1. The Bernoulli distribution
Discrete distributions
The Bernoulli distribution

2. The Binomial distribution
p(x) x
X = the number of successes in n repetitions of a Bernoulli trial p = the probability of success

3. The Poisson distribution
Events are occurring randomly and uniformly in time.
X = the number of events occuring in a fixed period of time.

4. P[X = x] = p(x) = p(1 – p)x – 1 = pqx – 1
The Geometric distribution
the Bernoulli trials are repeated independently the first success occurs (,k = 1) and X = the trial on which the 1st success occurred.
P[X = x] = p(x) = p(1 – p)x – 1 = pqx – 1
The Negative Binomial distribution
the Bernoulli trials are repeated independently until a fixed number, k, of successes has occurred and X = the trial on which the kth success occurred.
Geometric ≡ Negative Binomial with k = 1

5. The Hypergeometric distribution
Suppose we have a population containing N objects.
The population are partitioned into two groups.
a = the number of elements in group A
b = the number of elements in the other group (group B).
Note N = a + b.
n elements are selected from the population at random.
X = the elements from group A. (n – X will be the number of elements from group B.)

6. Example: Hyper-geometric distribution
Suppose that N = 10 automobiles have just come off the production line. Also assume that a = 3 are defective (have serious defects). Thus b = 7 are defect-free.
A sample of n = 4 are selected and tested to see if they are defective. Let X = the number in the sample that are defective. Find the probability function of X.
From the above discussion X will have a hyper-geometric distribution i.e.

7. Table and Graph of p(x)

8. Sampling with and without replacement
Suppose we have a population containing N objects.
Suppose the elements of the population are partitioned into two groups. Let a = the number of elements in group A and let b = the number of elements in the other group (group B). Note N = a + b.
Now suppose that n elements are selected from the population at random. Let X denote the elements from group A. (n – X will be the number of elements from group B.)
Find the probability distribution of X.
If the sampling was done with replacement.
If the sampling was done without replacement

9. Solution: If the sampling was done with replacement.
Then the distribution of X is the Binomial distn.

10. If the sampling was done without replacement.
Then the distribution of X is the hyper-geometric distn.

11. Note:

12. for large values of N, a and b
Thus
for large values of N, a and b
Thus for large values of N, a and b sampling with replacement is equivalent to sampling without replacement.

13. Continuous Distributions

14. Continuous random variables
For a continuous random variable X the probability distribution is described by the probability density function f(x), which has the following properties :
f(x) ≥ 0

15. Graph: Continuous Random Variable probability density function, f(x)

16. The Uniform distribution from a to b

17. Definition: A random variable , X, is said to have a Uniform distribution from a to b if X is a continuous random variable with probability density function f(x):

18. Graph: the Uniform Distribution (from a to b)

19. The Cumulative Distribution function, F(x) (Uniform Distribution from a to b)

20. Cumulative Distribution function, F(x)

21. The Normal distribution

22. Definition: A random variable , X, is said to have a Normal distribution with mean m and standard deviation s if X is a continuous random variable with probability density function f(x):

23. Graph: the Normal Distribution (mean m, standard deviation s)

24. Note:
Thus the point m is an extremum point of f(x). (In this case a maximum)

25. Thus the points m – s, m + s are points of inflection of f(x)

26. Also
Proof:
To evaluate
Make the substitution

27. Consider evaluating
Note:
Make the change to polar coordinates (R, q)
z = R sin(q) and u = R cos(q)

28. Hence
and or
and
Using

29. and or

30. The Exponential distribution

31. Consider a continuous random variable, X with the following properties:
P[X ≥ 0] = 1, and
P[X ≥ a + b] = P[X ≥ a] P[X ≥ b] for all a > 0, b > 0.
These two properties are reasonable to assume if X = lifetime of an object that doesn’t age.
The second property implies:

32. The property:
models the non-aging property
i.e. Given the object has lived to age a, the probability that is lives a further b units is the same as if it was born at age a.

33. Let F(x) = P[X ≤ x] and G(x) = P[X ≥ x] .
Since X is a continuous RV then G(x) = 1 – F(x)
(P[X ≥ x] = 0 for all x.)
The two properties can be written in terms of G(x):
G(0) = 1, and
G(a + b) = G(a) G(b) for all a > 0, b > 0.
We can show that the only continuous function, G(x), that satisfies 1. and 2. is a exponential function

34. From property 2 we can conclude
Using induction
Hence putting a = 1.
Also putting a = 1/n.
Finally putting a = 1/m.

35. Since G(x) is continuous
for all x ≥ 0.
If G(1) = 0 then G(x) = 0 for all x > 0 and G(0) = 0 if G is continuous. (a contradiction)
If G(1) = 1 then G(x) = 1 for all x > 0 and G() = 1 if G is continuous. (a contradiction)
Thus G(1) ≠ 0, 1 and 0 < G(1) < 1
Let l = - ln(G(1)) then G(1) = e-l

36. To find the density of X we use:
A continuous random variable with this density function is said to have the exponential distribution with parameter l.

37. Graphs of f(x) and F(x)
f(x)
F(x)

38. Another derivation of the Exponential distribution
Consider a continuous random variable, X with the following properties:
P[X ≥ 0] = 1, and
P[x ≤ X ≤ x + dx|X ≥ x] = ldx
for all x > 0 and small dx.
These two properties are reasonable to assume if X = lifetime of an object that doesn’t age.
The second property implies that if the object has lived up to time x, the chance that it dies in the small interval x to x + dx depends only on the length of that interval, dx, and not on its age x.

39. Determination of the distribution of X
Let F (x ) = P[X ≤ x] = the cumulative distribution function of the random variable, X .
Then P[X ≥ 0] = 1 implies that F(0) = 0.
Also P[x ≤ X ≤ x + dx|X ≥ x] = ldx implies

40. We can now solve the differential equation
for the unknown F.

41. Now using the fact that F(0) = 0.
This shows that X has an exponential distribution with parameter l.