If $Z\sim\text{Exp}(\lambda)$ and $X$ is an indicator variable s.t. $X=1\iff Z\ge t$ for a given $t\ge 0$, I'm trying to find the likelihood $p(X,Z|\lambda)$.
My calculations were: $p(x,z|\lambda)=p(x|z,\lambda)p(z|\lambda)=e^{-\lambda tx}(1-e^{-\lambda t})^{1-x}p(z|\lambda)=e^{-\lambda tx}(1-e^{-\lambda t})^{1-x}\lambda e^{-\lambda z}$.
However, the solution claims that it is : $p(x,z|\lambda)=\lambda e^{-\lambda z}$
Is this because X is dependent on Z? Which mistake am I making?
Your $p(x\mid z,\lambda)=e^{-\lambda tx}(1-e^{-\lambda t})^{1-x}$ is wrong.
It should not depend on $\lambda$ ($X$ is determined by $Z$ and $t$) and should be $1$ in any possible case $(x=1, z\ge t$ or $x=0,z<t)$ and $0$ otherwise.
This means that in any possible case, if you know the value of $Z$ then knowing $X$ gives you no additional information about $λ$, and so $X$ does not affect the likelihood of $λ$ given $X$ and $Z$. Thus the likelihood of $\lambda$ is proportional to $\lambda e^{-\lambda z}$