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Let $\left\{ X_i \right \}$ be an $n$-sample with pdf $f$. Show that the likelihood ratio test statistic for \begin{align} H_0 &: f(x) = 2x \\ H_1 &: f(x)= 3x^2 \end{align} has a $\chi^2$ distribution with $2n$ degrees of freedom.

The likelihood ratio test statistic is $$ \lambda = \frac{\prod f_0(x_i)}{\prod f_1(x_i)} = \frac{(2/3)^n}{t^n}, $$ with $t=\prod x_i$. Then I should show $$ \lim_{n\to\infty} -2\ln \lambda \stackrel{d}= \chi^2(2n). $$ How can the degree of freedom be $2n$?
I don't see any parameter in $H_0$ and $H_1$, to apply the theorem which says $$ \lim_{n\to\infty} -2\ln \lambda \stackrel{d}= \chi^2(\#\Theta_1-\#\Theta_0). $$

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$\lambda$ does not seem to have $\chi^2$ distribution under either $H_0$ or $H_1$. Probably you want to show that statement asymptotically. –  passerby51 Feb 24 '13 at 16:56
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@passerby51 Thanks, I meant asymptotically but didn't write clearly. I tried to copy a proof on the asymptotics of $\lambda$, but all the proof I know makes reference to the parameter space $\Theta_1$ and $\Theta_0$. –  Nicolas Essis-Breton Feb 24 '13 at 19:41
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The statement is still not quite correct. The RHS of your limit ($\chi^2(2n)$) depends on $n$ which shouldn't. Probably you want to show that $-2 \ln \lambda_n - Y_n$ goes to zero in probability, where $Y_n \sim \chi^2(2n)$ is some sequence of random variables. I am still not quite sure this is true. Are you sure you are not missing anything in the problem? –  passerby51 Feb 27 '13 at 16:27
    
@passerby51 I think everything of the problem is here. I will try to find such a sequence $Y_n$. I tried to view the LRT as a goodness-of-fit test with $n$ cells in $H_0$ and $2n$ cells in $H_1$ ($max(f_0(X_i),f_1(X_i))$). Then the degree of freedom is $2n-n=n$. –  Nicolas Essis-Breton Feb 27 '13 at 18:55
    
Asked and answer on statSE. @passerby51 the trick is that $-ln X_i$ has an exponential distribution under $H_0$, that is a $\chi^2$ with $2$ degrees of freedom. –  Nicolas Essis-Breton Mar 4 '13 at 13:46
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