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EXAMPLE 3.1 Suppose that a game is to be played with a single die assumed fair. In this game a player wins $20 if a 2 turns up, $40 if a 4 turns up; loses $30 if a 6 turns up; while the player neither wins nor loses if any other face turns up. Find the expected sum of money to be won. Let X be the random variable giving the amount of money won on any toss. The possible amounts won when the die turns up 1, 2, c, 6 are x1, x2, c, x6, respectively, while the probabilities of these are f(x1), f (x2), . . . , f (x6). The probability function for X is displayed in Table 3-1. Therefore, the expected value or expectation is

E(X)

Fig. 9-31

1 (0) 6

Use the node voltage method to obtain the current I in the network of Fig. 9-31.

1 (20) 6

"part 3"

1 (0) 6

There are three principal nodes in the network. The reference and node 1 are selected so that the node 1 voltage is the voltage across the j2- reactance. 21 32 3 2 3 1 1 1 V1 50 08 6 5 j2 4 76 7 6 7 4 5 6 76 7 6 7 4 1 1 1 1 54 5 4 50 908 5 V2 4 4 j2 2 2 from which

1 (40) 6

SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS 0:250 13:52 56:318 0:750 j0:500 24:76 V1 0:450 j0:500 0:546 15:948 0:250 0:250 0:750 j0:500 and I 24:76 72:258 12:38 2 908 17:758 A 10 j25

1 (0) 6

1 ( 30) 6

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Find the input impedance at terminals ab for the network of Fig. 9-32.

It follows that the player can expect to win $5. In a fair game, therefore, the player should be expected to pay $5 in order to play the game.

"part 2"

Fig. 9-32 With mesh current I1 selected as shown on the diagram. 8 j2 3 0 3 8 j5 5 0 5 7 j2 315:5 16:198 6:98 Zinput;1 z 8 j5 5 11 45:2 24:868 5 7 j2

E(X)

x3 2 2 6 0

For the network in Fig. 9-32, obtain the current in the inductor, Ix , by rst obtaining the transfer impedance. Let V 10 308 V.

Let X be a discrete random variable with probability function f (x). Then Y able, and the probability function of Y is h(y) P(Y y) a P(X x) a

Ztransfer;12 z 315:5 16:198 14:45 3 5 12 0 7 j2 308 0:692 32:148 32:148

g(X) is also a discrete random vari-

If X takes on the values x1, x2, c , xn, and Y the values y1, y2, c , ym (m ymh(ym ) g(x1)f (x1) g(x2) f (x2) c g(xn)f(xn ). Therefore, E[g(X)] g(x1) f (x1)

V 10 Ztransfer;12 14:45

"exiting"

g(x2)f(x2)

For the network in Fig. 9-32, nd the value of the source voltage V which results in V0 5:0 08 V.

g(xn)f(xn ) (5)

a g(xj) f(xj)

The transfer impedance can be used to compute the current in the 2 j2 impedance, from which V0 is readily obtained. Ztransfer;13 z 315:5 16:198 21:0 13 15 08 16:198

a g(x)f(x)

V 2 j2 V 0:135 Ztransfer;13

Similarly, if X is a continuous random variable having probability density f(x), then it can be shown that

Fig. 5-8.

E[g(X)]

61:198

Note that (5) and (6) do not involve, respectively, the probability function and the probability density function of Y g(X). Generalizations are easily made to functions of two or more random variables. For example, if X and Y are two continuous random variables having joint density function f(x, y), then the expectation of g(X, Y) is given by

CHAP. 9]

E[g(X, Y)]

EXAMPLE 3.3 If X is the random variable of Example 3.2,

Thus, if V0 5:0

10 3

Alternate Method The node voltage method may be used. V0 is the node voltage V2 for the selection of nodes indicated in Fig. 9-32. 1 1 1 V 5 j2 3 j5 5 j2 1 0 j5 V 0:134 61:158 V0 V2 1 1 1 1 5 j2 3 j5 j5 1 1 1 1 j5 j5 5 2 j2 For V0 5:0 08 V, V 37:3 61:158 V, which agrees with the previous answer to within roundo errors.

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