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let $X_{1}$, $X_{2}$, $X_3$, $X_{4}$ be iid bernoulli rvs with $\mathbb{P}(0)=0.5$, $\mathbb{P}(1)=0.5$.

$Y_{1} = X_{1}+X_{2}+X_{3}$ and $Y_{2}=X_{1}+X_{2}+X_{4}$

$Y_{1}$, $Y_{2}$ are dependent binomial rvs by definition. I need to find max likelihood estimate of $X_{1}$ given I have observed $Y_{1}$ and $Y_{2}$.

how should I go about it?

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    Welcome to math.SE! Please see [here](http://meta.math.stackexchange.com/questions/463/put-a-link-to-a-tutorial-on-the-tex-in-the-faq/464#464) and [here](http://meta.stackexchange.com/questions/68388/there-should-be-universal-latex-mathjax-guide-for-sites-supporting-it/70559#70559) for how to format your mathematics expressions with LaTeX, and see [here](http://meta.stackoverflow.com/editing-help) for how to use Markdown formatting.2012-11-08

1 Answers 1

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There are two different cases.

If the observed values differ by $1$, this implies known values for $X_3$ and $X_4$, and the lower of the two observed values is the value of $X_1+X_2$. If this is $0$ or $2$, it implies the value $0$ or $1$, respectively, for $X_1$; if it is $1$, both values of $X_1$ are equally likely.

On the other hand, if the observed values are equal, this implies that the values of $X_3$ and $X_4$ are equal. Given that they are equal, the problem effectively becomes finding the maximum likelihood estimate for $X_1$ given $X_1+X_2+X_3$. If the observed value is $0$ or $3$, that determines $X_1$ with value $0$ or $1$, respectively. If the observed value is $1$ or $2$, then the value of $X_1$ is $0$ or $1$, respectively, with probability $2/3$, and $1$ or $0$, respectively, with probability $1/3$, so the maximum likelihood estimate in this case is $0$ or $1$, respectively.

In summary, the maximum likelihood estimate for $X_1$ given the two observed values is given by

$ \begin{array}{c|cccc} &0&1&2&3\\\hline 0&0&0&-&-\\ 1&0&0&=&-\\ 2&-&=&1&1&\\ 3&-&-&1&1 \end{array} $

where $-$ indicates a case that cannot occur and $=$ indicates that both values of $X_1$ are equally likely.

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    thankyou for guiding me.....I was actually trying to find a generalized solution. I ll post a more targeted question in new post and ll link it to this one. I appreciate your suggestions.2012-11-10