Sufficient Statistic for exponential family Let $\mu, \nu$ be positive and suppose that $X_1,...,X_n$ are exponential with parameter $\mu$, and $Y_1,...,Y_m$ are exponential with paramter $\nu$ and that all variables are independent. Construct the generalized likelihood ratio test for the hypothesis $H_0:  \mu = \nu$ verses $H_1: \mu \neq \nu$ and show that it can be based on the statistic
\begin{equation}
        T(X_1,...,X_n, Y_1,...,Y_m) = \frac{X_1+ \cdots + X_n}{X_1 + \cdots X_n + Y_1 + \cdots Y_m}
    \end{equation}
My approach:
The Likelihood function of the exponential distribution is
\begin{equation}
              L(\mu|x) = \begin{cases}
              e^{-\sum_{i=1}^n x_i + n \mu} & \text{if }\mu \leq \min x_i\\
              0 & \text{if } \mu > \min x_i
              \end{cases}
          \end{equation}
For the denominator of the likelihood ratio, note that $L(\mu|x)$ is an increasing function with respect to $\mu$ as long as $\mu \leq \min x_i$. Thus, the unrestricted supremum of the likelihood function is
\begin{equation}
              L(\min x_i | x) = e^{-\sum_{i=1}^n x_i + n \min x_i}
          \end{equation}
On the other hand, for the numerator of the likelihood ration, since $\mu = \nu$, the supremum is the singular possible value for $\mu$, which is
\begin{equation}
              L(\mu|x) = \begin{cases}
              e^{-\sum_{i=1}^n x_i + n \nu} & \text{if }\nu \leq \min x_i\\
              e^{-\sum_{i=1}^n x_i + n \min x_i} & \text{if } \nu > \min x_i
              \end{cases}
          \end{equation}
Therefore, the likelihood ratio test statistic is
\begin{equation}
              \lambda(x) = \begin{cases}
              e^{n(\nu - \min x_i)} & \text{if }\nu \leq \min x_i\\
              1 & \text{if } \nu > \min x_i
              \end{cases}
          \end{equation}
I know that a sufficient statistic is one whose pdf doesn't depend on the parameter $\mu$. However, I'm not sure how to calculate the pdf of $T$.
 A: There are several issues with your attempt, such as (a) for some reason the $y_j$ do not appear in your work at all, and (b) you seem to not be using the density of the exponential distribution.
You need to consider both $\mu$ and $\nu$ as parameters, and all observations $x_i$ and $y_j$ as your data Specifically,
$$\Lambda(x_{1:n}, y_{1:m}) = \frac{\sup_{\mu, \nu : \mu = \nu} L(\mu, \nu \mid x_{1:n}, y_{1:m})}{\sup_{\mu, \nu} L(\mu, \nu \mid x_{1:n}, y_{1:m})}.$$

Let $S_x = \sum_{i=1}^n x_i$ and $S_y = \sum_{j=1}^m y_j$.
The likelihood of seeing $x_1, \ldots, x_n, y_1, \ldots, y_m$ (for arbitrary $\mu, \nu$) is
$$\prod_{i=1}^n \mu e^{-\mu x_i} \prod_{j=1}^m \nu e^{-\nu y_i} = \mu^n \nu^m e^{-\mu S_x - \nu S_y}.$$
The log likelihood is
$n \log \mu + m \log \nu - \mu S_x - \nu S_y$
which is maximized when $\mu = n/S_x$ and $\nu = m / S_y$. So, the denominator of the likelihood ratio is
$$(S_x/n)^{-n} (S_y/m)^{-m} e^{-(n+m)}.$$
For the numerator where $\mu=\nu$, the likelihood simplifies to $$\mu^{n+m}e^{-\mu(S_x + S_y)}$$ which is maximized when $\mu = \frac{n+m}{S_x + S_y}$. So, the numerator of the likelihood ratio is
$$\left(\frac{S_x + S_y}{n+m}\right)^{-(n+m)} e^{-(n+m)}.$$
Finally, the ratio is
$$n^n m^m (n+m)^{n+m} \left(\frac{S_x}{S_x + S_y}\right)^{-n} \left(\frac{S_y}{S_x + S_y}\right)^{-m}
= n^n m^m (n+m)^{n+m} T_{x,y}^{-n} (1-T_{x,y})^{-m},$$
where $T_{x,y} := \frac{S_x}{S_x+  S_y}$ is the sufficient statistic.
