1. SOME CONTINUOUS PROBABILITY DISTRIBUTIONS
Uniform, Normal, Exponential, Gamma and Chi-Square Distributions

2. Uniform Distribution
A random variable X is said to be uniformly distributed if its density function is
The expected value and the variance are

3. Indicator functions
It is sometimes convenient to express the p.m.f. or p.d.f. by using indicator functions. This is especially true when the range of random variable depends on a parameter.
Ex: Uniform distribution

4. Uniform Distribution Example 1
The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are:
Between 2,500 and 3,000 gallons
More than 4,000 gallons
Exactly 2,500 gallons
f(x) = 1/( ) = 1/3000 for x: [2000,5000]
P(2500£X£3000) = ( )(1/3000) = .1667
1/3000 x
2000
2500
3000
5000

5. Uniform Distribution Example 1
The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are:
Between 2,500 and 3,500 gallons
More than 4,000 gallons
Exactly 2,500 gallons
f(x) = 1/( ) = 1/3000 for x: [2000,5000]
P(X³4000) = ( )(1/3000) = .333
1/3000 x
2000
4000
5000

6. Uniform Distribution Example 1
The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are:
Between 2,500 and 3,500 gallons
More than 4,000 gallons
Exactly 2,500 gallons
f(x) = 1/( ) = 1/3000 for x: [2000,5000]
P(X=2500) = ( )(1/3000) = 0
1/3000 x
2000
2500
5000

7. Normal Distribution This is the most popular continuous distribution.
Many distributions can be approximated by a normal distribution.
The normal distribution is the cornerstone distribution of statistical inference.

8. Normal Distribution
A random variable X with mean m and variance s2 is normally distributed if its probability density function is given by

9. The Shape of the Normal Distribution
The normal distribution is bell shaped, and symmetrical around m. 90 m
110
Why symmetrical? Let m = Suppose x = 110.
Now suppose x = 90

10. The Effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
How does the expected value affect the location of f(x)?
m = 10
m = 11
m = 12

11. Finding Normal Probabilities
Two facts help calculate normal probabilities:
The normal distribution is symmetrical.
Any normal distribution can be transformed into a specific normal distribution called…
“STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes?

12. STANDARD NORMAL DISTRIBUTION
NORMAL DISTRIBUTION WITH MEAN 0 AND VARIANCE 1.
IF X~N( , 2), THEN
NOTE: Z IS KNOWN AS Z SCORES.
“ ~ “ MEANS “DISTRIBUTED AS”

13. Finding Normal Probabilities
Solution
If X denotes the assembly time of a computer, we seek the probability P(45<X<60).
This probability can be calculated by creating a new normal variable the standard normal variable.
Every normal variable
with some m and s, can
be transformed into this Z.
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
E(Z) = m = 0
V(Z) = s2 = 1

14. Standard normal probabilities
Copied from Walck, C (2007) Handbook on Statistical Distributions for experimentalists

15. Standard normal table 1

16. Standard normal table 2

17. Standard normal table 3

18. Finding Normal Probabilities
Example - continued 45
- 50 X
- m 60
- 50
P(45<X<60) = P( < < ) 10 s 10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution

19. Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
P(0<Z<z0)
The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0
Z = 0
Z = z0

20. Finding Normal Probabilities
Example - continued
P(45<X<60) = P( < < ) 45 X 60
- m
- 50 s 10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = 1
z0 = -.5

21. Finding Normal Probabilities
Example - continued
P(45<X<60) = P( < < ) 45 X 60
- m
- 50 s 10
= P(-.5<Z<1) =
P(-.5<Z<0)+ P(0<Z<1)
P(0<Z<1
.3413
z=0
z0 = 1
z0 =-.5

22. Finding Normal Probabilities
The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows:
-z0
+z0
P(-z0<Z<0) = P(0<Z<z0)

23. Finding Normal Probabilities
Example - continued
.1915
.3413
-.5 .5

24. Finding Normal Probabilities
Example - continued
.1915
.1915
.1915
.1915
.3413
-.5 .5
1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = = .5328

25. Finding Normal Probabilities
Example - continued
This table provides probabilities from -∞ to z0
.3413
P(Z<-0.5)=1-P(Z>-0.5)= =0.3085
By Symmetry
P(Z<0.5)

