Thus takes two values and with equal prob in 0

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Unformatted text preview: È ´ ¿¡ ¾¡µ Ê Ô¾½ ¿¡ ÜÔ´ ´Ü¾ ѵ µ Ü ¿¡ È ´ ¿¡µ Ô¾½ ½ ÜÔ´ ´Ü¾ ѵ µ Ü ¾ ¼ ¡ ¾¡ ¿¡ Æ Ê Ô¾½ ¼¡ ÜÔ´ ´Ü¾ ѵ¾ µ Ü ¾ Ê ¾¡ ´Ü ѵ¾ ½ Ô¾ ¡ ÜÔ´ ¾ ¾ µ Ü ¾¡µ Ê ¿¡ Ô¾½ ¾¡ ÜÔ´ ´Ü¾ ѵ¾ µ Ü ¿¡µ ¾ ʽ ´Ü ѵ¾ ½ Ô¾ ¿¡ ÜÔ´ ¾ ¾ µ Ü ¡µ ¾ ¾ ¾ ¾ ¾ ¾ ¾ É (2) Case 2: ½. Let ´ µ. Applying the similar approach as Case 1 where there is no the ¼ ¦½ ¦¾ overflow case or the underflow case, i.e., for any integer È ¾ ·½ ¾ ¡ È´ ¡ ´ · ½µ¡µ 1 Ô½ ¾ ´Ü ѵ µ Ü ¾¾ ´ ·½µ¡ ¡ ¾ ÜÔ´ Or equivalently, we may write this probability density function in the following form ½ Ý ´µ where Ü ´µ is the cdf of ´´ ½ Æ Ý ´¾ · ½µ¡µ ´ ¡µ ´ ·...
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This document was uploaded on 02/28/2014 for the course ECE 411 at University of Waterloo, Waterloo.

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