Chapter 3
EXPECTED VALUE OF A DISCRETE RANDOM VARIABLE
3.1 Expectation
One of the most expected notions of probability theory is that of the expectation of a
random variable. If X is a discrete random variable taking values x1, x2, …, then the
expectation or the expected value of X, denoted by E[X], is defined by
E[X]=  xi P( X  xi ) .
i
In words, the expected value of a random variable is a weighted sum of the values it
takes, weighted by their likelihood of occurrence. For example, if X is taking the
values 0 and 1 with probability ½, then,
E[X]=0 (1/2)+1 (1/2)=1/2,
which is the average of the values that X can take. Now, suppose X takes the value 1
with probability 2/3 and 0 with probability 1/3. In this case,
E[X]=0 (1/3)+ 1 (2/3)= 2/3,
is the weighted average of possible values where the value 1 is given twice as much
weight as the value 0.
Another motivation for the definition of the expected value is the frequency
interpretation. Suppose the random experiment of interest is repeated N times where
N is very large. Then, N P(X=xi) of time will result in the outcome xi, thus, the
average of all the possible values after N repeatings is
 NP( X  x ) x
i
i
i
,
N
which coincides with the definition of E[X].
Example: Find E[X] where X is the outcome when we roll a fair die.
Since p(i)=1/6, for all i=1, 2, ..., 6, we obtain
E[X]=1(1/6)+2(1/6)+3(1/6)+4(1/6)+5(1/6)+6(1/6)=7/2.
Example: Consider the Bernoulli random variable taking values 0 in case of failure
and 1 in case of success. If the probability of success is given by p, then the expected
value of a Bernoulli random variable is equal to 0 (1-p) + 1 p=p.
3.2 Properties of the expected value
1. If a and b are constants and X is a random variable, then
E[a X +b]= a E[X] + b
2. If X and Y are two random variables then
E[X+Y]=E[X]+E[Y]
3. In general, if X1, X2, ..., Xn are n random variables then
E[X1+X2+...+Xn]=E[X1]+E[X2]+... E[Xn].
4. If g is a function, then
E[g(X)]=
 g ( x ) P( X  x ) .
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i
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5. In particular, if g(x)=x^n, the n-th moment of X is given by
E[Xn]=  xin P( X  xi ) .
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Example: Consider a Binomial random variable consisting of n Bernoulli trials.
Then,
X=X1+X2+...+Xn,
where each Xi is a Bernoulli trial taking values 0 or 1. Then using Property 3,
E[X]= E[X1]+E[X2]+... E[Xn]=n p.
3.3 Variance
Expected value, E[X], of a random variable X is only the weighted average of the
possible values of X, so X also takes values around E[X]. One possible way of
measuring the variation of X is through its variance. Variance measures the deviation
of the random variable from its expected value ( or mean). If X is a random variable
with mean  , then the variance of X, denoted by Var(X), is defined by
Var(X)=E[(X-  )2].
Alternatively,
Var(X) =E[(X-  )2]
=E[X2-2  X  2]
=E[X2]-2  E[X] + E[  2]
=E[X2]-  2.
Example: Compute the Var(X) when X represents the outcome of a fair die.
Since P(X=i)=1/6,
6
E[X2]=  i 2 P( X  i ) =91/6.
i 1
We had already computed  =E[X]=7/2. Thus, Var(X)=91/6-(7/2)2=35/12.
3.3.1 Covariance
We showed earlier that the expectation of a sum is the sum of the expectation. The
same does not hold for the variance of a sum. In order to find the variance of a sum,
we first need to introduce the concept of covariance.
Given two random variables, X and Y, the Cov(X,Y) is defined by
Cov(X,Y)=E[(X-  x)(Y-  y)]
where  x and  y are the means of X and Y, respectively. Alternatively,
Cov(X,Y)=E[XY]-E[X]E[Y].
Thus, Cov(aX,Y)=aCov(X,Y). and Cov(X,X)=Var(X). Moreover,
Cov(X+Z,Y)=Cov(X,Y)+Cov(Z,Y).
Indepence: If X and Y are independent, Cov(X,Y)=0 (E[XY]=E[X]E[Y]).
Formula for the variance of a sum: Var(X+Y)=Var(X)+Var(Y)+2 Cov(X,Y)
Example: Let X be a binomial random variable consisting of n independent Bernoulli
trials: Then
n
n
Var(X)= Var ( X i )  2 Cov( X i , X j )  Var ( X i )
i 1
i j
i 1
since Xi and Xj are independent when i  j. Now, Var(Xi)=E[(Xi)2]-(E[Xi])2=12 pp=p(1-p). So, Var(X)= np(1-p).