Let $Y_k,k=1,\dotsc,n$ be iid RVs. Given $\vartheta$ each $Y_k$ has pdf $f_{Y_k|\vartheta}(y_k|\theta)=\frac{\theta}{2}e^{-\theta |y_k|}$. $\vartheta$ has the marginal pdf $f_\vartheta(\theta)=\frac{1}{\theta}$ on $[1,e]$ and $0$ otherwise. Prove or disprove: $Y_k|\vartheta,k=1,\dotsc,n$ are independent.
I am not sure if there is enough information here. I have found out that $Y_k$ is not independent of $\vartheta$ since with some multiplication and integration: $f_{Y_k,\vartheta}(y,\theta)=\frac{1}{2}e^{-\theta|y|}$ but $f_{Y_k}(y)= \frac{e^{-|y|}-e^{-e|y|}}{2|y|}$. This precludes taking an easy route.
My difficulty is that I don't know how to find for example $f_{Y_1,Y_2|\vartheta}(y_1,y_2|\theta)$ to compare it to $f_{Y_1|\vartheta}(y_1|\theta)f_{Y_2|\vartheta}(y_2|\theta)$.
If some additional minor assumptions need to be put in place to proceed that is OK.
Consider that if they were independent conditional on $\theta$, it would be impossible that they were unconditionally independent.
Basically, knowing the values of a subset of the $Y_i$ would allow us to do inference on $\theta$ via the conditional distribution. Our knowledge of that subset certainly changes our distributional outlook for the unobserved subset, since if we estimate a large $\theta\approx e,$ we would expect smaller values for the unobserved $Y$ then we would had we not observed the other $Y$ (since we know they're i.i.d. conditional on theta, which we have a distribution for from our inference). This means that the $Y_i$ are unconditionally dependent on each other.