Lecture 3: Probability Distributions and Probability...

Random Variables Discrete Probability Distributions Continuous Random Variables

Lecture 3: Probability Distributions andProbability Densities - 1

Assist. Prof. Dr. Emel YAVUZ DUMAN

MCB1007 Introduction to Probability and StatisticsIstanbul Kultur University

Outline

1 Random Variables

2 Discrete Probability Distributions

3 Continuous Random Variables

Outline

1 Random Variables

Definition 1

If S is a sample space with a probability measure and X is areal-valued function defined over the elements of S, then X iscalled a random variable (or stochastic variable).In this course we shall always denote random variables by capitalletters such as X , Y etc., and their values by the correspondinglowercase letters such as x and y , respectively.

Example 2

Suppose that a coin is tossed twice so that the sample space isS = {HH,HT ,TH,TT}. Let X represent the number of headsthat can come up. With each sample point we can associate anumber for X as shown in the table:

Sample Point HH HT TH TT

Probability 14

x 2 1 1 0

Thus, for example, in the case of HH (i.e., 2 heads), X = 2 whilefor TH (1 head), X = 1. It follows that X is a random variable.Also, we can write P(X = 2) = 1

4 , P(X = 1) = 14 + 1

4 = 12 , and

P(X = 0) = 14 .

Example 3

A balanced coin is tossed four times. List the elements of thesample space that are presumed to be equally likely, as this is whatwe mean by a coin being balanced, and the corresponding values xof the random variable X , the total number of heads.

Solution. If H and T stand for heads and tails, the results are asshown in the following table:

Elements of Proba- Elements of Proba-sample space bility x sample space bility x

HHHH 116 4 THHT 1

16 2HHHT 1

16 3 THTH 116 2

HHTH 116 3 TTHH 1

16 2HTHH 1

16 3 HTTT 116 1

THHH 116 3 THTT 1

16 1HHTT 1

16 2 TTHT 116 1

HTHT 116 2 TTTH 1

16 1HTTH 1

16 2 TTTT 116 0

Thus, we can write P(X = 0) = 116 , P(X = 1) = 4

16 ,P(X = 2) = 6

16 , P(X = 3) = 416 and P(X = 4) = 1

Example 4

Two socks are selected at random and removed in succession froma drawer containing five brown socks and three green socks. Listthe elements of the sample space, the corresponding probabilities,and the corresponding values x of the random variable X is thenumber of brown socks selected.

Solution. If B and G stand for brown and green, then we havefollowing probabilities

P(BB) =5

8· 47=

56,P(BG ) =

8· 37=

P(GB) =3

8· 57=

56, and P(GG ) =

8· 27=

and the results are shown in the following table:

Elements of Sample Space BB BG GB GG

Probability 20/56 15/56 15/56 6/56

x 2 1 1 0

In all of the examples of this section we have limited our discussionto discrete sample space, and hence to discrete random variable,namely, random variables whose range is finite or countablyinfinite. Continuous random variables defined over continuoussample spaces will be taken up in the third section.

Outline

1 Random Variables

Definition 5

If X is a discrete random variable, the function given by

f (x) = P(X = x)

for each x within the range of X is called the probabilitydistribution (or probability function) of X .

Based on the postulates of probability, it immediately follows that

Theorem 6

A function can serve as the probability distribution of a discreterandom variable X if and only if its values, f (x), satisfy theconditions

1 f (x) ≥ 0 for each value within its domains;

x f (x) = 1, where the summation extends over all thevalues within its domain.

Example 7

Find a formula for the probability distribution of the total numberof heads obtained in four tosses of a balanced coin.

Solution. We know that P(X = 0) = 116 , P(X = 1) = 4

16 ,P(X = 2) = 6

16 , P(X = 3) = 416 and P(X = 4) = 1

16 . Observingthat the numerators of these five fractions, 1, 4, 6, 4, and 1, arethe binomial coefficients

), and

), we find that

the formula for the probability distribution can be written as

f (x) =

for x = 0, 1, 2, 3, 4.

