
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 23rd, 2017, 06:18 PM  #11 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574  Now it's perfectly clear. Thanks. They're looking at a ternary input string with probabilities as given, and they're asking about the asymptotic behavior of the output string. The voltages and even the input string itself are not relevant. We're building a bitstring by appending 0, 10, or 11 with the given probabilities. Or we can think of the input as being a ternary string. I'm thinking about it like this. With probability .4 we append one bit; and with probability .6 we append two. So the expected number of output bits for an input string of length n is .4(1) + .6(2) = 1.6n. This seems reasonable to me but I can see I'm hiding the limit computation. There's no reason that an input string of length n must generate an output stream of exactly 1.6n. It's only true statistically. Likewise we add a single 1 with probability .3, and two 1's with probability .3. So the expected number of 1's is .3(1) + .3(2) = .9n. Now .9n/1.6n = .5625 as required. But there is something a little fishy with my thinking. I made the limits disappear and I'm not exactly sure what allows me to do that. Can anyone who knows probability theory clarify my confusion? ps  The probability that an output bit is 0 is just 1 minus .5625, you don't need to do the second computation. Last edited by Maschke; October 23rd, 2017 at 06:26 PM. 
October 23rd, 2017, 07:46 PM  #12 
Senior Member Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093  

Tags 
binary, bit, messages, pdf, probability 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Binary Probability Formula  Karma  Probability and Statistics  9  July 17th, 2017 01:28 PM 
Teachers Day Messages  Saginogdu7422  Art  0  September 4th, 2016 03:02 AM 
Why do I get these Error Messages?  vagulus  Elementary Math  4  February 16th, 2015 04:34 AM 