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For this sequence, the first few steps of the Berlekamp-Massey Algorithm are illustrated in Table 3.2. For the periodic sequence ( 3 . 2 ) , the linear complexity is L = 6 (Problem 1 asks for the connection polynomial). Therefore, if we let 10011 be the initial fill of the LFSR corresponding to the connection polynomial determined by the Berlekamp-Massey Algorithm, the LFSR generates the sequencc s in (3.2). Here, we do not attempt to prove the validity of the Berlekarnp-Massey Algorithm. but we note in passing that the algorithm is closely related to the extended Euclidean Algorithm and continued fraction algorithms. We also note one important- but non-obvious--fact, namely, that any 2L consecutive bits can be used to completely determine a sequence that has linear complexity L. That is, after processing 2 L bits through the Berlekamp-

t2 46 12 22

11 223x

t 2242

Table 3.2: Berlekamp--Massey Example sequence: s = (SO, s1,. . , s7) = 10011100 . initialize: C ( 5 )= B ( z )= 1, L = N = 0, m = -1

5-11

1 e 12

N=O d=so=l T ( 5 )= 1, C ( 5 )= 1 + z L = 1, m = 0, B ( z )= 1 N=l d = SI CB cis0 = 1 T ( z )= 1 5 , C ( 2 )= 1

u2 2

Now the integral is just the total area under a standard normal density, which is 1, so the moment generating function of a normal random variable is MX 1t2 dMX 1t2 ` dt t e

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Moment generating functions have many important and useful properties. One of the most important of these is the uniqueness property. That is, the moment generating function of a random variable is unique when it exists, so if we have two random variables X and Y, say, with moment generating functions MX (t) and MY (t), then if MX (t) MY (t) for all values of t, both X and Y have the same probability distribution. Some of the other useful properties of the moment generating function are summarized as follows.

=0 0 ~

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If X1, X2, p , Xn are independent random variables with moment generating functions MX1(t), MX2 (t), . . . , MXn(t), respectively, and if Y X1 X2 p Xn, then the moment generating function of Y is (3) MY 1t2 MX1 1t2 MX2 1t2 p MX 1t2 n (S5-10)

Property (1) follows from MX a 1t2 E3et 1X a2 4 eatE1etX 2 eatMX 1t2 . Property (2) t 1aX2 1at2X follows from MaX 1t2 E3e 4 E3e 4 MX 1at2 . Consider property (3) for the case where the X s are continuous random variables: MY 1t2 E1etY 2 p E3et 1X1 et 1x1

N=3 d = s3 CB c i s 2 C ~ 2 CB~CQSO = 1 1 T ( 5 )= 1, C ( 5 )= 1 + 5 3 L = 3 , m = 3, B ( z )= 1 N=4

5-12

Since the X s are independent, f 1x1, x2, p , xn 2 and one may write MY 1t2 etx1 fX1 1x1 2 dx1 MX1 1t2 MX2 1t2 etx2 fX2 1x2 2 dx2 p p MX 1t2 n etxn fXn 1xn 2 dxn fX1 1x1 2 fX2 1x2 2 p fX 1xn 2 n

Massey Algorithm, the minimal LFSR will have been obtained. Below, we see that this property has implications for stream cipher design. It is not too difficult to show that the Berlekamp-Massey Algorithm requires on the order of n2 operations [62], where n is the number of bits processed and the operations are XOR. This is the most efficient known general algorithm for solving the shift register synthesis problem. However, there are more efficient algorithms for certain special cases; see Problem 4 for an example of such an algorithm.

For the case when the X s are discrete, we would use the same approach replacing integrals with summations. Equation S5-10 is particularly useful. In many situations we need to nd the distribution of the sum of two or more independent random variables, and often this result makes the problem very easy. This is illustrated in the following example. EXAMPLE S5-7 Suppose that X1 and X2 are two independent Poisson random variables with parameters 1 and X1 X2. 2, respectively. Find the probability distribution of Y The moment generating function of a Poisson random variable with parameter is MX 1t2 e

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