general continuous distributions lecture 17, 18huang251/slides18.pdf · general continuous...

General ContinuousDistributions

From Discrete toContinuous RandomVariable

Cumulative DistributionFunctions

Expected Value andVariance

Percentiles andQuartiles

17, 18.1

Lecture 17, 18General Continuous DistributionsText: A Course in Probability by Weiss 8.1 ∼ 8.3

STAT 225 Introduction to Probability ModelsMarch 10, 2014

Whitney HuangPurdue University

17, 18.2

Agenda

1 From Discrete to Continuous Random Variable

2 Cumulative Distribution Functions

3 Expected Value and Variance

4 Percentiles and Quartiles

17, 18.3

Probability Mass Functions v.s. Probability DensityFunctions

0 2 4 6 8 10

Pmf for Binomial(n=10,p=0.3)

0 2 4 6 8 10

Pdf for Normal(mean=5, sd=1.5)

Remarks:pmf assign probabilities to each possible values of adiscrete distributionpdf describes the relative likelihood for this randomvariable to take on a given value

17, 18.4

Probability Mass Functions v.s. Probability DensityFunctions cont’d

Recall the properties of discrete probability mass functions(Pmfs):

0 ≤ pX (x) ≤ 1 for all possible values of x

∑x pX (x) = 1

P(a ≤ X ≤ b) =∑x=b

x=a pX (x)For continuous distributions, the properties for probabilitydensity functions (Pdfs) are similar:

fX (x) ≥ 0 for all possible values of x∫∞−∞ fX (x)dx = 1

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

0 ≤ pX (x) ≤ 1 for all possible values of x∑x pX (x) = 1

P(a ≤ X ≤ b) =∑x=b

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

P(a ≤ X ≤ b) =∑x=b

x=a pX (x)

For continuous distributions, the properties for probabilitydensity functions (Pdfs) are similar:

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

P(a ≤ X ≤ b) =∑x=b

x=a pX (x)

For continuous distributions, the properties for probabilitydensity functions (Pdfs) are similar:

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

P(a ≤ X ≤ b) =∑x=b

fX (x) ≥ 0 for all possible values of x

∫∞−∞ fX (x)dx = 1

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

P(a ≤ X ≤ b) =∑x=b

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.4

P(a ≤ X ≤ b) =∑x=b

P(a ≤ X ≤ b) =∫ b

a fX (x)dx

17, 18.5

Cumulative Distribution Functions (cdfs) for ContinuousDistribution

The cdf FX (x) is defined asFX (x) = P(X ≤ x) =

∫ x−∞ fX (x)dx

Some properties of the cdf:1 It is non-decreasing

2 It is everywhere right–continuous3 It has the value of 0 for x = −∞, i.e. FX (−∞) = 04 It has the value of 1 for x =∞, i.e. FX (∞) = 1

we use cdf to calculate probabilities of a continuousrandom variable within an interval, i.e.P(a ≤ X ≤ b) =

∫ ba fX (x)dx =∫ b

−∞ fX (x)dx −∫ a−∞ fX (x)dx = FX (b)− FX (a)

Some identities for cdf1 P(a < X < b) = FX (b)− FX (a) =

∫ ba fX (x) dx

2 P(a ≤ X ≤ b) = FX (b)− FX (a) =∫ b

a fX (x) dx

Remark: P(X = x) =∫ x

x fX (x) dx = 0 for all possible values ofx

17, 18.5

Some properties of the cdf:1 It is non-decreasing2 It is everywhere right–continuous

3 It has the value of 0 for x = −∞, i.e. FX (−∞) = 04 It has the value of 1 for x =∞, i.e. FX (∞) = 1

∫ ba fX (x) dx

2 P(a ≤ X ≤ b) = FX (b)− FX (a) =∫ b

a fX (x) dx

17, 18.5

Some properties of the cdf:1 It is non-decreasing2 It is everywhere right–continuous3 It has the value of 0 for x = −∞, i.e. FX (−∞) = 0

4 It has the value of 1 for x =∞, i.e. FX (∞) = 1

∫ ba fX (x) dx

2 P(a ≤ X ≤ b) = FX (b)− FX (a) =∫ b

a fX (x) dx

17, 18.5

Some properties of the cdf:1 It is non-decreasing2 It is everywhere right–continuous3 It has the value of 0 for x = −∞, i.e. FX (−∞) = 04 It has the value of 1 for x =∞, i.e. FX (∞) = 1

∫ ba fX (x) dx

2 P(a ≤ X ≤ b) = FX (b)− FX (a) =∫ b

a fX (x) dx

17, 18.6

Example 44

Let us find the cdf of a coin tossing example1 Let n = 4, p = .3, and X be the number of heads in the

sample. Find the cdf for X .2 Keep the above set-up, but use p = .5 instead. What is the

cdf for this r.v.?

