Solutions2

# Solutions2 - EE464 Homework#2 Solutions Problem#1 Let us...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE464 Homework #2 Solutions Problem #1 Let us define the following events: A â‰¡ { N heads } , B 1 â‰¡ { Fair Coin } and B 2 â‰¡ { Loaded Coin } . Then we want to plot P ( B 1 | A ) vs p l . Using Bayesâ€™ Theorem we can write P ( B 1 | A ) = P ( A | B 1 ) P ( B 1 ) P ( A | B 1 ) P ( B 1 )+ P ( A | B 2 ) P ( B 2 ) . Now note that P ( B 1 ) = P ( B 2 ) = 1 / 2. Furthermore, since the flips are independent we see that P ( A | B 1 ) = (1 / 2) n and that P ( A | B 2 ) = p n l . Therefore, we can write P ( B 1 | A ) = (1 / 2) n +1 (1 / 2) n +1 +(1 / 2) p n l = (1 / 2) n (1 / 2) n + p n l . A plot for the case N = 10 is shown below. Problem #2 We first use independence to prove the equality. We can write P ( A | B ) = P ( A âˆ© B ) P ( B ) , P ( B ) > 0 by definition, and similarly for P ( A | B C ), so that we will only consider the cases for which P ( B ) 6 = 0 and P ( B ) 6 = 1. Now by inde- pendence we can simplify and write P ( A | B ) = P ( A ) P ( B ) P ( B ) = P ( A ). So now it suffices to show that P ( A | B C ) = P ( A ). To do this we use total probability to write P ( A ) = P ( A âˆ© B ) + P ( A âˆ© B C ) = P ( A âˆ© B ) + P ( A | B C ) P ( B C ). By independence we rewrite as P ( A ) = P ( A ) P ( B ) + P ( A | B C ) P ( B C ). Now we note that P ( B ) = 1- P ( B C ) so that P ( A ) = P ( A )(1- P ( B C )) + P ( A | B C ) P ( B C ). Rearranging gives P ( A ) = P ( A )+ P ( B C )( P ( A | B C )- P ( A ) so that P ( B C )( P ( A | B C )- P ( A )) = 0. Now since we have assumed that P ( B C ) 6 = 0, it must be true that P ( A | B C ) = P ( A ), thus proving the forward 1 implication. QED For the reverse implication we assume that P ( A | B ) = P ( A | B C ) = p . Then we can write as before P ( A ) = P ( A | B ) P ( B ) + P ( A | B C ) P ( B C ) = pP ( B ) + pP ( B C ) = p ( P ( B ) + P ( B C )) = p . Therefore P ( A ) = P ( A | B ) = P ( A âˆ© B ) P ( B ) which implies that P ( A âˆ© B ) = P ( A ) P ( B ). QED Problem #3 (a) We first note that there are 81 possible outcomes ( N 1 ,N 2 ) each hav- ing probability = 1 / 81. Let B â‰¡ { Î£ is odd } . Now we are looking for p â‰¡ P (Î£ = 7 | B ). Using conditional probability we can write p = P (Î£=7 âˆ© B ) P ( B ) . Note that since { Î£ = 7 } âŠ† B , we can write p = P (Î£=7) P ( B ) ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

Solutions2 - EE464 Homework#2 Solutions Problem#1 Let us...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online