Example 8

Check whether the function given by

f (x) =x + 2

25for x = 1, 2, 3, 4, 5

can serve as the probability distribution of a discrete randomvariable.

Solution. Substituting the different values of x , we get f (1) = 325 ,

f (2) = 425 , f (3) =

525 , f (4) =

625 , and f (5) = 7

25 . Since thesevalues are all nonnegative, the first condition of Theorem 6 issatisfied, and since

f (1) + f (2) + f (3) + f (4) + f (5) =3

the second conditions of Theorem 6 is satisfied. Thus, the givenfunction can serve as the probability distribution of a randomvariable having the range {1, 2, 3, 4, 5}.

Example 9

Suppose that a pair of fair dice are to be tossed, and let therandom variable X denote the sum of the points. Obtain theprobability distribution for X .

Solution. The random variable X is the sum of the coordinates foreach point. Thus for (3, 2) we have X = 5. Using the fact that all36 sample points are equally probable, so that each sample pointhas probability 1/36.

��

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

f (x) 136

Figure 1 : Probability Bar Chart

Definition 10

If X is a discrete random variable, the function given by

F (x) = P(X ≤ x) =∑t≤x

f (t) for −∞ < x < ∞

where f (t) is the value of the probability distribution of X at t, iscalled the distribution function, or the cumulative distribution, ofX .

Based on the postulates of probability and some of their immediateconsequences, it follows that

Theorem 11

The values F (x) of the distribution function of a discrete randomvariable X satisfy the conditions

1 F (−∞) = 0 and F (∞) = 1;

2 if a < b, then F (a) ≤ F (b) for any real numbers a and b.

If we are given the probability distribution of a discrete randomvariable, the corresponding distribution function is generally easyto find.

Example 12

Find the distribution function of the total of heads obtained in fourtosses of a balanced coin.

Solution. Given f (0) = 116 , f (1) =

416 , f (2) =

616 , f (3) =

416 , and

f (4) = 116 from Example 3, it follows that

F (0) = f (0) =1

F (1) = f (0) + f (1) =5

F (2) = f (0) + f (1) + f (2) =11

F (3) = f (0) + f (1) + f (2) + f (3) =15

F (4) = f (0) + f (1) + f (2) + f (3) + f (4) = 1.

Hence, the distribution function is given by

F (x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 for x < 0,116 for 0 ≤ x < 1,516 for 1 ≤ x < 2,1116 for 2 ≤ x < 3,1516 for 3 ≤ x < 4,

1 for x ≥ 4.

Observe that this distribution function is defined not only for thevalues taken on by the given random variable, but for all realnumbers. For instance, we can write F (1.7) = 5

16 and F (100) = 1,although the probabilities of getting at most 1.7 heads or at most100 heads in four tosses of a balanced coin may not be of any realsignificance.

Example 13

Find the distribution function of the random variable X of Example4 and plot its graph.

Solution. Based on the probabilities given in the following table

Elements of Sample Space BB BG GB GG

Probability 20/56 15/56 15/56 6/56

x 2 1 1 0

we can write f (0) = 656 , f (1) =

1556 + 15

56 = 3056 , and f (2) = 20

56 , sothat

F (0) = f (0) =6

F (1) = f (0) + f (1) =36

F (2) = f (0) + f (1) + f (2) = 1.

Hence, the distribution function of X is given by

F (x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 for x < 0,656 for 0 ≤ x < 1,3656 for 1 ≤ x < 2,

1 for x ≥ 2.

Example 14

Find the distribution function of the random variable that has theprobability distribution

f (x) =x

15for x = 1, 2, 3, 4, 5.

