The shifted exponential distribution is given by
$$f(t) = \mu \exp [-\mu (t-\theta )]$$
with domain of support given by $t \in [\theta, \infty),\ \theta \geq 0$.
I want to do a change of variable to eliminate the $t - \theta$.
So, Let $x = t - \theta$, then $dx = dt$, and the limits of integration become $0$ and $\infty$. That is, the transformed function becomes
$$f(x) = \mu \exp [-\mu\ x ]$$
with domain of support $x \in [0, \infty)$, which is the exponential distribution.
Is this correct? It seems like I'm made a mistake somewhere...I.e., the $\theta$ disappeared...
Community Wiki answer to close this.
Why, yes, if you shift to eliminate the shift parameter, $\theta$, then the $\theta$ will vanish. That is the point.
So indeed, since this is a linear shift on the axis, whith no distortion of the curve, the required probability density function is obtained via an inline replacement: $t\gets (t_\star+\theta)$ .
$$~f_{T_\star}(t_\star)~{=~f_T(t_\star+\theta) \\[1ex] =~\mu \exp(-\mu(t_\star+\theta-\theta))~\mathbf 1_{(t_\star+\theta)\in[\theta;\infty)}\\[1ex]=~ \mu\exp(-\mu t_\star)~\mathbf 1_{t_\star\in[0;\infty)}}$$
Which is an unshifted exponential distribution, as should be anticipated.
Oh, no, not at all. Since the function is transformed, a transformation should occur.
Well, to be precise, in a linear shift, only the mean changes (it is shifted!), while variance and other centred moments remain untouched, as they describe shape of the curve, which is unchanged by a shift.
$$\begin{align}\mathsf E(T_\star)~&=~\mathsf E(T)-\theta \\[1ex] \mathsf{Var}(T_\star)~&{=~ \mathsf E((T-\theta-\mathsf E(T-\theta))^2)\\[1ex] =~ \mathsf {Var}(T)}\\ \textit{et cetera}~~&\ldots\end{align}$$
That is all.