1
$\begingroup$

Let $F=(X/n_1)/(Y/n_2)$ where $X$ and $Y$ are independent, $X$ is $\chi^2$ with noncentrality parameter $\lambda_1\geq 0$, and $Y$ $\chi^2$ with NCP $\lambda_2\geq 0$. According to this Wolfram article, the CDF of $F$ can be written as: $$ D(n_1,n_2;\lambda_1,\lambda_2;x)=\sum_{k,\,\ell\,\geq\, 0} f(\lambda_1,\lambda_2;k,\ell) g(n_1,n_2;k,\ell;x) $$ where $$ f(\lambda_1,\lambda_2;k,\ell)=\frac{\lambda_1^k\lambda_2^\ell}{\exp[(\lambda_1+\lambda_2)/2]} $$ and $g$ captures the remaining terms of $D$, as described explicitly in the linked article. Now, I note that $f\geq 0$ depends on $\lambda_1$, $\lambda_2$ but not on $x$ whereas $g$ does not depend on $\lambda_1$ nor $\lambda_2$. Moreover, as $(\lambda_1,\lambda_2)\to (0,0)$, we have $f\to 0$ unless $k=\ell=0$, then $f\to 1$. Then: $$ D(n_1,n_2;\lambda_1,\lambda_2;x)\geq f(0,0;0,0)g(n_1,n_2;0,0;x)=D(n_1,n_2;0,0;x).\tag{$*$} $$ So $D(n_1,n_2;\lambda_1,\lambda_2;x)$ is first-order stochastically dominated by $D(n_1,n_2;0,0;x)$. Now, I'm pretty sure this is false.

For example, $D(n_1,n_2;\lambda_1,0;x)$ can represent the distribution of the Fisher statistic when a null is false in a linear model whereas $D(n_1,n_2;0,0;x)$ is the distribution of the same statistic when the null is true. Then, it is typically the case that $D(n_1,n_2;\lambda_1,0;x)$ is not stochastically dominated by $D(n_1,n_2;0,0;x)$. (Think of the power curve.)

So where have I erred please in arriving at $(*)$?

p.s. I originally posted this on the Mathematics SE; the question received no answer and I was suggested to post it here. To avoid duplicates, I have deleted the MSE version.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.