26. Finding Normal Probabilities
Example
The rate of return (X) on an investment is normally distributed with a mean of 10% and standard deviation of (i) 5%, (ii) 10%.
What is the probability of losing money? X
10% 0%
0 - 10 5
(i) P(X< 0 ) = P(Z< ) = P(Z< - 2)
.4772 Z -2 2
=P(Z>2) =
0.5 - P(0<Z<2) = = .0228

27. Finding Normal Probabilities
Example
The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%.
What is the probability of losing money? X
10% 0% 1
0 - 10 10
(ii) P(X< 0 ) = P(Z< )
.3413 Z -1
= P(Z< - 1) = P(Z>1) =
0.5 - P(0<Z<1) = = .1587

28. AREAS UNDER THE STANDARD NORMAL DENSITY
P(0<Z<1)=.3413 Z 1

29. AREAS UNDER THE STANDARD NORMAL DENSITY
.3413
.4772
P(1<Z<2)= =.1359

30. EXAMPLES P( Z < 0.94 ) = 0.5 + P( 0 < Z < 0.94 )
= =
0.8264
0.94

31. EXAMPLES P( Z > 1.76 ) = 0.5 – P( 0 < Z < 1.76 )
= 0.5 – =
0.0392
1.76

32. EXAMPLES P( -1.56 < Z < 2.13 ) =
= P( < Z < 0 ) + P( 0 < Z < 2.13 )
= =
Because of symmetry
P(0 < Z < 1.56)
0.9240
-1.56
2.13

33. STANDARDIZATION FORMULA
If X~N( , 2), then the standardized value Z of any ‘X-score’ associated with calculating probabilities for the X distribution is:
The standardized value Z of any ‘X-score’ associated with calculating probabilities for the X distribution is:
(Converse Formula)

34. Finding Values of Z
Sometimes we need to find the value of Z for a given probability
We use the notation zA to express a Z value for which P(Z > zA) = A A zA

35. PERCENTILE
The pth percentile of a set of measurements is the value for which at most p% of the measurements are less than that value.
80th percentile means P( Z < a ) = 0.80
If Z ~ N(0,1) and A is any probability, then
P( Z > zA) = A A zA

36. Finding Values of Z Example Solution
Determine z exceeded by 5% of the population
Determine z such that 5% of the population is below
Solution
z.05 is defined as the z value for which the area on its right under the standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
Z0.05
1.645

37. Standardization formula
EXAMPLES
Let X be rate of return on a proposed investment. Mean is 0.30 and standard deviation is 0.1.
a) P(X>.55)=?
b) P(X<.22)=?
c) P(.25<X<.35)=?
d) 80th Percentile of X is?
e) 30th Percentile of X is?
Standardization formula
Converse Formula

38. ANSWERS a) b) c) d) e)
80th Percentile of X is
30th Percentile of X is

39. The Normal Approximation to the Binomial Distribution
The normal distribution provides a close approximation to the Binomial distribution when n (number of trials) is large and p (success probability) is close to 0.5.
The approximation is used only when
np  5 and
n(1-p)  5

40. The Normal Approximation to the Binomial Distribution
If the assumptions are satisfied, the Binomial random variable X can be approximated by normal distribution with mean  = np and 2 = np(1-p).
In probability calculations, the continuity correction improves the results. For example, if X is Binomial random variable, then
P(X  a) ≈ P(X<a+0.5)
P(X  a) ≈ P(X>a-0.5)

41. EXAMPLE Let X ~ Binomial(25,0.6), and want to find P(X ≤ 13).
Exact binomial calculation:
Normal approximation (w/o correction): Y~N(15,2.45²)
25*0.6=15; 25*0.6*0.4=2.45**2
Normal approximation is good, but not great!

42. EXAMPLE, cont. Bars – Bin(25,0.6); line – N(15, 2.45²)
Copied from Casella & Berger (1990)

43. Much better approximation to the exact value: 0.267
EXAMPLE, cont.
Normal approximation (w correction):
Y~N(15,2.45²)
Much better approximation to the exact value: 0.267

44. Exponential Distribution
The exponential distribution can be used to model
the length of time between telephone calls
the length of time between arrivals at a service station
the lifetime of electronic components.
When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.