Solution. Since f (1) = 115 , f (2) =

215 , f (3) =

315 , f (4) =

415 , and

f (5) = 515 , then

F (x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

0 for x < 1,115 for 1 ≤ x < 2,315 for 2 ≤ x < 3,615 for 3 ≤ x < 4,1015 for 4 ≤ x < 5,

1 for x ≥ 5

Theorem 15

If the range of a random variable X consists of the valuesx1 < x2 < x3 < · · · < xn, then f (x1) = F (x1) and

f (xi ) = F (xi )− F (xi−1) for i = 2, 3, · · · , n.

Example 16

If X has the distribution function F (1) = 0.25, F (2) = 0.61,F (3) = 0.83, and F (4) = 1 for x = 1, 2, 3, 4, find the probabilitydistribution of X .

Solution. We have

f (1) = F (1) = 0.25,

f (2) = F (2)− F (1) = 0.61 − 0.25 = 0.36,

f (3) = F (3)− F (2) = 0.83 − 0.61 = 0.22,

f (4) = F (4)− F (3) = 1− 0.83 = 0.17.

Example 17

If X has the distribution function

F (x) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

0 for x < −1,14 for − 1 ≤ x < 1,12 for 1 ≤ x < 3,34 for 3 ≤ x < 5,

1 for x ≥ 5.

1 P(X ≤ 3), P(X = 3), P(X < 3);

2 P(X ≥ 1);

3 P(−0.4 < X < 4);

4 P(X = 5);

5 the probability distribution of X .

F (x) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

1 for x ≥ 5.

Solution.

P(X ≤ 3) =3

P(X = 3) =3

4− 1

P(X < 3) =1

F (x) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

1 for x ≥ 5.

2 P(X ≥ 1) = 1− P(X < 1) = 1− 14 = 3

3 P(−0.4 < X < 4) = 34 − 1

4 = 12 .

4 P(X = 5) = 1− 34 = 1

5 f (−1) = 14 , f (1) =

12 − 1

4 = 14 , f (3) =

34 − 1

2 = 14 ,

f (5) = 1− 34 = 1

4 , and 0 elsewhere.

Outline

1 Random Variables

Continuous Random Variables

So far we have considered discrete random variables that can takeon a finite or countably infinite number of values. In applications,we are often interested in random variables that can take on anuncountable continuum of values; we call these continuous randomvariables.Consider modeling the distribution of the age that a person dies at.Age of death, measured perfectly with all the decimals and norounding, is a continuous random variable (e.g., age of death couldbe 87.3248583585642 years). Other examples of continuousrandom variables include: time until the occurrence of the nextearthquake in Istanbul; the lifetime of a battery; the annual rainfallin Ankara.

Because it can take on so many different values, each value of acontinuous random variable winds up having probability zero. If Iask you to guess someone’s age of death perfectly, notapproximately to the nearest millionth year, but rather exactly toall the decimals, there is no way to guess correctly - each valuewith all decimals has probability zero. But for an interval, say thenearest half year, there is a nonzero chance you can guesscorrectly. So, we have the following definition:

Definition 18

A random variable X is called a continuous random variable ifits distribution function F is a continuous function on R, orequivalently, if

P(X = x) = 0, for every x ∈ R.

The definition of probability in the continuous case presumes foreach random variable the existence of a function, called aprobability density function (pdf), such that areas under the curvegive the probabilities associated with the corresponding intervalsalong the horizontal axis. In other words, a probability densityfunction, integrated form a to b (with a ≤ b), gives the probabilitythat the corresponding random variable will take on a value on theinterval from a to b.

Definition 19

A function with values f (x), defined over the set of all realnumbers, is called a probability density function (pdf) of thecontinuous random variable X if and only if

P(a ≤ X ≤ b) =

af (x)dx

for any real constants a and b with a ≤ b.

Figure 2 : P(a ≤ X ≤ b) = area under the density curve between a andb.

Theorem 20

If X is a continuous random variable and a and b are realconstants with a ≤ b, then

P(a ≤ X ≤ b) = P(a ≤ X < b) = P(a < X ≤ b) = P(a < X < b).

