econ 482 lecture 1 i. administration: introduction syllabus thursday, jan 16 th, “lab” class is...

Econ 482

Lecture 1

I. Administration:• Introduction• Syllabus• Thursday, Jan 16th, “Lab” class is from 5-6pm in

Savery 117

II. Material:• Start of Statistical Review:

• Discrete and continuous random variables• Expected value, variance

Random variables:A random variable is any variable whose value cannot be predicted exactly

1. Discrete random variables:Random variable which has a specific (countable set)

of possible valuesExamples?

2. Continuous random variables:Random variable which can take any value of a

continuous range of values

Properties of discrete random variables

© Christopher Dougherty 1999–2006

PROBABILITY DISTRIBUTION EXAMPLE: X IS THE SUM OF TWO DICE

red 1 2 3 4 5 6

This sequence provides an example of a discrete random variable. Suppose that you have a red die which, when thrown, takes the numbers from 1 to 6 with equal probability.


red 1 2 3 4 5 6 green

1

2

3

4

5

6

Suppose that you also have a green die that can take the numbers from 1 to 6 with equal probability.



red 1 2 3 4 5 6 green

1

2

3

4

5

6

We will define a random variable X as the sum of the numbers when the dice are thrown.



For example, if the red die is 4 and the green one is 6, X is equal to 10.

red 1 2 3 4 5 6 green

1

2

3

4

5

6 10



red 1 2 3 4 5 6 green

1

2

3

4

5 7

6

Similarly, if the red die is 2 and the green one is 5, X is equal to 7.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

The table shows all the possible outcomes.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X 23456789

101112

If you look at the table, you can see that X can be any of the numbers from 2 to 12.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f 23456789

101112

We will now define f, the frequencies associated with the possible values of X.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f 2345 46789

101112

For example, there are four outcomes which make X equal to 5.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f 2 13 24 35 46 57 68 59 4

10 311 212 1

Similarly you can work out the frequencies for all the other values of X.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f p 2 13 24 35 46 57 68 59 4

10 311 212 1

Finally we will derive the probability of obtaining each value of X.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f p 2 13 24 35 46 57 68 59 4

10 311 212 1

If there is 1/6 probability of obtaining each number on the red die, and the same on the green die, each outcome in the table will occur with 1/36 probability.



red 1 2 3 4 5 6 green

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

X f p 2 1 1/363 2 2/364 3 3/365 4 4/366 5 5/367 6 6/368 5 5/369 4 4/36

10 3 3/3611 2 2/3612 1 1/36

Hence to obtain the probabilities associated with the different values of X, we divide the frequencies by 36.



The distribution is shown graphically. in this example it is symmetrical, highest for X equal to 7 and declining on either side.

6__36

5__36

4__36

3__36

2__36

2__36

3__36

5__36

4__36

probability

2 3 4 5 6 7 8 9 10 11 12 X


1 36

1 36


Definition of E(X), the expected value of X:

EXPECTED VALUE OF A RANDOM VARIABLE

The expected value of a random variable, also known as its population mean, is the weighted average of its possible values, the weights being the probabilities attached to the values.

n

iiinn pxpxpxXE

111 ...)(


Definition of E(X), the expected value of X:


Note that the sum of the probabilities must be unity, so there is no need to divide by the sum of the weights.

n

iiinn pxpxpxXE

111 ...)(


xi

x1

x2

x3

x4

x5

x6

x7

x8

x9

x10

x11


This sequence shows how the expected value is calculated, first in abstract and then with the random variable defined in the first sequence. We begin by listing the possible values of X.


xi pi

x1 p1

x2 p2

x3 p3

x4 p4

x5 p5

x6 p6

x7 p7

x8 p8

x9 p9

x10 p10

x11 p11


Next we list the probabilities attached to the different possible values of X.


xi pi xi pi

x1 p1 x1 p1

x2 p2

x3 p3

x4 p4

x5 p5

x6 p6

x7 p7

x8 p8

x9 p9

x10 p10

x11 p11


Then we define a column in which the values are weighted by the corresponding probabilities.


xi pi xi pi

x1 p1 x1 p1

x2 p2 x2 p2

x3 p3

x4 p4

x5 p5

x6 p6

x7 p7

x8 p8

x9 p9

x10 p10

x11 p11


We do this for each value separately.


