This question is about hypothesis testing, where we want to use the likelihood ratio statistic with permutations test.

Suppose we sample $n$ observations from the distribution $F_{XY}$, which is the uniform over the triangle defined by (0,0), (0,1), and (1,0) to get $S=\{(x_1,y_1),\ldots,(x_n,y_n)\}$. Note that $X$ and $Y$ are the regular Cartesian axes. Then we know that $$f_{XY}(x,y)=1/\text{area of triangle}=2$$ if $(x,y)$ is within our triangle. Let $H_1$ be the hypotehsis that $S$ was sampled from $F_{XY}$, which we know is true. Let $H_0$ be the hypothesis that $S$ was sampled in the following way: $(X_i,Y_i)\sim F_XF_Y$, where $F_X,F_Y$ are the marginals of $F_{XY}$. Compute the likelihood ratio test statistic, defined as $\lambda=\frac{L_{H_1}(S)}{L_{H_0}(S)}$. Now we want to compute the empirical p-value of this $\lambda$ using permutation test. This means (I think): we permuate $S$ several times (say $M$ times) to get $S_1,\ldots,S_m$, compute $\lambda_1, \ldots, \lambda_m$, and then $\hat{p}=\frac{1}{M}\sum_{i=1}^{M}1_{\{\lambda_i>\lambda\}}$.

Here's what I did, and I may be wrong.

$$f_X(x)=\int_0^{1-x} f_{XY}(x,y)\ dy=2(1-x)$$ and similarly $f_Y(y)=2(1-y)$. Then $$L_{H_0}(S)=\prod_{i=1}^n f_X(x_i)f_Y(y_i)=\prod_{i=1}^n 2(1-x_i)\cdot 2(1-y_i)$$ and $$L_{H_1}(S)=\prod_{i=1}^n f_{XY}(x_i,y_i)=\prod_{i=1}^n 2=2^n$$ So $\lambda=\dfrac{2^n}{\prod_{i=1}^n 2(1-x_i)\cdot 2(1-y_i)}$.

Here's my problem: I permute $S$ in the following way: we have $2n$ points, I permute those $2n$ points, and simply assign the first $n$ numbers to be the $x_i$'s and the next $n$ numbers to be the $y_i$'s to form $S_i$ for each permutation. I then compute $\lambda_i$ as I did before. The problem is that I get the same $\lambda$ every time! that is $\lambda=\lambda_1= \ldots=\lambda_m$, since $X,Y$ are symmetrical in my calculations. I'm sure I'm doing something wrong. I'd appreciate if you could point out my mistake. Since we know $H_1$ is true, we're trying to show that the empirical p-value is small.


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