Expectation of a Function of a Random Variable

Expectation of a Function of a Random Variable 

As we have discussed previously, if X is a random variable and if Y = H(X) is a function of X, then Y is also a random variable with some probability distribution. Hence it will be of interest and meaningful to evaluate E( Y). There are two ways of evaluating E( Y) which turn out to be equivalent. To show that they are in general equivalent is not trivial, and we shall only prove a special case. However, it is important that the reader understands the two approaches discussed below.

Definition. 

Let X be a random variable and let Y = H(X).
(a) If Y is a discrete random variable with possible values y₁ , y₂, ••• and if q(y_i) = P( Y = y_i), we define

                              E( Y) = ∑_lim i→∞  y_iq(y_i).                                        (i)
(b) If Y is a continuous random variable with pdf g, we define

                              E( Y) = ∫_{lim -∞→∞}yg(y)dy                                       (ii)

Note: 

Of course, these definitions are completely consistent with the previous definition given for the expected value of a random variable. In fact, the above simply represents a restatement in terms of Y. One "disadvantage" of applying the above definition in order to obtain E(Y) is that the probability distribution of Y (that is, the probability distribution over the range space Ry) is required. We discussed, in the previous chapter, methods by which we may obtain either the point probabilities q(y;) or g, the pdf of Y. However, the question arises as to whether we can obtain E(Y) without first finding the probability distribution of Y, simply from the knowledge of the probability distribution of X. The answer is in the affirmative as the following theorem indicates.

Theorem 1. 

Let X be a random variable and let Y = H(X).

(a) If X is a discrete random variable and p(x_i) = P(X = x_i), we have

                E( Y) = E(H(X)) = ∑_{lim j=1→∞}H(x_j)p(x_j).                                (iii)
(b) If X is a continuous random variable with pdf f, we have
                 
                      E( Y) = E(H(X)) = ∫_{lim -∞→∞} H(x)f(x) dx.                         (iv)

Note: 

This theorem makes the evaluation of E(Y) much simpler, for it means that we need not find the probability distribution of Y in order to evaluate E( Y). The knowledge of the probability distribution of X suffices.

Proof· 

[We shall only prove 'Eq. (iii). The proof of Eq. (iv) is somewhat more intricate.] Consider the sum  ∑_{lim j=1→∞} H(x_i)p(x_i) =  ∑_{lim j=1→∞}( ∑_i H(x_i)p(x_i)), where the inner sum is taken over all indices i for which H(x_i) = y_j for some fixed y_j· Hence all the terms H(x_i) are constant in the inner sum. Hence                       
                                             ∑_{lim j=1→∞} H(x_i)p(x_i) =  ∑_{lim j=1→∞} p(x_i). 
 However,
                                             ∑_{lim j=1→∞} p(x_i) =  ∑_{lim j=1→∞} P[x | H(x_i) = Yi] = q(y_j). 
              Therefore,              ∑_{lim j=1→∞}H(x_i)p(x_i) =  ∑_{lim j=1→∞}y_jq(y_i), which establishes Eq. (iii). 

Note: 

The method of proof is essentially equivalent to the method of counting in which we put together all items having the same value. Thus, if we want to find the total sum of the values 1, 1, 2, 3, 5, 3, 2, 1, 2, 2, 3, we could either add directly or point out that since there are 3 one's, 4 two's, 2 three's, and 1 five, the total sum equals 3(1) + 4(2) + 3(3) + 1 (5) = 25.