17, 18.7

Example 44 cont’d

Solution.

1 X ∼ Bin(n = 4,p = 0.3)

FX (x) =

0 if x < 0P(X = 0) = .2401 if 0 ≤ x < 1∑1

i=0 P(X = i) = .6517 if 1 ≤ x < 2∑2i=0 P(X = i) = .9163 if 2 ≤ x < 3∑3i=0 P(X = i) = .9919 if 3 ≤ x < 4∑4i=0 P(X = i) = 1 if x ≥ 4

2 X ∼ Bin(n = 4,p = 0.5)

FX (x) =

0 if x < 0P(X = 0) = .0625 if 0 ≤ x < 1∑1

i=0 P(X = i) = .3125 if 1 ≤ x < 2∑2i=0 P(X = i) = .6875 if 2 ≤ x < 3∑3i=0 P(X = i) = .9375 if 3 ≤ x < 4∑4i=0 P(X = i) = 1 if x ≥ 4

17, 18.8

Example 44 cont’d

Solution.

0 1 2 3 4

Cdf for Binomial(n=4,p=0.3)

0 1 2 3 4

Cdf for Binomial(n=4, p=0.5)

17, 18.9

Example 45

Let us find the cdf of a random experiment over an interval1 Let X denote a number selected at random from the

interval (0,1), what is the cdf of X?2 Let X denote a number selected at random from the

interval (0,10), what is the cdf of X?

17, 18.10

Example 45 cont’d

1 FX (x) =

0 x < 0x 0 ≤ x ≤ 11 x > 1

2 FX (x) =

0 x < 0x10 0 ≤ x ≤ 101 x > 10

17, 18.11

Example 45 cont’d

Solution.

0.0 0.2 0.4 0.6 0.8 1.0

Cdf for Uniform(0,1)

0 2 4 6 8 10

Cdf for Uniform(0,10)

17, 18.12

Expected Value and Variance

Recall the expected value formula for the discrete randomvariable: E[X ] =

∑x xpX (x)

For continuous random variables, we have similar formulas:Let a, b, and c are constant real numbers

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

(∫∞−∞ xfX (x)dx

Var(cX ) = c2Var(X )

Var(X − c) = Var(X )

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.12

∑x xpX (x)

E[X ] =∫∞−∞ xfX (x)dx

E[g(X )] =∫∞−∞ g(x)fX (x)dx

E[X + Y ] = E[X ] + E[Y ]

E[cX ] = cE[X ]

E[aX + bY ] = aE[X ] + bE[Y ]

Var(X ) = E[X 2]− (E[X ])2 =∫∞−∞ x2fX (x)dx −

17, 18.13

Percentiles and Quartiles

A percentile represent the lower percent of the distributionFor example, the 10th percentile, written as X10, meansthat 10% of the distribution is less than or equal to thatvalue. It is the x value such that FX (x) = .10

The first quartile, Q1, represents the bottom (lower) 25% ofthe dataThe second quartile, Q2, aka the median, represents thebottom (lower) 50% of the dataThe third quartile, Q3, represents the bottom (lower) 75%of the data

17, 18.13

A percentile represent the lower percent of the distributionFor example, the 10th percentile, written as X10, meansthat 10% of the distribution is less than or equal to thatvalue. It is the x value such that FX (x) = .10The first quartile, Q1, represents the bottom (lower) 25% ofthe data

The second quartile, Q2, aka the median, represents thebottom (lower) 50% of the dataThe third quartile, Q3, represents the bottom (lower) 75%of the data

17, 18.13

A percentile represent the lower percent of the distributionFor example, the 10th percentile, written as X10, meansthat 10% of the distribution is less than or equal to thatvalue. It is the x value such that FX (x) = .10The first quartile, Q1, represents the bottom (lower) 25% ofthe dataThe second quartile, Q2, aka the median, represents thebottom (lower) 50% of the data

The third quartile, Q3, represents the bottom (lower) 75%of the data

17, 18.13

A percentile represent the lower percent of the distributionFor example, the 10th percentile, written as X10, meansthat 10% of the distribution is less than or equal to thatvalue. It is the x value such that FX (x) = .10The first quartile, Q1, represents the bottom (lower) 25% ofthe dataThe second quartile, Q2, aka the median, represents thebottom (lower) 50% of the dataThe third quartile, Q3, represents the bottom (lower) 75%of the data

17, 18.14

Example 46

Let X represent the diameter in inches of a circular disk cut bya machine. Let fX (x) = c(4x − x2) for 1 ≤ x ≤ 4 and be 0otherwise. Answer the following questions:

1 Find the value of c that makes this a valid pdf

2 Find the expected value and variance of X3 What is the probability that X is within .5 inches of the

expected diameter?4 Find FX (x)5 What is the 33rd percentile of X?