45. Exponential Distribution
A random variable is exponentially distributed if its probability density function is given by f(x) = e-x, x>=0.
 is the distribution parameter >0).
E(X) =  V(X) = 2
 is a distribution parameter.
The cumulative distribution function is
F(x) =1e-x/, x0

46. Exponential distribution for l-1 = .5, 1, 2
f(x) = 2e-2x
f(x) = 1e-1x
f(x) = .5e-.5x
P(a<X<b) = e-a/ e-b/ a b

47. Exponential Distribution
Finding exponential probabilities is relatively easy:
P(X < a) = P(X ≤ a)=F(a)=1 – e –a/ 
P(X > a) = e–a/ 
P(a< X < b) = e – a/ – e – b/ 

48. Exponential Distribution
Example
The lifetime of an alkaline battery is exponentially distributed with mean 20 hours.
Find the following probabilities:
The battery will last between 10 and 15 hours.
The battery will last for more than 20 hours.

49. Exponential Distribution
Solution
The mean = standard deviation =  = 20 hours.
Let X denote the lifetime.
P(10<X<15) = e-.05(10) – e-.05(15) = .1341
P(X > 20) = e-.05(20) = .3679

50. Exponential Distribution
Example
The service rate at a supermarket checkout is 6 customers per hour.
If the service time is exponential, find the following probabilities:
A service is completed in 5 minutes,
A customer leaves the counter more than 10 minutes after arriving
A service is completed between 5 and 8 minutes.

51. Exponential Distribution
Solution
A service rate of 6 per hour = A service rate of .1 per minute (l-1 = .1/minute).
P(X < 5) = 1-e-.lx = 1 – e-.1(5) = .3935
P(X >10) = e-.lx = e-.1(10) = .3679
P(5 < X < 8) = e-.1(5) – e-.1(8) =

52. Exponential Distribution
If pdf of lifetime of fluorescent lamp is exponential with mean 0.10, find the life for 95% reliability?
The reliability function = R(t) = 1F(t) = e-t/
R(t) = 0.95  t = ?
R(t)=P(T>t)=exp(-t/0.1)=0.95 solve for t. t=

53. GAMMA DISTRIBUTION
X~ Gamma(,)
In graph, k=alpha, theta=beta

54. GAMMA DISTRIBUTION Gamma Function: where  is a positive integer.
Properties:

55. GAMMA DISTRIBUTION
since the last integral is the expectation of a Gamma distribution with parameters alpha and 1.

56. GAMMA DISTRIBUTION
Let X1,X2,…,Xn be independent rvs with Xi~Gamma(i, ). Then,
Let X be an rv with X~Gamma(, ). Then,
Let X1,X2,…,Xn be a random sample with Xi~Gamma(, ). Then,

57. GAMMA DISTRIBUTION Special cases: Suppose X~Gamma(α,β)
If α=1, then X~ Exponential(β)
If α=p/2, β=2, then X~ 2 (p) (will come back in a min.)
If Y=1/X, then Y ~ inverted gamma.
Gamma approximates to Normal distribution on the limit as alpha goes to infinity.

58. Integral tricks Recall: h(t) is an odd function if h(-t)=-h(t);
It is an even function if h(-t)=h(t).
Ex: h(x)=x exp{-x²/2} odd;
h(x)= exp{-x²/2-x} neither odd nor even
odd function =0
even function = 2* even function

59. CHI-SQUARE DISTRIBUTION
Chi-square with  degrees of freedom
X~ 2()= Gamma(/2,2)
In graphs, k=alpha.

60. DEGREES OF FREEDOM
In statistics, the phrase degrees of freedom is used to describe the number of values in the final calculation of a statistic that are free to vary.
The number of independent pieces of information that go into the estimate of a parameter is called the degrees of freedom (df) .
How many components need to be known before the vector is fully determined?

61. CHI-SQUARE DISTRIBUTION
Chi-square (2) ≡ Exponential (2)

62. CHI-SQUARE DISTRIBUTION
If rv X has Gamma(,) distribution, then Y=2X/ has Gamma(,2) distribution. If 2 is positive integer, then Y has
distribution.
Let X be an rv with X~N(0, 1). Then,
Let X1,X2,…,Xn be a r.s. with Xi~N(0,1). Then,

63. BETA DISTRIBUTION
The Beta family of distributions is a continuous family on (0,1) and often used to model proportions.
where

64. WEIBULL DISTRIBUTION
To model the failure time data or hazard functions.
Hazard function:
Rate of change in prob that subject survives a little
past time t, given that subject survives upto time t.
If X~Exp(), then Y=X1/ has Weibull(, ) distribution.

