Let $X_1,\ldots,X_n$ be i.i.d random variables with common PDF $f(x\mid\theta)$ of the shifted exponential distribution with parameter $\theta$. (a) Show that if $Z\sim\mathrm{Exp}(1)$ and if $X = Z + \theta$ for some constant $\theta$, then the distribution of $X$ is a shifted exponential.
What I did was this:
$f(x\mid\theta) = f(z+\mid\theta)= e^{[-(z+\theta-\theta)]}= e^{-z}$. Since $X=Z+\theta$, this implies that $Z=X-\theta$. So, we have $e^{[-(x-\theta)]}$ which the shifted exponential since the support is the same also.
Is this correct? Thanks in advance!
For $t>0$ we have \begin{align} \mathbb P(X>t) &= \mathbb P(Z+\theta>t)\\ &= \mathbb P(Z>t-\theta)\\ &=e^{-(t-\theta)}\mathsf1_{\{t>\theta\}}, \end{align} so that $X$ has a shifted exponential distribution. Note that the inclusion of the indicator $\mathsf 1_{\{t>\theta\}}$ is essential, as otherwise we would have $\mathbb P(X>t)>1$ for $t<\theta$, which is not possible for a probability distribution.