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!

  • $\begingroup$ Here's a tutorial and reference for typesetting math on this site. $\endgroup$
    – joriki
    Jan 8, 2020 at 22:05
  • 1
    $\begingroup$ Why are you introducing the $X_i$? It seems that you never use them afterwards? $\endgroup$
    – joriki
    Jan 8, 2020 at 22:06
  • $\begingroup$ I didn't introduce this. It is in the question. $\endgroup$
    – lj_growl
    Jan 9, 2020 at 0:29
  • $\begingroup$ Yes, in the question that you posted. If this is not your own question and you copied it from another source, you should state that and state the source. $\endgroup$
    – joriki
    Jan 9, 2020 at 1:18

1 Answer 1


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.


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