Suppose that \(X\) is a random variable with that follows a binomial distribution with parameters \(n\) and \( p_x\). And let \(Y\) be a random variable independent of \(X\) with a binomial distributions with parameters \(m\) and \(p_y\). I would like to find the generalized likelihood ratio test statistic \(T\) given that \(H_0: p_x=p_y\) vs \(H_a_x \neq p_y\).

Now here is the solution to the problem step by step.

We know that the MLE for \( p_x\) is just \(\hat{p_x}=\dfrac{x}{n}\) and for \( p_y\) is just \(\hat{p_y}=\dfrac{y}{m}\) I already found the MLE if \(p=p_x=p_y\) which was \( \hat{p_0}=\dfrac{x+y}{m+n}\). Now lastly we need to find the GLRT test statistic \( T[ /math] which is

[1]: http://i.stack.imgur.com/7LzGF.jpg

From the picture then \(\hat{\theta}=\hat{p}\) and \(\hat{\theta_0}=\hat{p_0}\)(under the \(H_0\).

Therefore

\(L(x,y|\hat{p_x}\hat{p_y})={n\choose x }{m\choose y}\hat{p_x}^{x}\hat{p_y}^y(1-\hat{p_x})^{n-x}(1-\hat{p_y})^{m-y}\)

\(lnL(x,y|\hat{p_x},\hat{p_y})=ln({n\choose x })+ln({m\choose y})\)

\(+xln(\hat{p_x})+yln\hat{p_y}+(n-x)ln(1-\hat{p_x})+(m-y)ln(1-\hat{p_y})\)

Now for the second piece we get

\( L(x,y|\hat{p_0})={m\choose y}{n\choose x} \hat{p_0}^{x+y}(1-\hat{p_0})^{m+n-x-y}\)

\(ln L(x,y|\hat{p_0})=ln({m\choose y})+ln({n\choose x})\) \(+(x+y)ln(\hat{p_0})+(m+n-x-y)ln(1-\hat{p_0})\)

Thus \(T=2(xln(\hat{p_x})+yln\hat{p_y}+(n-x)ln(1-\hat{p_x})\)

\(+(m-y)ln(1-\hat{p_y})-(x+y)ln(p_0)-(m+n-x-y)ln(1-\hat{p_0})) \)

I don't know if all this is correct and I hope it is. I am lost with this problem and this is my solution so far. Thanks.

Update:

I noticed there are alot of views but no replies. Is there parts that are confusing that need clarification? or is it right?\)

Now here is the solution to the problem step by step.

We know that the MLE for \( p_x\) is just \(\hat{p_x}=\dfrac{x}{n}\) and for \( p_y\) is just \(\hat{p_y}=\dfrac{y}{m}\) I already found the MLE if \(p=p_x=p_y\) which was \( \hat{p_0}=\dfrac{x+y}{m+n}\). Now lastly we need to find the GLRT test statistic \( T[ /math] which is

[1]: http://i.stack.imgur.com/7LzGF.jpg

From the picture then \(\hat{\theta}=\hat{p}\) and \(\hat{\theta_0}=\hat{p_0}\)(under the \(H_0\).

Therefore

\(L(x,y|\hat{p_x}\hat{p_y})={n\choose x }{m\choose y}\hat{p_x}^{x}\hat{p_y}^y(1-\hat{p_x})^{n-x}(1-\hat{p_y})^{m-y}\)

\(lnL(x,y|\hat{p_x},\hat{p_y})=ln({n\choose x })+ln({m\choose y})\)

\(+xln(\hat{p_x})+yln\hat{p_y}+(n-x)ln(1-\hat{p_x})+(m-y)ln(1-\hat{p_y})\)

Now for the second piece we get

\( L(x,y|\hat{p_0})={m\choose y}{n\choose x} \hat{p_0}^{x+y}(1-\hat{p_0})^{m+n-x-y}\)

\(ln L(x,y|\hat{p_0})=ln({m\choose y})+ln({n\choose x})\) \(+(x+y)ln(\hat{p_0})+(m+n-x-y)ln(1-\hat{p_0})\)

Thus \(T=2(xln(\hat{p_x})+yln\hat{p_y}+(n-x)ln(1-\hat{p_x})\)

\(+(m-y)ln(1-\hat{p_y})-(x+y)ln(p_0)-(m+n-x-y)ln(1-\hat{p_0})) \)

I don't know if all this is correct and I hope it is. I am lost with this problem and this is my solution so far. Thanks.

Update:

I noticed there are alot of views but no replies. Is there parts that are confusing that need clarification? or is it right?\)

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