xi pi xi pi

x1 p1 x1 p1

x2 p2 x2 p2

x3 p3 x3 p3

x4 p4 x4 p4

x5 p5 x5 p5

x6 p6 x6 p6

x7 p7 x7 p7

x8 p8 x8 p8

x9 p9 x9 p9

x10 p10 x10 p10

x11 p11 x11 p11


Here we are assuming that n, the number of possible values, is equal to 11, but it could be any number.


xi pi xi pi

x1 p1 x1 p1

x2 p2 x2 p2

x3 p3 x3 p3

x4 p4 x4 p4

x5 p5 x5 p5

x6 p6 x6 p6

x7 p7 x7 p7

x8 p8 x8 p8

x9 p9 x9 p9

x10 p10 x10 p10

x11 p11 x11 p11

S xi pi = E(X)


The expected value is the sum of the entries in the third column.


xi pi xi pi xi pi

x1 p1 x1 p1 2 1/36

x2 p2 x2 p2 3 2/36

x3 p3 x3 p3 4 3/36

x4 p4 x4 p4 5 4/36

x5 p5 x5 p5 6 5/36

x6 p6 x6 p6 7 6/36

x7 p7 x7 p7 8 5/36

x8 p8 x8 p8 9 4/36

x9 p9 x9 p9 10 3/36

x10 p10 x10 p10 11 2/36

x11 p11 x11 p11 12 1/36

S xi pi = E(X)


The random variable X defined in the previous sequence could be any of the integers from 2 to 12 with probabilities as shown.


xi pi xi pi xi pi xi pi

x1 p1 x1 p1 2 1/36 2/36

x2 p2 x2 p2 3 2/36

x3 p3 x3 p3 4 3/36

x4 p4 x4 p4 5 4/36

x5 p5 x5 p5 6 5/36

x6 p6 x6 p6 7 6/36

x7 p7 x7 p7 8 5/36

x8 p8 x8 p8 9 4/36

x9 p9 x9 p9 10 3/36

x10 p10 x10 p10 11 2/36

x11 p11 x11 p11 12 1/36

S xi pi = E(X)


X could be equal to 2 with probability 1/36, so the first entry in the calculation of the expected value is 2/36.



x1 p1 x1 p1 2 1/36 2/36

x2 p2 x2 p2 3 2/36 6/36

x3 p3 x3 p3 4 3/36

x4 p4 x4 p4 5 4/36

x5 p5 x5 p5 6 5/36

x6 p6 x6 p6 7 6/36

x7 p7 x7 p7 8 5/36

x8 p8 x8 p8 9 4/36

x9 p9 x9 p9 10 3/36

x10 p10 x10 p10 11 2/36

x11 p11 x11 p11 12 1/36

S xi pi = E(X)


The probability of x being equal to 3 was 2/36, so the second entry is 6/36.



x1 p1 x1 p1 2 1/36 2/36

x2 p2 x2 p2 3 2/36 6/36

x3 p3 x3 p3 4 3/36 12/36

x4 p4 x4 p4 5 4/36 20/36

x5 p5 x5 p5 6 5/36 30/36

x6 p6 x6 p6 7 6/36 42/36

x7 p7 x7 p7 8 5/36 40/36

x8 p8 x8 p8 9 4/36 36/36

x9 p9 x9 p9 10 3/36 30/36

x10 p10 x10 p10 11 2/36 22/36

x11 p11 x11 p11 12 1/36 12/36

S xi pi = E(X)


Similarly for the other 9 possible values.



x1 p1 x1 p1 2 1/36 2/36

x2 p2 x2 p2 3 2/36 6/36

x3 p3 x3 p3 4 3/36 12/36

x4 p4 x4 p4 5 4/36 20/36

x5 p5 x5 p5 6 5/36 30/36

x6 p6 x6 p6 7 6/36 42/36

x7 p7 x7 p7 8 5/36 40/36

x8 p8 x8 p8 9 4/36 36/36

x9 p9 x9 p9 10 3/36 30/36

x10 p10 x10 p10 11 2/36 22/36

x11 p11 x11 p11 12 1/36 12/36

S xi pi = E(X) 252/36To obtain the expected value, we sum the entries in this column.