Analogous to Theorem 6, let us now state the following propertiesof probability densities, which again follow directly from thepostulates of probability.

Theorem 21

A function can serve as a probability density of a continuousrandom variable X if its values, f (x), satisfy the conditions

1 f (x) ≥ 0 for −∞ < x < ∞,

2∫∞−∞ f (x)dx = 1.

Example 22

The probability density of the continuous random variable X isgiven by

f (x) =

{15 for 2 < x < 7,

0 elsewhere.

1 Draw its graph and verify that the total area under the curve(above the x-axis) is equal to 1.

2 Find P(3 < X < 5).

Solution.

f (x) =

{15 for 2 < x < 7,

0 elsewhere.

1 It is clear that f (x) ≥ 0 for all x and∫ ∞

−∞f (x)dx =

−∞f (x)︸︷︷︸

2f (x)︸︷︷︸

∫ ∞

7f (x)︸︷︷︸

∣∣∣∣72

5(7− 2) = 1.

P(3 < X < 5) =

3f (x)dx =

∣∣∣∣53

5(5− 3)

Example 23

The pdf of the random variable X is given by

f (x) =

{c√x

for 0 < x < 4,

0 elsewhere.

1 the value of c ,

2 P(X < 0.25) and P(X > 1).

f (x) =

{c√x

for 0 < x < 4,

0 elsewhere.

Solution.

1 To satisfy the second condition of Theorem 21, we must have

∫ ∞

−∞f (x)dx =

−∞f (x)︸︷︷︸

0f (x)︸︷︷︸

∫ ∞

4f (x)︸︷︷︸

c√xdx =

0cx−

12dx = c

x−12+1

−12 + 1

∣∣∣∣∣4

= 2c√x |40 = 4c = 1

and it follows that c = 1/4.

f (x) =

for 0 < x < 4,

0 elsewhere.

P(X < 0.25) =

∫ 0.25

−∞f (x)dx =

−∞f (x)︸︷︷︸

∫ 0.25

0f (x)︸︷︷︸

∫ 0.25

4√xdx =

42√x

∣∣∣∣0.250

2(√0.25 − 0)

f (x) =

for 0 < x < 4,

0 elsewhere.

P(X < 1) =

∫ ∞

1f (x)dx =

1f (x)︸︷︷︸

∫ ∞

4f (x)︸︷︷︸0

4√xdx =

42√x

∣∣∣∣41

2(2− 1)

Example 24

If X has the probability density

f (x) =

{ke−3x for x > 0,

0 elsewhere,

find k and P(0.5 ≤ X ≤ 1).

Solution. To satisfy the second condition of Theorem 21, we musthave

∫ ∞

−∞f (x)dx =

∫ ∞

0ke−3xdx = k lim

t→∞e−3x

∣∣∣∣t0

= −k

3limt→∞(e−3t − e0) = −k

3(0− 1) =

and it follows that k = 3.

f (x) =

{3e−3x for x > 0,

0 elsewhere,

For the probability we get

P(0.5 ≤ X ≤ 1) =

0.53e−3xdx = 3

e−3x

∣∣∣∣10.5

= −e−3 + e−1.5

≈ 0.1733430918.

Definition 25

If X is a continuous random variable and the value of itsprobability density at t is f (t), then the function given by

F (x) = P(X ≤ x) =

−∞f (t)dt for −∞ < x < ∞

is called the distribution function, or the cumulativedistribution, of X .

The values F (x) of the distribution function of a continuousrandom variable X satisfies the conditions: F (−∞) = 0,F (∞) = 1, and F (a) ≤ F (b) when a < b.

Theorem 26

If f (x) and F (x) are the values of the probability density and thedistribution function of X at x, then

P(a ≤ X ≤ b) = F (b)− F (a)

for any real constants a and b with a ≤ b, and

f (x) =dF (x)

where the derivative exists.

