Download Chapter 2 : Discrete random variables 1 General properties

L2-PC-C
Probabilités - Statistiques
2011/2012
Chapter 2 : Discrete random variables
1
General properties
1.1 Denition
Let (Ω, A, P) be a probability space associated to an experiment. We call realvalued random
variable a map X from Ω to R. X is said a discrete random variable if X(Ω) is a nite or countable
set, where X(Ω) is the set of the values assumed by the map X .
Example : We throw a dice twice and we call X the maximum number obtained. We have
Ω{(i, j), i, j ∈ [1, 2, . . . 6]} and X(Ω) = {1, 2, 3, 4, 5, 6}
Random variables can be added and multiplied.
1.2 Law of a random variable
The law of a random variable is the knowledge of X(Ω) and ∀a ∈ X(Ω) the probability P (X = a)
associated to the possible value a.
1
3
In the previous example, we have P (X = 1) = 36
, P (X = 2) = 36
and so. We note that
X
P (X = a) = 1
a∈X(Ω)
1.3 Independence
Let X and Y be two given random variables. X and Y are said independent if ∀(a, b) ∈ X(Ω)×Y (Ω),
the events (X = a) and (Y = b) are independent, or equivalently if P (X = a et Y = b) = P (X =
a) × P (Y = b)
1.4 Distribution function
The distribution function of the random variable X is dened as
FX : R → [0, 1]
x 7→ FX (x) = P (X < x)
Example : Uniform distribution
Properties
1. The distribution function of a discrete probability law is an increasing, nonegative, simple function, left-continuous at each point. Discontinuities correspond to the values taken by the random
variable where the probability is non-null.
2. limx→−∞ FX (x) = 0, et limx→+∞ FX (x) = 1
3. If a < b, FX (b) − FX (a) = P (a ≤ X < b).
4. If x is a value assumed by the random variable X , then
P (X = x) = lim FX (x + ) − FX (x)
→0
1.5 Expectation - Variance - Standard deviation - Covariance
Denition
:
P
E(X) =
index.
The expectation (or expected value) of the random variable X is the quantity :
x∈X(Ω) x · P (X = x). It represents the mean value of the random variable. It is a centrality
Properties
1. Expectation is linear. Let X and Y be two random variables. It holds
E(X + Y ) = E(X) + E(Y ) and ∀λ ∈ R, E(λX) = λE(X)
2. If X is a random variable such that X ≥ 0, then E(X) ≥ 0. If X and Y are random variables
such that Y ≥ X , then E(Y ) ≥ E(X).
3. If X = a, then E(X) = a.
4. Let f : R → R. If X is a random variable, then Y = f ◦ X also is a random variable and it holds
X
E(Y ) =
f (x) · P (X = x).
x∈X(Ω)
For instance this can be used to compute E(X 2 ).
5. If X and Y are independent, then E(X · Y ) = E(X) · E(Y ). The reciprocal is false.
Denition :
The variance of a randompvariable X is given by V (X) = E[(X − E(X))2 ]. The
standard deviation is the quantity σ(X) = V (X). The standard deviation is the average deviation
at the expected value. It is a dispersion index.
Properties
1. V (X) ≥ 0 and if V (X) = 0, then X = E(X).
2. If a is a constant, then V (X + a) = V (X)
3. V (aX) = a2 V (X).
4. If X and Y are independent, then V (X + Y ) = V (X) + V (Y ).
5. V (X) = E(X 2 ) − (E(X))2 .
Denition : The covariance is dened as cov(X, Y ) = E[(X − E(X))(Y
properties hold :
− E(Y ))]. The following
1. cov(X, Y ) = E(XY ) − E(X)E(Y )
2. V (X + Y ) = V (X) + V (Y ) + 2cov(X, Y )
3. If X and Y are independent, then cov(X, Y ) = 0.
2
Some example of distributions
2.1 Bernouilli distribution with parameter p ∈]0, 1[
We want to describe a phenomenon for which two possible states are possible (Ω = {S, F }). The
rst state is called "success". It occurs with probability p. The second state is called "failure" and
it occurs with probability 1 − p. Then we introduce the following random variable : X = 1 in case
of a success and X = 0 in case od failure, i.e. X(Ω) = {0, 1}. Then we have P (X = 1) = p and
P (X = 0) = 1 − p. We say that X follows a Bernoulli distribution with parameter p, denoted by B(p).
It holds :
E(X) = p, V (X) = p(1 − p)
2.2 Binomial distribution with parameter (n, p)
We repeat n times an experiment in which, each time, two results are possible : the success (with
probability p) and the failure (with probability 1 − p). The random variable X counts thenumber
of
n
successes. Then, we have Ω = {S, F }n and X(Ω) = {0, 1, . . . , n}. Moreover P (X = k) =
pk (1 −
k
p)n−k . We say that X follows the binomial distribution with parameters (n, p), denoted B(n, p). We
observe that a binomial distribution B(n, p) is the sum of n Bernouilli distributions B(p). We have
E(X) = np,
V (X) = np(1 − p)
Example : 5 balanced coins are tossed. The outcomes are supposed to be independent. Give the
probability distribution of the random variable which gives the number of "heads" obtained.
2.3 Hypergeometric distribution
We consider an experiment with N possible events, among which there are N1 successes and N2
failures. We choose

n,
 n≤N
 1 elements and the random variable X counts the number of successes. Then
N
N
 1 2

k
n−k
 
P (X = k) =
. We say that X follows the hypergeometric distribution with paramenter
N
 
n
2
1
and q = 1 − p = N1N+N
. Then it holds
(n, N1 , N2 ) denoted H(n, N1 , N2 ). Set p = N1N+N
2
2
E(X) = np,
V (X) = npq
N − n
N −1
Example : An urn contains 15 marbles, among which 10 are green. We draw randomly 8 marbles
from the urn without replacement. Give the probability distribution of the random variable which gives
the number of green marbles among the 8 drawn marbles.
2.4 Pascal distribution
We consider the following model : we do some samplings with replacement from an urn containing
two types of marbles : white marbles with rate p (0 < p < 1) and black marbles with rate q = 1 − p.
Let X be thenumber
of samplings needed to obtain r white marbles : X(Ω) = {r, r + 1, . . . , k, . . .}.
k − 1 r k−r
P (X = k) =
p q . Then we have E(X) = pr and V (X) = prq2 .
r−1
Remark : If r = 1, the Pascal distribution is called geometric distribution. It corresponds to
sampling with replacement untill the rst white marble is drawn. The Pascal distribution is also called
negative binomial distribution.
2.5 Poisson distribution with paramenter λ
This is a probability distribution which permits, for instance, to model the number of phone call
per unit time or the number of persons in a queue per unit time. We say that the random variable X
follows a Poisson distribution with parameter λ, denoted P(λ), if
X(Ω) = N,
P (X = k) =
λk −λ
e
k!
It can be checked that this is a well-dened probability and it holds :
E(X) = λ,
V (X) = λ
3
Appendix
3.1 Relations to be known
We will use the following relations
1.
n
X
k=
k=1
2.
n
X
k2 =
k=1
3.
n
X
n(n + 1)(2n + 1)
6
k3 =
k=1
n(n + 1)
2
n2 (n + 1)2
4
3.2 Series used in probability theory
If |p| < 1, it holds
1.
+∞
X
pk =
k=0
2.
+∞
X
k=1
3.
+∞
X
k=2
kpk−1 =
1
1−p
1
(1 − p)2
k(k − 1)pk−2 =
2
(1 − p)3