65. CAUCHY DISTRIBUTION
It is a symmetric and bell-shaped distribution on (,) with pdf
Since
, the mean does not exist.
The mgf does not exist.
 measures the center of the distribution and it is the median.
If X and Y have N(0,1) distribution, then Z=X/Y has a Cauchy distribution with =0 and σ=1.

66. CAUCHY DISTRIBUTION
When studying hypothesis tests that assume normality, seeing how the tests perform on data from a Cauchy distribution is a good indicator of how sensitive the tests are to heavy-tail departures from normality.
Likewise, it is a good check for robust techniques that are designed to work well under a wide variety of distributional assumptions.

67. LOG-NORMAL DISTRIBUTION
An rv X is said to have the lognormal distribution, with parameters µ and 2, if Y=ln(X) has the N(µ, 2) distribution.
The lognormal distribution is used to model continuous random quantities when the distribution is believed to be skewed, such as certain income and lifetime variables.

68. STUDENT’S T DISTRIBUTION
This distribution will arise in the study of population mean when the underlying distribution is normal.
Let Z be a standard normal rv and let U be a chi-square distributed rv independent of Z, with  degrees of freedom. Then,
When n, XN(0,1).

69. F DISTRIBUTION
Let U and V be independent rvs with chi-square distributions with 1 and 2 degrees of freedom. Then,

70. MULTIVARIATE DISTRIBUTIONS

71. BIVARIATE NORMAL DISTRIBUTION
A pair of continuous rvs X and Y is said to have a bivariate normal distribution if it has a joint pdf of the form

72. BIVARIATE NORMAL DISTRIBUTIONIf
, then
and  is the correlation coefficient btw X and Y.
1. Conditional on X=x,
2. Conditional on Y=y,

73. Mixture of distributions
Let f1(y) and f2(y) be density functions, and let a be a constant such that 0≤ a ≤1. Consider the function
f(y)=a f1(y) + (1-a) f2(y)
Such a density function is often called a mixture distribution. Here are some examples of real life applications for such distributions.

74. Mixture of distributions
Example 1: Financial returns often behave differently in normal situations and during crisis times. A mixture of two normal distributions with different means and variances can be assumed for returns, one for the returns during normal situations, and another for during crises.

75. Mixture of distributions
Example 2: The prices of houses in a particular neighborhood will tend to be similar, while prices in a different neighborhood may be extremely different. For instance, if we are to collect prices in both Cukurambar and Sincan, we will need a mixture of two distributions to describe these prices.

76. Mixture of distributions
Note that f(y) is a valid density function.
Suppose that Y1 is a random variable with density function f1(y), and E(Y1)=μ1 and Var(Y1)= . Similarly, suppose that Y2 is a random variable with density function f2(y), and E(Y2)=μ2 and Var(Y2)= Assume that Y is random variable whose density is a mixture of the densities corresponding to Y1 and Y2.
i) It can be shown that E(Y)= a μ1 + (1-a) μ2
ii) It can be shown that
Var(Y)= a (1-a) a(1-a) (μ1 - μ2)2

77. Problems
1. Mensa (from the Latin word for “mind”) is an international society devoted to intellectual pursuits. Any person who has an IQ in the upper 2% of the general population is eligible to join. If we assume that IQs are normally distributed with μ = 100 and σ = 16, what is the lowest IQ that will qualify a person for membership?

78. Problems
2. Suppose that numerical grades in a statistics class are values of a r.v. X which is Normally distributed with mean μ = 65 and standard deviation 15. Suppose that letter grades are assigned according to the following rule: student receives an A if X ≥ 85; B if 70 ≤ X < 85; C if 55 ≤ X < 70; D if 45 ≤ X < 55; and F if X ≤ 45. If a student is chosen at random from this class, calculate the probability that the student will earn i) A; ii) B; iii) F.

79. Problems
3. Economic conditions cause fluctuations in the prices of raw commodities as well as in finished products. Let X denote the price paid for a barrel of crude oil by the initial carrier, and let Y denote the price paid by the refinery purchasing the product from the carrier. Assume that the joint density for (X,Y) is given by f(x,y)=c 20< x < y < 40

80. Problems
a) Find the value of c that makes this a joint density for a two-dimensional random variable.
b) Find the probability that the carrier will pay at least $25 per barrel and the refinery will pay at most $30 per barrel for the oil.
c) Find the probability that the price paid by the refinery exceeds that of the carrier by at least $10 per barrel.

81. Problems
d) Are X and Y independent? Explain. e) Find E(Y-X). Interpret this expectation in a practical sense.