17, 18.14

Example 46

1 Find the value of c that makes this a valid pdf2 Find the expected value and variance of X

3 What is the probability that X is within .5 inches of theexpected diameter?

4 Find FX (x)5 What is the 33rd percentile of X?

17, 18.14

Example 46

1 Find the value of c that makes this a valid pdf2 Find the expected value and variance of X3 What is the probability that X is within .5 inches of the

expected diameter?

4 Find FX (x)5 What is the 33rd percentile of X?

17, 18.14

Example 46

expected diameter?4 Find FX (x)

5 What is the 33rd percentile of X?

17, 18.14

Example 46

expected diameter?4 Find FX (x)5 What is the 33rd percentile of X?

17, 18.15

Example 46 cont’d

Solution.

1∫ 4

1 fX (x) dx =∫ 4

1 c(4x − x2) dx = 9c = 1⇒ c = 19

2 E[X ] =∫ 4

1 xfX (x) dx = 2.25E[X 2] =

∫ 41 x2fX (x) dx = 5.6

⇒ Var(X ) = E[X 2]− (E[X ])2 = .53753 P(1.75 ≤ X ≤ 2.75) =

∫ 2.751.75 fX (x) dx = .4282

4 FX (x) =∫ x

1 fX (x) dx = 19 (2x2 − x3

3 )− 527 for 1 ≤ X ≤ 4 and

0 if x < 1 and 1 if x > 45 X.33 is the x value such that∫ x

1 fX (x) dx = .33⇒ X.33 = 1.8254

17, 18.15

Example 46 cont’d

Solution.

1∫ 4

1 fX (x) dx =∫ 4

1 c(4x − x2) dx = 9c = 1⇒ c = 19

2 E[X ] =∫ 4

1 xfX (x) dx = 2.25E[X 2] =

∫ 41 x2fX (x) dx = 5.6

⇒ Var(X ) = E[X 2]− (E[X ])2 = .5375

3 P(1.75 ≤ X ≤ 2.75) =∫ 2.75

1.75 fX (x) dx = .4282

4 FX (x) =∫ x

1 fX (x) dx = 19 (2x2 − x3

3 )− 527 for 1 ≤ X ≤ 4 and

1 fX (x) dx = .33⇒ X.33 = 1.8254

17, 18.15

Example 46 cont’d

Solution.

1∫ 4

1 fX (x) dx =∫ 4

1 c(4x − x2) dx = 9c = 1⇒ c = 19

2 E[X ] =∫ 4

1 xfX (x) dx = 2.25E[X 2] =

∫ 41 x2fX (x) dx = 5.6

⇒ Var(X ) = E[X 2]− (E[X ])2 = .53753 P(1.75 ≤ X ≤ 2.75) =

∫ 2.751.75 fX (x) dx = .4282

4 FX (x) =∫ x

1 fX (x) dx = 19 (2x2 − x3

3 )− 527 for 1 ≤ X ≤ 4 and

1 fX (x) dx = .33⇒ X.33 = 1.8254

17, 18.15

Example 46 cont’d

Solution.

1∫ 4

1 fX (x) dx =∫ 4

1 c(4x − x2) dx = 9c = 1⇒ c = 19

2 E[X ] =∫ 4

1 xfX (x) dx = 2.25E[X 2] =

∫ 41 x2fX (x) dx = 5.6

⇒ Var(X ) = E[X 2]− (E[X ])2 = .53753 P(1.75 ≤ X ≤ 2.75) =

∫ 2.751.75 fX (x) dx = .4282

4 FX (x) =∫ x

1 fX (x) dx = 19 (2x2 − x3

3 )− 527 for 1 ≤ X ≤ 4 and

0 if x < 1 and 1 if x > 4

5 X.33 is the x value such that∫ x1 fX (x) dx = .33⇒ X.33 = 1.8254

17, 18.15

Example 46 cont’d

Solution.

1∫ 4

1 fX (x) dx =∫ 4

1 c(4x − x2) dx = 9c = 1⇒ c = 19

2 E[X ] =∫ 4

1 xfX (x) dx = 2.25E[X 2] =

∫ 41 x2fX (x) dx = 5.6

⇒ Var(X ) = E[X 2]− (E[X ])2 = .53753 P(1.75 ≤ X ≤ 2.75) =

∫ 2.751.75 fX (x) dx = .4282

4 FX (x) =∫ x

1 fX (x) dx = 19 (2x2 − x3

3 )− 527 for 1 ≤ X ≤ 4 and

1 fX (x) dx = .33⇒ X.33 = 1.8254

17, 18.16

Example 47

Let fX (x) = .25x for 1 ≤ x ≤ 3 and 0 otherwise:1 Is X more likely to be within [1,2] or within [2,3]? First

answer this question using logic. Check your answer bycalculating the probabilities

2 What is the probability that X is more than 2.2?3 Find the mean and standard deviation of X4 Find FX (x)5 What value of X represents the top 15% of the

distribution?