The expected value turns out to be 7. Actually, this was obvious anyway. We saw in the previous sequence that the distribution is symmetrical about 7.



x1 p1 x1 p1 2 1/36 2/36

x2 p2 x2 p2 3 2/36 6/36

x3 p3 x3 p3 4 3/36 12/36

x4 p4 x4 p4 5 4/36 20/36

x5 p5 x5 p5 6 5/36 30/36

x6 p6 x6 p6 7 6/36 42/36

x7 p7 x7 p7 8 5/36 40/36

x8 p8 x8 p8 9 4/36 36/36

x9 p9 x9 p9 10 3/36 30/36

x10 p10 x10 p10 11 2/36 22/36

x11 p11 x11 p11 12 1/36 12/36

S xi pi = E(X) 252/36 = 7


Alternative notation for E(X):

E(X) = mX

Very often the expected value of a random variable is represented by m, the Greek m. If there is more than one random variable, their expected values are differentiated by adding subscripts to m.



Definition of E[g(X)], the expected value of a function of X:

To find the expected value of a function of a random variable, you calculate all the possible values of the function, weight them by the corresponding probabilities, and sum the results.

EXPECTED VALUE OF A FUNCTION OF A RANDOM VARIABLE

n

iiinn pxgpxgpxgXgE

111 )()(...)()(


Definition of E[g(X)], the expected value of a function of X:

Example:

For example, the expected value of X2 is found by calculating all its possible values, multiplying them by the corresponding probabilities, and summing.


n

iiinn pxpxpxXE

1

221

21

2 ...)(

n

iiinn pxgpxgpxgXgE

111 )()(...)()(


xi pi

x1 p1

x2 p2

x3 p3

… …… …… …… …… …… …… …xn pn

First you list the possible values of X and the corresponding probabilities.



xi pi g(xi)

x1 p1 g(x1)x2 p2 g(x2) x3 p3 g(x3) … … …...… … …...… … …...… … …...… … …...… … …...… … …...xn pn g(xn)

Next you calculate the function of X for each possible value of X.



xi pi g(xi) g(xi ) pi

x1 p1 g(x1) g(x1) p1

x2 p2 g(x2) x3 p3 g(x3) … … …...… … …...… … …...… … …...… … …...… … …...… … …...xn pn g(xn)

Then, one at a time, you weight the value of the function by its corresponding probability.




x1 p1 g(x1) g(x1) p1

x2 p2 g(x2) g(x2) p2

x3 p3 g(x3) g(x3) p3

… … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... xn pn g(xn) g(xn) pn

You do this individually for each possible value of X.




x1 p1 g(x1) g(x1) p1

x2 p2 g(x2) g(x2) p2

x3 p3 g(x3) g(x3) p3

… … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... … … …... ……... xn pn g(xn) g(xn) pn

S g(xi) pi The sum of the weighted values is the expected value of the function of X.



xi pi g(xi) g(xi ) pi xi pi

x1 p1 g(x1) g(x1) p1 2 1/36

x2 p2 g(x2) g(x2) p2 3 2/36

x3 p3 g(x3) g(x3) p3 4 3/36

… … …... ……... 5 4/36

… … …... ……... 6 5/36

… … …... ……... 7 6/36

… … …... ……... 8 5/36

… … …... ……... 9 4/36

… … …... ……... 10 3/36

… … …... ……... 11 2/36

xn pn g(xn) g(xn) pn 12 1/36

S g(xi) pi The process will be illustrated for X2, where X is the random variable defined in the first sequence. The 11 possible values of X and the corresponding probabilities are listed.



xi pi g(xi) g(xi ) pi xi pi xi2

x1 p1 g(x1) g(x1) p1 2 1/36 4

x2 p2 g(x2) g(x2) p2 3 2/36 9

x3 p3 g(x3) g(x3) p3 4 3/36 16

… … …... ……... 5 4/36 25

… … …... ……... 6 5/36 36

… … …... ……... 7 6/36 49

… … …... ……... 8 5/36 64

… … …... ……... 9 4/36 81

… … …... ……... 10 3/36 100

… … …... ……... 11 2/36 121

xn pn g(xn) g(xn) pn 12 1/36 144

S g(xi) pi First you calculate the possible values of X2.



