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September 11th, 2017, 06:30 AM   #1
Joined: Mar 2013

Posts: 24
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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:

$p(Y_{2} = y_{1}|Y_{1} = y_{1}) = \sum{p(Y_{2} = y_{1}|Y_{1} = y_{1},X)p(X|Y_{1} = y_{1})}$

This simplifies to:

$\sum{p(Y_{2} = y_{1}|X)p(X|Y_{1} = y_{1})}$

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:

$p(Y_{3} = y_{1}|Y_{2} = y_{1},Y_{1} = y_{1}) = \sum{p(Y_{3} = y_{1}|Y_{2} = y_{1},Y_{1} = y_{1},X)p(X|Y_{2} = y_{1},Y_{1} = y_{1})} = \sum{p(Y_{3} = y_{1}|X)p(X|Y_{2} = y_{1},Y_{1} = y_{1})}$

Is this even correct?

If so, do I then use Bayes' Rule to compute:

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 07:04 AM.
Polaris84 is offline  

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