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Suppose that X1 and X2 are continuous random variables with joint probability distribution fX1 X2 1x1, x2 2, and let Y1 h1(X1, X2) and Y2 h2(X1, X2) de ne a one-to-one transformation between the points (x1, x2) and (y1, y2). Let the equations y1 h1(x1, x2) and y2 h2(x1, x2) be uniquely solved for x1 and x2 in terms of y1 and y2 as x1 u1(y1, y2) and x2 u2(y1, y2). Then the joint probability of Y1 and Y2 is fY1Y2 1 y1, y2 2 fX1X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 0 J 0 (S5-4)

If the key can be securely distributed, why not send the message by the same means, in which case there is no need to encrypt However, it is important to note that there are some cases where a onetime pad is practical In some situations it may be easy to send the key at a particular time, and then use it at a later time when it would be difficult or impossible to communicate securely For example, in the 1930s and 1940s, the Soviet Union used a one-time pad cipher to t,ransmit intelligence gathered from spies in the United States Soviet agents would simply bring their onetime pads with them when entering the United States, then use these to encrypt sensitive messages as necessary In fact, these one-time pads were often used more than once and, as a result, many of the messages were eventually broken by United States cryptanalysts.

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This result can be used to nd fY1Y2 1 y1, y2 2, the joint probability distribution of Y1 and Y2. Then the probability distribution of Y1 is fY1 1 y1 2 fY1Y2 1 y1, y2 2 dy2

That is, fY1 ( y1) is the marginal probability distribution of Y1. EXAMPLE S5-4 Suppose that X1 and X2 are independent exponential random variables with fX1 1x1 2 and fX2 1x2 2 2e 2x2. Find the probability distribution of Y X1 X2 . 2e

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because X1 and X2 are independent. Let Y1 h1(X1, X2) X1 X2. The inverse solutions of y1 x1 x2 and y2 x1 x2 y2 11 y1 2 , and it follows that x1 y1 x2 y1 Therefore y2 11 y1 2 2 y2 11 y1 2 2 11 11 y1 1 y1 2 y1 2 y2 c y2 c 11 11 1 y1 2 2 d, x1 y2 x2 y2 c c 11 11

The famous Project VENONA [151] details this impressive cryptanalytic success The VENONA decrypts provide tremendous insight into Soviet spying in general, and nuclear espionage in particular Modern stream ciphers are a generalization of the one-time pad, where provable security is traded for practicality In a stream cipher a short secret key is stretched into a long pseudo-random string of bits, which is then used just like a one-time pad The provable security is lost since the number of possible keys is much smaller than the number of possible messages Stream ciphers are discussed in 3..

and from Equation S5-4 the joint probability distribution of Y1 and Y2 is fY1Y2 1 y1, y2 2 fX1X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 0 J 0 y2 4e 23 y1 y2 11 y12 y2 11 y124 ` ` 11 y1 2 2 4e

y2 11

....................

y1 2 2

for y1 0, y2 0. We need to nd the distribution of Y1 ability distribution of Y1, or fY1 1 y1 2 fY1Y2 1y1, y2 2 dy2 4e

....................

X1 X2 . This is the marginal prob-

y1 2 2

An important application of Equation S5-4 is to obtain the distribution of the sum of two independent random variables X1 and X2. Let Y1 X1 X2 and let Y2 X2. The inverse solutions are x1 y1 y2 and x2 y2. Therefore, x1 y1 x2 y1 1 0 x1 y2 x2 y2 1 1

At the price of $276,495 Alf Blowfish sells his house to Ann Bonidea ... Figure 5.4: MD4 collision [42].

and J 1. From Equation S5-4, the joint probability density function of Y1 and Y2 is fY1Y2 1 y1, y2 2 f X1 1 y1 y2 2 f X2 1 y2 2

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