HW06 - University of Illinois Fall 2009 ECE 313: Problem...

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University of Illinois Fall 2009 ECE 313: Problem Set 6: Solutions Bayes’ Formula and Midterm Review Problems 1. [Conditional Probabilities] (a) Yes, A gives positive information about B . Note that by the definition of conditional probability, P ( A | B ) = P ( A B ) /P ( B ) > P ( A ) is equivalent to P ( A B ) > P ( A ) P ( B ), and so it is also true that P ( B | A ) = P ( A B ) /P ( A ) > P ( B ). (b) Yes, B c give negative information about A , that is, it is true that P ( A | B c ) < P ( A ). Notice from part (a) that P ( A B ) > P ( A ) P ( B ) and so P ( A ) = P ( A B ) + P ( A B c ) > P ( A ) P ( B ) + P ( A B c ) . Therefore, it must be that P ( A )(1 - P ( B )) = P ( A ) P ( B c ) is larger than P ( A B c ), and so P ( A ) > P ( A B c ) P ( B c ) = P ( A | B c ) , that is, if B gives positive information about A , then B c gives negative information about A . (c) B c gives positive information about A c . Since part (b) gives us that P ( A | B c ) < P ( A ), it follows that P ( A c | B c ) = 1 - P ( A | B c ) > 1 - P ( A ) = P ( A c ) and so P ( A c | B c ) > P ( A c ). 2. [Hindsight] Let an experiment consist of picking a bulb at random (say from a box containing bulbs manufactured by Transylvania and Eastinghouse) and using it till the bulb burns out. Let T denote the event that the bulb picked was made by Transylvania Corp. and E = T c the event that the bulb was made by Eastinghouse Corp. We are told that P ( E ) = P ( T ) = 0 . 5, that is, we can suppose that the box contains equal numbers of bulbs made by the two companies. If X denotes the lifetime (rounded up to the next full hour) of the bulb, then we are also told that P { X = n | T } = p 1 ( n ) and P { X = n | E } = p 2 ( n ). (a) By the law of total probability (LOTP), P { X = M } = P { X = M | T } P ( T ) + P { X = M | E } P ( E ) = p 1 ( M ) P ( T ) + p 2 ( M ) P ( E ) = p 1 ( M ) + p 2 ( M ) 2 . Be sure that you understand that you cannot just add p 1 ( M ) and p 2 ( M ) to find P { X = M } , nor can you in general simply compute the arithmetic average of p 1 ( M ) and p 2 ( M ); you have to compute a weighted sum of p 1 ( M ) and p 2 ( M ) as specified by the law of total probability. In this case , the LOTP calculation turns out to be the arithmetic average because it so happens that P ( E ) = P ( T ) = 0 . 5, that is, the weights are 1 2 for both
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This note was uploaded on 10/25/2010 for the course ECE 313 taught by Professor Milenkovic,o during the Spring '08 term at University of Illinois, Urbana Champaign.

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HW06 - University of Illinois Fall 2009 ECE 313: Problem...

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