell has

( )

^{1 3} ² ⁼^{1 9} chance of occurring.

We convert the above table into the new table below:

To understand the above table, let’s look at pX

( )

³ =^{2 9}.This is how we get

( )

³ ^{2 9}

pX = . There are two ways to have X =3: you get a 1 from the 1^st die and 2 from the 2^nd die (with probability of 1 9); you get 2 from the 1^st die and 1 from the 2^nd die (with probability of 1 9). So the total probability of having X =3 is 2 9.

Example 12. Claim payment, X , has the following PMF:

x $0 $50 $80 $135 $250 $329

( )

pX x 0.32 0.2 0.18 0.1 0.15 0.05

Calculate

1. ^{P X}

(

^>¹²⁰

)

2. ^{P X}

(

³⁰⁰ ^X ^>¹²⁰

)

Solution

outcome of 2nd throw outcome of the 1st throw 1 2 3

1 2 3 4

2 3 4 5

3 4 5 6

x 2 3 4 5 6

( )

pX x 1 9 2 9 3 9 2 9 1 9

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©Yufeng Guo, Deeper Understanding: Exam P

Cumulative probability function (CDF)

The cumulative function is defined as FX

( )

x =P X

(

)

©Yufeng Guo, Deeper Understanding: Exam P

( )

⁴

(

⁴

) ( )

( )

⁴ 1 9 2 9 3 9 2 3

F =P X = p + p +p = + + =

( )

⁵

(

⁵

) ( )

( )

⁴

( )

⁵ 1 9 2 9 3 9 2 9 8 9

F =P X = p +p + p + p = + + + =

( )

⁶

(

⁶

) ( )

( )

⁴

( )

⁵

( )

⁶ 1 9 2 9 3 9 2 9 1 9 1

F =P X = p +p + p + p + p = + + + + =

( )

⁷

(

⁷

) ( )

( )

⁴

( )

⁵

( )

⁶

( )

⁷

F =P X = p + p + p +p +p +p

1 9 2 9 3 9 2 9 1 9 0 1

= + + + + + = because ^p

( )

⁷ ⁼⁰

( ) ( ) (

⁶

)

F + =P X + =P X =

PDF and CDF for continuous random variables

For a continuous random variable X, the probability density function (PDF), ^{f x}

( )

^{, is}

defined as:

( )

P a x b = f x dx

( )

P a x b is the area under the graph ^{f x}

( )

. Because including or excluding the end points doesn’t affect the area, including or excluding the end points doesn’t affect the probability:

( ) ( ) ( ) ( )

( )

P a< X <b =P a X <b =P a X b =P a< X b = f x dx

The CDF (cumulative probability function) of the continuous random variable X is defined as:

( ) ( )

F x =P X x . This is the same definition when X is discrete.

If a random variable is discrete, we say PMF (probability mass function); if a random variable is continuous, we say PDF (probability density function).

Whether a random variable is discrete or continuous, we always say CDF (cumulative probability function).

Please note that often for the sake of convenience, people use ^{f x}

( )

to refer to either PMF pX

( )

x or PDF ^{f x}

( )

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Properties of CDF

Rule 1 ^{F x}

( )

⁼^{P X}

(

)

^{for all}x -- this is just the definition.

Rule 2 CDF can never be decreasing. If a b, then ^{F a}

( )

^{F b .}

( )

To see why, notice ^{F x}

( )

⁼^{P X}

(

)

^{. If}^a ^b^{, then}

{

^x ^a

} {

^x b . In other words,

}

x b contains x a. So ^{P x}

(

)

^{P x}

(

a . This gives us

)

^{F b}

( )

^{F a .}

( )

Rule 3 ^F

( )

^{= and}⁰ ^F

( )

⁺ ^{= .}¹

They are true for both discrete and continuous random variables. To see why, notice

< X < + . There’s zero chance that X can be smaller or equal to ;

( ) ( )

⁰

F =P X = . On the other hand, we are 100% certain that Xcannot exceed + . So ^F

( )

⁺ ⁼^{P X}

(

⁺

)

^{= .}¹

Rule 4 If X is discrete and takes integer values, the PMF and CDF can be obtained from each other by summing or differencing:

( )

^k X

( )

F k p i

= -- this is the definition of ^{F k}

( )

( ) ( ) (

) ( ) (

)

pX k =P X k P X k =F k F k

Rule 5 If X is continuous, the PDF and CDF can be obtained from each other by integration or differentiation:

( )

F x = f t dt, ^{f x}

( )

^d ^{F x}

( )

= dx .

By definition, ^{F x}

( )

⁼^{P X}

(

)

⁼^P

(

^X ^x

)

⁼ ^x ^{f t dt}

( )

. Taking the derivative at both sides of ^{F x}

( )

⁼ ^x ^{f t dt}

( )

^{gives us} ^{f x}

( )

^d ^{F x}

( )

= dx .

Example 14. X has the following density: ^{f x}

( )

⁼³^x² ^where⁰ ^x ¹^.

Then,

( ) ( ) ( ) ( )

² ³

0 0

x x x

F x =P X x = f t dt= f t dt= t dt=x .

(

^0.2 ^0.6

) ( )

^0.6

( )

^0.2 ^0.6³ ^0.2³ ^0.208

P X =F F = =

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Example 15. A real number is randomly chosen from [0,1]. Then this number is squared. Let X represent the result.

Find the PDF and CDF forX. Solution

We’ll find the CDF first. Let U represent the # randomly drawn from [0,1]. Then X =U2.

( ) ( ) (

) ( )

FX x =P X x =P U x =P U x

Because any number in the interval [0,1] has an equal chance of being drawn,

( )

P U x must be proportional to the length of the interval 0, x . The total probability that ^{P U}

(

)

= -- we are 100% certain that any number taken from [0,1] ¹ must not exceed 1. Consequently,

( )

Please note that the following key difference between PMF for a discrete random variable and PDF for a continuous random variable:

PMF is a real probability and its value must not exceed one; PDF is a fake probability and can take on any non-negative value. PDF itself doesn’t have any meaning. For PDF to be useful, we must integrate it over a range.

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( )

2 f x

= for 0<x 1. When x 0,

( )

f x 2

= x + .

( )

f x 2

= x is not a probability. To get a probability, we must integrate

( )

2 f x

= over a range. For example, if we integrate ^{f x over}

( ) [ ]

a b , we’ll ^, get a real probability:

( )

P a< X b = f x dx

Mean and variance of a random variable

You just have to memorize a series of formulas:

If X is discrete, then

mean

( )

E X = x p x

variance

( ) ( )

( )

² ^X

( ) ( )

² ²

^{( )}

Var X =E X E X = x E X p x =E X E X

If X is continuous, then the mean ^{E X}

( )

⁼⁺ ^{xf x dx}

( )

variance ^{Var X}

( )

⁼^{E X} ^{E X}

( )

² ⁼⁺ ^x ^{E X}

( )

²^{f x dx}

( )

⁼^{E X}

( )

² ^E²

^{( )}

Standard deviation of X - no matter X is continuous or discrete

( )

X Var X

" =

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Mean of a function

Many times we need to find ^{E Y} ⁼^{g X}

( )

. One way to find E Y is to find the pdf

[ ]

Don’t worry about how to prove it. Just memorize it.

Example 18. Y =X² 1, where X has the following distribution:

e x is the exponential pdf (probability density function) with mean ) =1. Consequently,

( ) ^{( )} ^{( )}

Example 19. X has the following distribution:

2 1

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