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September 11th, 2017, 07:30 AM  #1 
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Need help setting up this problem.
The problem states we have two random variables, $X$ and $Y$. There are only two likely outcomes for both random variables: (don't know if this is correct 'syntax') $X = \{x_{1}, x_{2}\}$ $Y = \{y_{1}, y_{2}\}$ $Y_{1}$, $Y_{2}$, and $Y_{3}$ are 3 consecutive samples of the random variable $Y$ and are assumed to be conditionally independent given $X$. It states to find the probability that all 3 have the value $y_{1}$. I'm having trouble figuring out how to set this up. In a previous part of the problem, I needed to calculate the below: $p(Y_{2} = y_{1}Y_{1} = y_{1}) = \sum{p(Y_{2} = y_{1}Y_{1} = y_{1},X)p(XY_{1} = y_{1})}$ This simplifies to: $\sum{p(Y_{2} = y_{1}X)p(XY_{1} = y_{1})}$ The reason my professor gave us in our examples is because "$Y_{2}$ is conditionally independent of $Y_{1}$ given $X$, and therefore $Y_{1}$ doesn't give any additional information about the $Y_{2}$". Which makes perfect sense in itself and is easy enough to calculate, only I don't fully understand the concepts at work here. What happens then if you have 3 consecutive samples from the random variable $Y$? Currently, through my intuition alone, I have the aforementioned problem of "find the probability that all 3 have the value $y_{1}$" set ups as: $p(Y_{3} = y_{1}Y_{2} = y_{1},Y_{1} = y_{1}) = \sum{p(Y_{3} = y_{1}Y_{2} = y_{1},Y_{1} = y_{1},X)p(XY_{2} = y_{1},Y_{1} = y_{1})} = \sum{p(Y_{3} = y_{1}X)p(XY_{2} = y_{1},Y_{1} = y_{1})}$ Is this even correct? If so, do I then use Bayes' Rule to compute: $p(XY_{2}=y{1},Y_{1}=y{1})$ I'm an undergrad in computer science that hasn't had a statistics course in 2 years, so sorry if this is poorly described. Hopefully someone can help anyway. EDIT: Made a mistake on the first equation Last edited by Polaris84; September 11th, 2017 at 08:04 AM. 

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