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October 23rd, 2017, 06:18 PM   #11
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Quote:
 Originally Posted by romsek Ok, I've got it. Tricky little devil but really not that bad.
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
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Quote:
 Originally Posted by Maschke It's only true statistically.
yes

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