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Unformatted text preview: ECE 302: Homework 2 Prof. Saul Gelfand Spring 2010 Due February 5, 2010 Problem 1 a. The error process described is Bernoulli. Let K be the number of errors out of n operations. We have then P [ K = k ] = parenleftbigg n k parenrightbigg p k (1 p ) n k , for k = 1 ,...,n . b. We have that type 1 errors occur with probability p . Thus, type 1 errors also form a Bernoulli process with parameter p . If we let K 1 be the number of type 1 errors out of n operations, then P [ K 1 = k 1 ] = parenleftbigg n k 1 parenrightbigg ( p ) k (1 p ) n k , for k 1 = 1 ,...,n . c. Similar to above. Let K 2 be the number of type 2 errors out of n operations. P [ K 2 = k 2 ] = parenleftbigg n k 2 parenrightbigg ((1 ) p ) k (1 (1 ) p ) n k , for k 2 = 1 ,...,n . d. We first rewrite the joint probability P [ K 1 = k 1 K 2 = k 2 ] = P [ K 1 = k 1 K = k 1 + k 2 ] = P [ K = k 1  K = k 1 + k 2 ] P [ K = k 1 + k 2 ] . 1 As mentioned in the hint, we will have that if we are given that K = k = k 1 + k 2 errors occur in n operations, then each of the k errors are either type 1 (with probability ) or type 2 (with probability 1 ). Thus, type 1 errors amoung errors are Bernoulli trials. We then have that: P [ K = k 1  K = k 1 + k 2 ] = parenleftbigg k 1 + k 2 k 1 parenrightbigg k 1 (1 ) k 2 , and P [ K 1 = k 1 K 2 = k 2 ] = P [ K = k 1  K = k 1 + k 2 ] P [ K = k 1 + k 2 ] = parenleftbigg k 1 + k 2 k 1 parenrightbigg k 1 (1 ) k 2 parenleftbigg n k 1 + k 2 parenrightbigg p k 1 + k 2 (1...
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 Fall '08
 GELFAND
 k2, Px, Typetoken distinction, cdf FX

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