Example 27

Find the distribution function of the random variable X of Example24, and use it to reevaluate P(0.5 ≤ x ≤ 1).

Solution. For x > 0,

F (x) =

−∞f (t)dt =

03e−3tdt = −e−3t

∣∣x0= 1− e−3x

and since F (x) = 0 for x ≤ 0, we can write

F (x) =

{0 for x ≤ 0,

1− e−3x for x > 0.

F (x) =

{0 for x ≤ 0,

1− e−3x for x > 0.

To determine the probability P(0.5 ≤ X ≤ 1), we use the first partof Theorem 26, getting

P(0.5 ≤ X ≤ 1) = F (1)− F (0.5)

= (1− e−3)− (1− e−1.5)

= −e−3 + e−1.5

≈ 0.1733430918.

This agrees with the result obtained by using the probabilitydensity directly in Example 24.

Example 28

Find the distribution function of the random variable X whoseprobability density is given by

f (x) =

⎧⎪⎨⎪⎩x for 0 < x < 1,

2− x for 1 ≤ x < 2,

0 elsewhere.

f (x) =

⎧⎪⎨⎪⎩x for 0 < x < 1,

2− x for 1 ≤ x < 2,

0 elsewhere.

Solution. For 0 < x < 1,

F (x) =

−∞f (t)dt =

0tdt =

∣∣∣∣x0

for 1 ≤ x < 2,

F (x) =

−∞f (t)dt =

0f (t)dt +

1f (t)dt =

0tdt +

1(2− t)dt

∣∣∣∣10

(2t − t2

2+ 2x − x2

2− 2 +

2= −x2

2+ 2x − 1,

and for F (x) = 1 where x ≥ 2.

So, we have

F (x) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩0 for x ≤ 0,x2

2 for 0 < x < 1,

− x2

2 + 2x − 1 for 1 ≤ x < 2,

1 for x ≥ 2.

Example 29

Find a probability density function for the random variable whosedistribution function is given by

F (x) =

⎧⎪⎨⎪⎩0 for x ≤ 0,

x for 0 < x < 1,

1 for x ≥ 1.

Solution. Since the given density function is differentiableeverywhere except at x = 0 and x = 1, we differentiate for x < 0,0 < x < 1, and x > 1, getting 0, 1, and 0. Thus, we can write

f (x) =

⎧⎪⎨⎪⎩0 for x ≤ 0,

1 for 0 < x < 1,

0 for x ≥ 1.

To fill the gaps at x = 0 and x = 1, we let f (0) and f (1) bothequal to zero. Actually, it does not matter how the probabilitydensity is defined at these points, but there are certain advantagesfor choosing the values in such a way that the probability density isnonzero over an open interval. Thus, we can write the probabilitydensity of the original random variable as

f (x) =

{1 for 0 < x < 1,

0 elsewhere.

Example 30

The distribution function of the random variable Y is given by

F (y) =

{1− 9

y2 for y > 3,

0 elsewhere.

Find P(Y ≤ 5), P(Y > 8) and a probability density function of Y .

Solution.

P(Y ≤ 5) = F (5) = 1− 952

= 1625 ,

P(Y > 8) = 1− P(Y ≤ 8) = 1− F (8) = 1− 1 + 964 = 9

f (y) =

(1− 9

y3 for y > 3,

0 elsewhere.

Example 31

The distribution function of the random variable X is given by

F (x) =

⎧⎪⎨⎪⎩0 for x < −1,x+12 for − 1 < x < 1,

1 for x ≥ 1.

Find P(−1

2 < X < 12

)and P(2 < X < 3).

Solution.

P(−1

2 < X < 12

)− F(−1

2 − − 12+1

2 =34 − 1

4 = 12 .

P(2 < X < 3) = F (3)− F (2) = 1− 1 = 0.

Thank You!!!

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