Harvard statistics, video 7 note(Gambler's ruin & random variable)

10:48 2014-10-06
start Harvard statistics, video 9


Gambler's ruin & random variables


10:49 2014-10-06
* conditioning is the soul of statistics


* random variables & their distributions


10:56 2014-10-06
Gambler's Ruin problem:


Two gamblers A & B, sequence of rounds bet $ 1,


P = P(A wins a certain round), q = 1-p // B wins round


repeat until one of them goes to bankrupt.


ruins means bankrupcy


assuming A starts with $ i, B starts with $ (N-i)


11:02 2014-10-06
that's the setup of the problem


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another way to think of this is a "random walk"


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random walk


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absorbing state at 0, N


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how many round the game will lasts?


mathematically it can last forever.


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strategy: condition on 1st step


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you have to pick something to condition on to 


break it up into manageable pieces.


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Let Pi = P(A wins game | A starts at $i)


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LOTP == Law Of Total Probability


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this is a recursive equation


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computer are very good at recursion


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difference equation, differential equation


11:17 2014-10-06
we would very quickly run out of notation.


11:50 2014-10-06
we need the concept of random variable


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we need the notion of a function


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What is a random variable?


It's a function from the sample sapce S to the R


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think of the random variable as a numerical "summary"


of an aspect of the experiment.


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each possbile outcomes maps to some number.


12:04 2014-10-06
Defn(Bernoulli distribution):


A r.v. X is said to have Bernoulli(p) distribution


if X has only 2 possible values, 0 & 1, and


P(X = 1) = p, P(X = 0) = 1-p


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then we say that this is a Bernoulli p r.v.


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before we do the experiment, we don't know 


whether it's going to be 0 or 1;


after we do this experiment, either it turns out


to be 0 or 1


12:08 2014-10-06
Binomial(n, p) The distribution of #successes in


n independent Bernoulli(p) trials is called Binomial(n,p)


its distribution is given by 


P(X = k) // X is going to an integer between 1 & n


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Bernoulli trials


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Bernoulli(p) => Binomial(n, p)


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PMF == Probability Mass Function


12:14 2014-10-06
one quick comment about the Binomial, suppose 


X ~ Binomial(n, p), Y ~ Bin(m, p), independent


then X + Y ~ Bin(n,p)