While trying to solve a problem on hypothesis testing, more specifically, computing a probability of false alarm, I've encountered an integral of a PDF. This PDF corresponds to the conditional PDF of the likelihood ratio test subject to a given hypothesis. In order to compute this PDF, I assume the likelihood ratio test to be a function applying a transformation on my original random variable. This is possible since I know the PDF of my original random variable.
I expect that after computing the PDF of the transformed variable, computing the integral will be straight forward, at least numerically since the resulting PDF is not simple.
Is is there other approaches preferable to this one, e.g., generating the transformed random variables samples numerically thus dispensing the need for the PDF, assuming that this can be done?
Below I state the problem and summarise the steps I've taken to obtain the PDF.
Problem statement
Consider a Cauchy distributed random variable $X$ with PDF \begin{align*} p_{x|H_0}(X|H_0)&=\frac{1}{\pi[1+X^2]},\\ p_{x|H_1}(X|H_1)&=\frac{1}{\pi[1+(X-1)^2]}. \end{align*} when conditioned to $H_0$, and $H_1$, respectively.
Assuming the Cauchy PDF \begin{equation*} p_x(X)=\frac{\gamma}{\pi[\gamma^2+(X-u)^2]}, \end{equation*} means that the above equations may correspond to two different sources: one $H_0$, generating samples $X$ with $u=u_0=0$, and the other $H_1$, generating samples $X$ with $u=u_1=1$.
In order to distinguish which source generated an observed sample I may proceed with a likelihood ratio test, \begin{equation*} \Lambda(X)=\frac{p_{x|H_1}(X|H_1)}{p_{x|H_0}(X|H_0)}=\frac{1+X^2}{1+(X-1)^2}\overset{H_1}{\underset{H_0}\gtrless}\eta. \end{equation*}
Naturally, if I'm to determine the probability of false detection I will have \begin{equation} P_F=\mathrm{Pr}\{\Lambda(X)>\eta|H_0\}=\int_{\eta}^{\infty}p_{\Lambda|H_0}(\Lambda(X)|H_0)\,d\Lambda(X). \end{equation}
This means that in order to determine $P_F$ with respect to the threshold $\eta$ I must know $p_{\Lambda|H_0}(\Lambda(X)|H_0)$.
Assuming $\Lambda(X)$ results from applying a transformation $g(X)$ to $X$ we have \begin{equation} p_\Lambda(\Lambda(X))=\frac{p_x(X_1)}{|g^\prime(X_1)|}+\frac{p_x(X_2)}{|g^\prime(X_2)|}+\cdots \end{equation} where $X_1$, $X_2$, $\ldots$, are the roots of $g(X)$, and $g^\prime(X)$ is its first order derivative. For $g(X)=\frac{1+X^2}{1+(X-1)^2}$, these are \begin{equation} X_{1,2}=\frac{\Lambda(X)\pm\sqrt{-\Lambda^2(X)+3\Lambda(X)-1}}{\Lambda(X)-1};\quad \frac{3-\sqrt{5}}{2}<\Lambda(X)<\frac{3+\sqrt{5}}{2},\quad \Lambda(X)\neq 1, \end{equation} and \begin{equation} g^\prime(X)=2X\frac{1}{1+(X-1)^2}-(1+X^2)\frac{2(X-1)}{[1+(X-1)^2]^2}, \end{equation} respectively.
Resulting, \begin{equation} \begin{split} p_\Lambda(\Lambda(X))&=\frac{1}{2}\frac{\gamma}{\pi[\gamma^2+(X_1-u)^2]}\left|\frac{[1+(X_1-1)^2]^2}{X_1[1+(X_1-1)]-(1+X_1^2)(X_1-1)}\right|\\ &\quad+\frac{1}{2}\frac{\gamma}{\pi[\gamma^2+(X_2-u)^2]}\left|\frac{[1+(X_2-1)^2]^2}{X_2[1+(X_2-1)]-(1+X_2^2)(X_2-1)}\right| \end{split} \end{equation}
Finally, \begin{equation} \begin{split} p_{\Lambda|H_0}(\Lambda(X)|H_0)&=\frac{1}{2}\frac{1}{\pi(1+X_1^2)}\left|\frac{[1+(X_1-1)^2]^2}{X_1[1+(X_1-1)]-(1+X_1^2)(X_1-1)}\right|\\ &\quad+\frac{1}{2}\frac{1}{\pi(1+X_2^2)}\left|\frac{[1+(X_2-1)^2]^2}{X_2[1+(X_2-1)]-(1+X_2^2)(X_2-1)}\right| \end{split} \end{equation}
Notice that I've yet to compute the integral...
