Suppose $x_1 \cdots x_n$ are iid sample from the double-exponential distribution with density: $$p_{\theta}(x) = (2\theta)^{-1}e^{-|x|/\theta}$$ It is obvious from the factorization theorem that $t(x) = \sum_i|x_i|$ is sufficient. How do I find the conditional distribution of the data given $T = t$?
I had thought of finding the gamma density that equates to the exponential distribution:
$$T = \frac{\beta^{\alpha}}{\Gamma(\alpha)}(t^{\alpha-1}e^{-|t|\beta})$$
Such that $\beta = \theta^{-1}$ and $\alpha=1$
$$\implies T = \theta^{-1}(e^{-|t|\theta^{-1}})$$
$$P(X_1=x_1 \cdots, X_n = x_n | T=t) = \frac{\prod_{i=1}^n(2\theta)^{-1}e^{-|x|/\theta}}{\theta^{-1}(e^{-|t|\theta^{-1}})}$$
However I cannot get $T=t$ free of $\theta$