Calculating the transformation of 1/X where the pdf of X is 1/x I feel very silly asking this, since it seems so simple but I'm almost certain I have done something wrong.
I have a random variable $A$ with the following pdf:
\begin{equation}
f_{A}(x_{A}) = 
\begin{cases}
\frac{1}{x_{A}}, & x\in[5, 6] \\
0, & \text{else}
\end{cases}
\end{equation}
$B$ is another random variable that equals $\frac{1}{A}$, and I want to find the pdf of B
\begin{align*}
f_{B}(x_{B}) 
&= \frac{d}{dx_{B}}(F_B(x_{B})) \\
&= \frac{d}{dx_{B}}(\mathbb{P}(B \leq x_{B})) \\
&= \frac{d}{dx_{B}}(\mathbb{P}(\frac{1}{A} \leq x_{B})) \\
&= \frac{d}{dx_{B}}(\mathbb{P}(A > \frac{1}{x_{B}})) \\
&= \frac{d}{dx_{B}}(1 - \mathbb{P}(A \leq \frac{1}{x_{B}})) \\
&= \frac{d}{dx_{B}}(1 - F_{A}(\frac{1}{x_{B}})) \\
&= -f_{A}(\frac{1}{x_{B}})\cdot \frac{-1}{x_{B}^{2}} \\
&= 
\begin{cases}
\frac{1}{x_{B}}, & \frac{1}{x_{B}} \in [5, 6] \\
0, & \text{else}
\end{cases} \\
&= 
\begin{cases}
\frac{1}{x_{B}}, & x_{B} \in [\frac{1}{6}, \frac{1}{5}] \\
0, & \text{else}
\end{cases} \\
\end{align*}
Have I done this correctly?
 A: You may if you like, but there is no particular need to subscript the free variable, $x$. It just distracts.
Otherwise, yes indeed your working is good: although I would add the caveats. $$\begin{align}f_{B}(x) &= \dfrac{\mathrm d \mathsf P(A^{-1}\leqslant x)}{\mathrm d x}\\[1ex]&=\dfrac{\mathrm d \mathsf P(A\geqslant x^{-1})}{\mathrm d x}&&\raise{1.5ex}{\text{because }A\in[5;6]\text{ a.s.}\\\text{so negatives are not an issue.}}\\[1ex]&=\dfrac{\mathrm d(1-\mathsf P(A\lt x^{-1}))}{\mathrm d x}\\[1ex]&=\dfrac{\mathrm d(1-\mathsf P(A\leqslant x^{-1}))}{\mathrm d x}&&{\small A\text{ is absolutely continuously distributed}}\\[1ex]&=(-x^{-2})\,(-f_A(x^{-1}))&&\\[1ex]&=x^{-2}\begin{cases}(x^{-1})^{-1}&:& x^{-1}\in[5;6]\\0&:&\text{else}\end{cases}\\[2ex]\therefore\qquad f_B(x)&=\begin{cases}x^{-1}\hspace{8ex}&:& x\in[1/6;1/5]\\0&:&\text{else}\end{cases}\end{align}$$

The only issue I have is that $f_A$ is not a valid probability density function. $$\int_5^6 \dfrac 1x\,\mathrm d x = \ln(6/5) \qquad\neq 1$$
So consequently, $f_B$ will not be a valid probability density function either.
GIGO
