# Finding Generalized Linear Ratio Test Statistic for binomial distrributions

#### raidernation

##### New Member
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?\)

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