xi pi g(xi) g(xi ) pi xi pi xi2 xi

2 pi

x1 p1 g(x1) g(x1) p1 2 1/36 4 0.11

x2 p2 g(x2) g(x2) p2 3 2/36 9

x3 p3 g(x3) g(x3) p3 4 3/36 16

… … …... ……... 5 4/36 25

… … …... ……... 6 5/36 36

… … …... ……... 7 6/36 49

… … …... ……... 8 5/36 64

… … …... ……... 9 4/36 81

… … …... ……... 10 3/36 100

… … …... ……... 11 2/36 121

xn pn g(xn) g(xn) pn 12 1/36 144

S g(xi) pi The first value is 4, which arises when X is equal to 2. The probability of X being equal to 2 is 1/36, so the weighted function is 4/36, which we shall write in decimal form as 0.11.




2 pi

x1 p1 g(x1) g(x1) p1 2 1/36 4 0.11

x2 p2 g(x2) g(x2) p2 3 2/36 9 0.50

x3 p3 g(x3) g(x3) p3 4 3/36 16 1.33

… … …... ……... 5 4/36 25 2.78

… … …... ……... 6 5/36 36 5.00

… … …... ……... 7 6/36 49 8.17

… … …... ……... 8 5/36 64 8.89

… … …... ……... 9 4/36 81 9.00

… … …... ……... 10 3/36 100 8.83

… … …... ……... 11 2/36 121 6.72

xn pn g(xn) g(xn) pn 12 1/36 144 4.00

S g(xi) pi Similarly for all the other possible values of X.




2 pi

x1 p1 g(x1) g(x1) p1 2 1/36 4 0.11

x2 p2 g(x2) g(x2) p2 3 2/36 9 0.50

x3 p3 g(x3) g(x3) p3 4 3/36 16 1.33

… … …... ……... 5 4/36 25 2.78

… … …... ……... 6 5/36 36 5.00

… … …... ……... 7 6/36 49 8.17

… … …... ……... 8 5/36 64 8.89

… … …... ……... 9 4/36 81 9.00

… … …... ……... 10 3/36 100 8.83

… … …... ……... 11 2/36 121 6.72

xn pn g(xn) g(xn) pn 12 1/36 144 4.00

S g(xi) pi

54.83The expected value of X2 is the sum of its weighted values in the final column. It is equal to 54.83. It is the average value of the figures in the previous column, taking the differing probabilities into account.




2 pi

x1 p1 g(x1) g(x1) p1 2 1/36 4 0.11

x2 p2 g(x2) g(x2) p2 3 2/36 9 0.50

x3 p3 g(x3) g(x3) p3 4 3/36 16 1.33

… … …... ……... 5 4/36 25 2.78

… … …... ……... 6 5/36 36 5.00

… … …... ……... 7 6/36 49 8.17

… … …... ……... 8 5/36 64 8.89

… … …... ……... 9 4/36 81 9.00

… … …... ……... 10 3/36 100 8.83

… … …... ……... 11 2/36 121 6.72

xn pn g(xn) g(xn) pn 12 1/36 144 4.00

S g(xi) pi

54.83Note that E(X2) is not the same thing as E(X), squared. In the previous sequence we saw that E(X) for this example was 7. Its square is 49.


Population variance of X:

The previous sequence defined the expected value of a function of a random variable X. There is only one function that is of much interest to us, at least initially: the squared deviation from the population mean.

The expected value of the squared deviation is known as the population variance of X. It is a measure of the dispersion of the distribution of X about its population mean.

POPULATION VARIANCE OF A DISCRETE RANDOM VARIABLE

i

n

iinn pxpxpxXE

1

221

21

2 )()(...)()(

2)( XE


We will calculate the population variance of the random variable X defined in the first sequence. We start as usual by listing the possible values of X and the corresponding probabilities.


xi pi xi – m (xi – m)2 (xi – m)2 pi

2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83


Next we need a column giving the deviations of the possible values of X about its population mean. In the second sequence we saw that the population mean of X was 7.



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83

7)( XEX


When X is equal to 2, the deviation is –5.


