Latest revision as of 10:52, 29 January 2013

Homework 2, ECE302, Spring 2013, Prof. Boutin

Due Monday January 28, 2013 (in class)

As for homework 1, this homework consists in handing in the problems that you were asked to solve after each lecture. For convenience, the problems are listed (again) below. Hand in a hard copy of your solutions in class. Make sure to include a cover page and to staple all the pages together. Write legibly and clearly. Put the problems in order. Do not write on the back of the pages. Do not use paper torn out of a spiral book. Thank you very much.

• Problem 2.62 from the textbook: Probability, Statistics, and Random Processes for Electrical Engineering, 3rd Edition, by Alberto Leon-Garcia, Pearson Education, Inc., 2008.

• Problems 14, 15, 16, 17, 18, 24, 25 30, 31, 32, 33 34 from Chapter 1 of "Introduction to Probability," by Dimitri P. Bertsekas and John N. Tsitsiklis. Athena Scientific, Belmont, Massachusetts, 2008.

Questions/comments/Discussion=

• For Problem 16, should we assume each of the 3 coins (double heads, double tails and heads-tails) are un-biased? or do we give them each a variable for their individual biases? - a. willats

EDIT - I suppose this only actually matters for the heads-tails coin

• • Yes, assume that the coins are unbiased. -pm

• For problem 25, it seems like the chances of X being larger than the amount in the envelope are incredibly low, especially as the value in the envelope rises. Wouldn't the final probability depend on the values that exist in the envelopes? If not, what is wrong with our thought process?

• • You are right. The final probability of getting the envelope with more money is depend on the values of the two envelopes. Here is the illustration:

We assume the amount of money in the two envelopes are a and b, which a < b.

X= 1/2 + (number of tosses to get the first head).

$ P(get\ b\ in\ the\ end) = P(switch \to\ b | a) P(a) + P(don't\ switch|b)P(b) $

$ P(switch \to\ b | a) = P(X>a) = \sum_{k=a}^{\infty} (\frac{1}{2})^{k} = \sum_{k=1}^{\infty}(\frac{1}{2})^{k} - \sum_{k=1}^{a-1}(\frac{1}{2})^{k} = (\frac{1}{2})^{a-1} $.

$ P(don't\ switch|b) = P(X<b) = 1-P(X>b) =1 - (\frac{1}{2})^{b-1} $

$ \therefore P(get\ b\ in\ the\ end) = (\frac{1}{2})^{a-1} \cdot (\frac{1}{2}) + (1 - (\frac{1}{2})^{b-1}) \cdot (\frac{1}{2}) = (\frac{1}{2}) + [(\frac{1}{2})^{a} - (\frac{1}{2})^{b}] $

$ \because a<b, the\ second\ term\ will\ be\ greater\ than\ 0. $

$ \therefore the\ probability\ is\ larger\ than\ \frac{1}{2}. $

I would like to make a note here. Tossing a coin to make the switch decision in this question can be any kind of decision method that gives you a hint of how relatively large the money in the envelop is. For example, if you say "I'm going to switch if the money in the envelop is smaller than 100." It will still give you probability of more than 0.5 to get the envelop that contains more money.

The point is, you have more information about the money in your hand. And that decision is always better than just randomly guess (0.5),right? -TA

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