It is known that the equivalent resistance of a parallel combination of two resistors is equal to \begin{align*} R = \frac{R_1R_2}{R_1+R_2} \end{align*} which could be also written as \begin{align*} R= (\tfrac{1}{R_1}+\tfrac{1}{R_2})^{-1} \end{align*} Consider the case where $R_1 = a+0.5\pm 0.5 \Omega$ and $R_2 = a+1.5 \pm 0.5 \Omega$
i.e. $R_1 \sim \mathcal{U}(a,a+1)$ and $R_2 \sim \mathcal{U}(a+1,a+2)$
I want to find the distribution of R
First, I've computed the pdf of $\frac{1}{R_1}$ and I've found it to be $\frac{1}{x^2}$ on the interval $[\frac{1}{a+1},\frac{1}{a}]$.
An analogous result could be deduced for the pdf of $\frac{1}{R_2}$
I tried then to use convolution for the sum $\tfrac{1}{R_1}+\tfrac{1}{R_2}$ but that's where things get a bit strenuous for me.
I know it should look something like this $h(s)=\int {\frac{dt}{t^2(s-t)^2}} $ but I failed to see on which domain I should integrate this expression (I succeeded in computing the indefinite integral)
Hint: Add the domain information about the density functions as unit step functions to the integration and take the integral on the whole real line. This will give you the correct result. E.g.: $f(x)=\frac{1}{x^2}{\bf 1} _{\{1/(a+1),1/a\}}$