# Show that the distribution of $X$ is shifted exponential

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.

$$f(x\mid\theta) = f(z+\mid\theta)= e^{[-(z+\theta-\theta)]}= e^{-z}$$. 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.
• Why are you introducing the $X_i$? It seems that you never use them afterwards? Jan 8, 2020 at 22:06
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.