ECE_313_SP2023_HW5_sol

University of Illinois Spring 2023 ECE 313: Problem Set 5: Problems and Solutions Due: Friday, February 23 at 5:00:00 p.m. Reading: ECE 313 Course Notes, Section 2.10, 2.11. Note on reading: For most sections of the course notes there are short answer questions at the end of the chapter. We recommend that after reading each section you try answering the short answer questions. Do not hand in; answers to the short answer questions are provided in the appendix of the notes. Note on turning in homework: You must upload handwritten homework to Gradescope. No typeset homework will be accepted. No late homework will be accepted. Please write on the top right corner of the first page: NAME AS IT APPEARS ON Canvas NETID SECTION PROBLEM SET # Page numbers are encouraged but not required. Five points will be deducted for improper headings. 1. [Election polls] You want to run a poll to estimate what fraction p of the population will vote for candidate A in the next elections. Suppose you interview n people, and out of those, X people say they will vote for candidate A. (a) What is the maximum likelihood estimate for p given X = k ? Solution: The number of people X that will vote for candidate A can be modelled as a Binomial( n, p ) random variable. The corresponding likelihood function is L ( p ) = n k p k (1 − p ) n − k Differentiating L ( p ) with respect to p, dL ( p ) dp = n k kp k − 1 (1 − p ) n − k − ( n − k ) p k (1 − p ) n − k − 1 = n k p k − 1 (1 − p ) n − k − 1 ( k − np ) Setting the right-hand side to zero, we obtain the ML estimate for p given X = k as ˆ p ML = k/n. Note that the derivative is positive for p < ˆ p ML and negative for p > ˆ p ML , hence the obtained value is indeed the ML estimate. (b) If you want to estimate p to within 0.01 with a confidence level of 96%, how large does n have to be? Justify your answer. Solution: We know that for any a > 0, P p ∈ ˆ p ML − a 2 √ n , ˆ p ML + a 2 √ n ≥ 1 − 1 a 2 .

To obtain a 96% confidence, 0 . 96 = 1 − 1 a 2 = ⇒ a = 5 . Now, in order to estimate p to within 0 . 01 of ˆ p ML , we need a 2 √ n ≤ 0 . 01 = ⇒ √ n ≥ 50 ∗ a = ⇒ n ≥ 62 , 500 . 2. [Drawing aces from a deck] Suppose I have a standard 52-card deck, and I ask Bob to draw two random cards from it. (a) Bob draws two cards and tells me that, out of the two cards he drew, he has at least one ace. What is the conditional probability that both cards drawn were aces? Solution: Let X be the event that both the drawn cards were aces and Y be the event that at least one of the two cards drawn is an ace. Note that X ⊆ Y, therefore X ∩ Y = X. There are ( 4 2 ) ways of choosing 2 aces out of the total 4 and ( 52 2 ) ways of selecting any 2 cards out of the total 52 . Hence, P ( X ) = ( 4 2 ) ( 52 2 ) = 4 52 · 3 51 If at least one card is ace, possible cases are that both cards are aces, which can happen in ( 4 2 ) ways, or one card is ace and the other is a non-ace, which can happen in ( 4 1 )( 48 1 ) ways. Note that these two cases are mutually exclusive. Hence, P ( Y ) = ( 4 2 ) + ( 4 1 )( 48 1 ) ( 52 2 ) Note that P ( Y ) can also be calculated as P ( Y ) = 1 − P (No aces) = 1 − ( 48 2 ) ( 52 2 ) = 1 − 48 52 · 47 51 Using the definition of conditional probability, P ( X | Y ) = P ( X ∩ Y ) P ( Y ) = P ( X ) P ( Y ) = ( 4 2 ) ( 4 2 ) + ( 4 1 )( 48 1 ) = 1 33 = 0 . 0303 (b) The cards are returned to the deck, which is shuffled, and Bob draws two cards again. This time, when Bob is not looking, I take a peek at one of the cards that he drew, and 2