By Fazlollah M. Reza

ISBN-10: 0486682102

ISBN-13: 9780486682105

Info conception

**Read Online or Download An Introduction to Information Theory PDF**

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**Extra info for An Introduction to Information Theory**

**Example text**

Probability space of two events. consider two particular events A and B among the events that result from the experiment. Let the OV1'''\O''r·I'II'''V\lf~n't. be repeated n times. Each observation can belong to one of the four following categories (Fig. 2-18): 1. A has occurred but not B, the event AB'. 2. B has occurred but not A, the event BA'. 3. Both A and B have occurred, the event AB. 4. Neither A nor B has occurred, the event A'B'.. 1"l>'ln"""l[7' of either ff AB}, relative frequency of A and or both = - - - - n occurring together = n f{AIR}, relative frequency of A under condition that B has occurred na n2 na f{BIA}, relative frequency of B under condition that A has occurred na (2-57) When the number of experiments tends to infinity, these simple relations with proper interpretation lead to the addition law and multiplication law: (2-58) AB} PtA U B} = PtA} + P{B} (2-59) P{AB} = P{A}P{BIA} (2-60) P{AB} = P{B}P{AIB} For the special case of mutually exclusive events, = 0, P{A+R}=P{A}+ Equation (2-61) shows the validity of the requirement (2-37) for the chosen set function termed the relative frequency of the event.

One assumes that an element x is a member of the set of the left side of each identity, and then one has to prove that x will necessarily be a member of the set of the right side of the same equation. For instance, in order to prove the distributive law let + x Then A x (B x + C) Then at least one of the following three cases must be true: x x A B (b) x x A C (c) x E A x x C These are in turn equivalent to (a) x AB (b) x AC (c) x E ABC but ABC CAB Therefore it is sufficient to require x E AB AC Similarly, one can show that x E AB + AC implies x A(B + The Venn is often a very useful visual aid.

What is the probability of having at least (a) one tail and (b) two tails? Solution. The main assumption in this and in similar problems is the concept of independence of successive trials and the equally probable outcomes. Let A and B be the events of getting no tail and exactly one tail, respectively. Then PIA} = PIB} nr = 10 The events of interest are or 1,~24 = 1,~~4 U -- A = A' (a) PIA'} (b) = U - (A PIA' 1 1 1,023 = - 1,024 = 1,024 + B) = (U - A) - B A' _ B} = 1,023 _ ~ = 1,013 =; 1,024 1,024 B 1,024 2-10.