Let $X$ and $T_1, \ldots, T_m$ be independent random variables following exponential distributions with parameters $\lambda_X, \lambda_T$, respectively. Let
$$ T_{\rho} = T_1 + \ldots + T_m $$
so that $T_{\rho} \sim \text{Gamma}(m, \lambda_T)$. Finally, define
$$ \rho(X, T) := \frac{Xm}{T_{\rho}} $$
I want to find the PDF (and the CDF) of $\rho$. I assumed the best course of action was to find the CDF and then compute the PDF by derivation. Now, if $F_{\rho}$ denotes the CDF of $\rho$,
$$ \begin{align*} F_{\rho}(y) &= P\left[ \frac{Xm}{T_{\rho}} \leq y \right] \\ &= P \left[ X \leq \frac{yT_{\rho}}{m} \right] \\ &= \int_{\mathbb{R}} P(X \leq \frac{yt}{m})f_{T_{\rho}}(t) ~ dt \\ &= \int_{\mathbb{R}} \left( 1-\exp(-\lambda_X \frac{yt}{m}) \right) \left( \frac{\lambda_{T}^m t^{m-1}\exp(-t\lambda_{T})}{(m-1)!} \right) ~ dt \\ &= \int_{\mathbb{R}^+} \left( \frac{\lambda_{T}^m t^{m-1}\exp(-t\lambda_{T})}{(m-1)!} \right) ~ dt - \int_{\mathbb{R}^+} \left( \frac{\lambda_{T}^m t^{m-1}\exp(-t\lambda_{T})\exp(-\lambda_X \frac{yt}{m})}{(m-1)!} \right) ~ dt \\ &= 1 - \frac{\lambda_T^{m}}{(m-1)!} \int_{\mathbb{R}^+} t^{m-1}e^{-t\lambda_T - \lambda_X \frac{yt}{m}} ~ dt \\ &= 1 - \frac{\lambda_T^{m}}{(m-1)!} \int_{\mathbb{R}^+} t^{m-1}e^{-t \left( \lambda_T + \frac{y\lambda_X}{m}\right) } ~ dt \\ &= 1 - \frac{\lambda_T^{m}e^{\lambda_T + \frac{y\lambda_X}{m}}}{(m-1)!} \int_{\mathbb{R}^+} t^{m-1}e^{-t} ~ dt \\ &= 1 - \frac{\lambda_T^{m}e^{\lambda_T + \frac{y\lambda_X}{m}}}{(m-1)!} m! \end{align*} $$
I have not been able to confirm that this solution is correct. Is this the correct approach?
Thanks in advance.
You have the right idea, but I think the computation can be simplified.
Let $$F_X(x) = 1 - e^{-\lambda_X x}, \quad x \ge 0$$ be the CDF of $X$. Then clearly, the variable $$Y = \lambda_X X \sim \operatorname{Exponential}(1),$$ and similarly, $$S = \lambda_T T_\rho \sim \operatorname{Gamma}(m,1),$$ hence $$\rho = \frac{m X}{T_\rho} = \frac{m \lambda_T}{\lambda_X} \frac{Y}{S}.$$ This lets us work with $Y$ and $S$ which have simpler densities since their rate parameters are now $1$. To this end, let $W = Y/S$. Then $$\begin{align} \Pr[W > w] &= \Pr[Y > wS] \\ &= \int_{s=0}^\infty \Pr[Y > ws]f_S(s) \, ds \\ &= \int_{s=0}^\infty e^{-ws} \frac{s^{m-1} e^{-s}}{\Gamma(m)} \, ds \\ &= \int_{s=0}^\infty \frac{s^{m-1} e^{-(w+1)s}}{\Gamma(m)} \, ds. \end{align}$$ The integrand almost looks like a gamma density, but there should be a factor of $(w+1)^m$. So we put it in: $$\Pr[W > w] = \frac{1}{(w+1)^m} \int_{s=0}^\infty \frac{(w+1)^m s^{m-1} e^{-(w+1)s}}{\Gamma(m)} \, ds = \frac{1}{(w+1)^m},$$ since now the integrand is the density of a gamma distributed random variable with shape parameter $m$ and rate $w+1$, so it integrates to $1$. So now we have the survival function of $W$, which has the support $W \ge 0$. Consequently, letting $c = m\lambda_T/\lambda_X$ be a constant function of the given parameters, $$\Pr[\rho > w] = \Pr[cW > w] = \Pr[W > w/c] = \frac{1}{(w/c + 1)^m},$$ from which we readily obtain the CDF $$F_\rho(w) = 1 - \frac{1}{(w/c + 1)^m}, \quad w \ge 0,$$ and the PDF $$f_\rho(w) = \frac{m}{c(w/c+1)^{m+1}} = \frac{\lambda_X}{\lambda_T} \left( \frac{\lambda_X}{m\lambda_T} w + 1 \right)^{-(m+1)}, w \ge 0.$$ Thus, $\rho$ follows a Lomax or Pareto Type II distribution with shape $m$ and scale $c$.