17, 18.16

Example 47

2 What is the probability that X is more than 2.2?

3 Find the mean and standard deviation of X4 Find FX (x)5 What value of X represents the top 15% of the

distribution?

17, 18.16

Example 47

2 What is the probability that X is more than 2.2?3 Find the mean and standard deviation of X

4 Find FX (x)5 What value of X represents the top 15% of the

distribution?

17, 18.16

Example 47

2 What is the probability that X is more than 2.2?3 Find the mean and standard deviation of X4 Find FX (x)

5 What value of X represents the top 15% of thedistribution?

17, 18.16

Example 47

2 What is the probability that X is more than 2.2?3 Find the mean and standard deviation of X4 Find FX (x)5 What value of X represents the top 15% of the

distribution?

17, 18.17

Example 47 cont’d

Solution.

1 X is more likely be within [2,3] because fX (x) is increaseswith xP(1 ≤ X ≤ 2) =

∫ 21 fX (x) dx = .375

P(2 ≤ X ≤ 3) =∫ 3

2 fX (x) dx = .625

2 P(X ≥ 2.2) =∫ 3

2.2 fX (x) dx = .52

3 E[X ] =∫ 3

1 xfX (x) dx = 2.1667E[X 2] =

∫ 31 x2fX (x) dx = 5

⇒ Var(X ) = E[X 2]− (E[X ])2 = .3056⇒ Sd(X ) = .55284 FX (x) =

∫ x1 fX (x) dx = 1

8 (x2 − 1) for 1 ≤ X ≤ 3 and 0 if

x < 1 and 1 if x > 35 X.85 is the x value such that∫ x

1 fX (x) dx = .85⇒ X.85 = 2.7928

17, 18.17

Example 47 cont’d

Solution.

∫ 21 fX (x) dx = .375

P(2 ≤ X ≤ 3) =∫ 3

2 fX (x) dx = .625

2 P(X ≥ 2.2) =∫ 3

2.2 fX (x) dx = .52

3 E[X ] =∫ 3

1 xfX (x) dx = 2.1667E[X 2] =

∫ 31 x2fX (x) dx = 5

⇒ Var(X ) = E[X 2]− (E[X ])2 = .3056⇒ Sd(X ) = .55284 FX (x) =

∫ x1 fX (x) dx = 1

8 (x2 − 1) for 1 ≤ X ≤ 3 and 0 if

1 fX (x) dx = .85⇒ X.85 = 2.7928

17, 18.17

Example 47 cont’d

Solution.

∫ 21 fX (x) dx = .375

P(2 ≤ X ≤ 3) =∫ 3

2 fX (x) dx = .625

2 P(X ≥ 2.2) =∫ 3

2.2 fX (x) dx = .52

3 E[X ] =∫ 3

1 xfX (x) dx = 2.1667E[X 2] =

∫ 31 x2fX (x) dx = 5

⇒ Var(X ) = E[X 2]− (E[X ])2 = .3056⇒ Sd(X ) = .5528

4 FX (x) =∫ x

1 fX (x) dx = 18 (x

2 − 1) for 1 ≤ X ≤ 3 and 0 ifx < 1 and 1 if x > 3

17, 18.17

Example 47 cont’d

Solution.

∫ 21 fX (x) dx = .375

P(2 ≤ X ≤ 3) =∫ 3

2 fX (x) dx = .625

2 P(X ≥ 2.2) =∫ 3

2.2 fX (x) dx = .52

3 E[X ] =∫ 3

1 xfX (x) dx = 2.1667E[X 2] =

∫ 31 x2fX (x) dx = 5

⇒ Var(X ) = E[X 2]− (E[X ])2 = .3056⇒ Sd(X ) = .55284 FX (x) =

∫ x1 fX (x) dx = 1

8 (x2 − 1) for 1 ≤ X ≤ 3 and 0 if

x < 1 and 1 if x > 3

17, 18.17

Example 47 cont’d

Solution.

∫ 21 fX (x) dx = .375

P(2 ≤ X ≤ 3) =∫ 3

2 fX (x) dx = .625

2 P(X ≥ 2.2) =∫ 3

2.2 fX (x) dx = .52

3 E[X ] =∫ 3

1 xfX (x) dx = 2.1667E[X 2] =

∫ 31 x2fX (x) dx = 5

⇒ Var(X ) = E[X 2]− (E[X ])2 = .3056⇒ Sd(X ) = .55284 FX (x) =

∫ x1 fX (x) dx = 1

8 (x2 − 1) for 1 ≤ X ≤ 3 and 0 if

1 fX (x) dx = .85⇒ X.85 = 2.7928

general continuous distributions lecture 17, 18huang251/slides18.pdf · general continuous...

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