)(XEX


2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83

7)( XEX




2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83

7)( XEX


Next we need a column giving the squared deviations. When X is equal to 2, the squared deviation is 25.



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83


Similarly for the other values of X.



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83




2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83

Now we start weighting the squared deviations by the corresponding probabilities. What do you think the weighted average will be? Have a guess.


A reason for making an initial guess is that it may help you to identify an arithmetical error, if you make one. If the initial guess and the outcome are very different, that is a warning.



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83


We calculate all the weighted squared deviations.



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83



2 1/36 –5 25 0.69

3 2/36 –4 16 0.89

4 3/36 –3 9 0.75

5 4/36 –2 4 0.44

6 5/36 –1 1 0.14

7 6/36 0 0 0.00

8 5/36 1 1 0.14

9 4/36 2 4 0.44

10 3/36 3 9 0.75

11 2/36 4 16 0.89

12 1/36 5 25 0.69

5.83

The sum is the population variance of X.



Population variance of X

In equations, the population variance of X is usually written sX2, s being the Greek s.


2)( XE

2X


Standard deviation of X

The standard deviation of X is the square root of its population variance. Usually written sx, it is an alternative measure of dispersion. It has the same units as X.


])[( 2XE

X

Properties of continuous random variables


CONTINUOUS RANDOM VARIABLES

A discrete random variable is one that can take only a finite set of values. The sum of the numbers when two dice are thrown is an example.

Each value has associated with it a finite probability, which you can think of as a ‘packet’ of probability. The packets sum to unity because the variable must take one of the values.

6__

36

5__

36

4__

36

3__

36

2__

36

2__

36

3__

36

5__

36

4__

36

probability

2 3 4 5 6 7 8 9 10 11 12 X

1

36

1

36


55 60 70 75 X65


height

However, most random variables encountered in econometrics are continuous. They can take any one of an infinite set of values defined over a range (or possibly, ranges).

As a simple example, take the temperature in a room. We will assume that it can be anywhere from 55 to 75 degrees Fahrenheit with equal probability within the range.


55 60 70 75 X65


height

In the case of a continuous random variable, the probability of it being equal to a given finite value (for example, temperature equal to 55.473927) is always infinitesimal.

For this reason, you can only talk about the probability of a continuous random variable lying between two given values. The probability is represented graphically as an area.


55 60 70 75 X65


7

height

56

For example, you could measure the probability of the temperature being between 55 and 56, both measured exactly.

Given that the temperature lies anywhere between 55 and 75 with equal probability, the probability of it lying between 55 and 56 must be 0.05.


55 60 70 75 X65


57

height

Similarly, the probability of the temperature lying between 56 and 57 is 0.05.

56

0.05


55 60 70 75 X65


57 58

height

And similarly for all the other one-degree intervals within the range.

0.05


55 60 70 75X

65

0.05


57 58

height

The probability per unit interval is 0.05 and accordingly the area of the rectangle representing the probability of the temperature lying in any given unit interval is 0.05.

The probability per unit interval is called the probability density and it is equal to the height of the unit-interval rectangle.


55 60 70 75 X65

0.05


f(X) = 0.05 for 55 X 75f(X) = 0 for X < 55 and X > 75

57 58

Mathematically, the probability density is written as a function of the variable, for example f(X). In this example, f(X) is 0.05 for 55 < X < 75 and it is zero elsewhere.

height


55 60 70 75 X65

0.05


probabilitydensity

f(X)


57 58

The vertical axis is given the label probability density, rather than height. f(X) is known as the probability density function and is shown graphically in the diagram as the thick black line.


55 60 70 75 X65

0.05


probabilitydensity

f(X)


Suppose that you wish to calculate the probability of the temperature lying between 65 and 70 degrees. To do this, you should calculate the area under the probability density function between 65 and 70. Typically you have to use the integral calculus to work out the area under a curve, but in this very simple example all you have to do is calculate the area of a rectangle.


55 60 70 75 X65

0.05


The height of the rectangle is 0.05 and its width is 5, so its area is 0.25.

probabilitydensity

f(X)

0.05

5

0.25


econ 482 lecture 1 i. administration: introduction syllabus thursday, jan 16 th, “lab” class is...

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