 EXAMPLE 1

 Let V be the wind velocity (mph) and suppose that V is uniformly distributed over the interval [O, 10]. The pressure, say W (in lb/ft²), on the surface of an airplane wing is given by the relationship: W = 0.003V²• To find the expected value of W, E( W), we can proceed in two ways:

(a) Using Theorem (1), we have
                 
                              E( W) = ∫_lim 0→100.003v²f(v) dv 
                               =∫_lim 0→100.003v²f(v) dv  
                               = 0.l lb/ft²

(b) Using the definition of E( W), we first need to find the pdf of W, say g, and then evaluate e: wg(w) dw. To find g(w), we note that w = 0.003V² is a monotone function of v, for v � 0. We may apply Theorem 5.1 and obtain

                      g(w) =1/10|dv/dw|
                     = 1/2√(10/3)w^-1/2                      ,  0≤w≤0.3
                     = 0, elsewhere.
Hence
                E(W) = ∫_lim 0→0.3wg(w) dw = 0.1

after a simple computation. Thus, as the theorem stated, the two evaluations of E( W) yield the same result.

EXAMPLE 2 

In many problems we are interested only in the magnitude of a random variable without regard to its algebraic sign. That is, we are concerned with |X|. Suppose that X is a continuous random variable with the following pdf
         
            f(x) = e^x//2                if x ≤0, 
                    =e^-x//2               if x >0

Let Y = |X|. To obtain E( Y) we may proceed in one of two ways.
(a) Using Theorem 1 we have

              E( Y) =∫_{lim -∞→∞}|x|f(x) dx 
                        =1/2[∫_{lim -∞→0}(-x)e^xdx + ∫_lim 0→∞(x)e^-x dx ] = l.
                        =1/2[1+1]=1

(b) To evaluate E( Y) using the definition, we need to obtain the pdf of Y = |X|, say g. Let G be the cdf of Y. Hence

            G(y) = P(Y ≤ y) = P[|X| ≤ y] = P[-y≤ X ≤ y] = 2P(0 ≤ X ≤ y), 

since the pdf of X is symmetric about zero. Therefore

          G(y) = 2 ∫_lim 0→yf(x) dx = 2∫_lim 0→ye^-x/2dx = -e^-y+ l. 

Thus we have for g, the pdf of Y,
g(y) = G'(y) = e^-y, y ≥ 0. Hence E(Y) = ∫_lim 0→∞yg(y) dy = ∫_lim 0→∞ ye^-y dy = 1, as above. 

EXAMPLE 3

In many problems we can use the expected value of a random variable in order to make a certain decision in an optimum way. Suppose that a manufacturer produces a certain type of lubricating oil which loses some of its special attributes if it is not used within a certain period of time. Let X be the number of units of oil ordered from the manufacturer during each year. (One unit equals 1000 gallons.) Suppose that X is a continuous random variable, uniformly distributed over [2, 4 ]. Hence the pdf f has the form,

                        f(x) = 1/2,        , 2 ≤ x ≤ 4, 
                               =0               elsewhere

Suppose that for each unit sold a profit of $300 is earned, while for·each unit not sold (during any specified year) a loss of $100 is taken, since a unit not used will have to be discarded. Assume that the manufacturer must decide a few months prior to the beginning of each year how much he will produce, and that he decides to manufacture Y units. ( Y is not a random variable; it is specified by the manufacturer.) Let Z be the profit per year (in dollars). Here Z is clearly a random variable since it is a function of the random variable X. Specifically, Z = H(X) where

                 H(X) = 300Y;                      if X ≥ Y, 
              = 300X + (-100)(Y - X)          if X < Y. 

(The last expression may be written as 400X - 100 Y.)
In order for us to obtain E(Z) we apply Theorem 1 and write

E(Z) = ∫_{lim -∞→∞} H(x)f(x) dx
         = 1/2∫_lim2→4 H(x) dx. 

To evaluate this integral we must consider three cases: Y < 2, 2 ≤. Y ≤ 4, E(Z) and Y > 4. With the aid of  and after some simplification we obtain

              E(Z) = 300Y if Y ≤ 2 
                        =-100Y² + 700 Y - 400                  if 2 < y < 4 
                        =1200 - 100 y                                  if y ≤ 4

The following question is of interest. How should the manufacturer choose the Y=2 Y=3.5 Y=4 value of Y in order to maximize his expected profit? We can answer this question easily by simply setting dE(Z)/dY = 0. This yields Y = 3.5.