Definition of $\text{Exp}(\theta,1)$ distribution

Question

For one of my assignments, I have to work with an exponential distribution “$\text{Exp}(\theta,1)$”, where the pdf is given as $f(x,\theta)=e^{-(x-\theta)}$, $x>\theta$. I recognize this as a location exponential distribution, but where does the number $1$ come in, in terms of the pdf, and why use the notation $\text{Exp}(\theta,1)$? More generally, what is the pdf of an $\text{Exp}(\theta,c)$ distribution, where $c$ is any constant?

Answer 1

There are two general conventions. When you take $\theta=0$ it's usual to say that if $X\sim \operatorname{Exp}(\lambda)$ (or $\mathcal{E}(\lambda)$), then a density for $X$ is $$f_X(x)=\lambda e^{-\lambda x}\quad (x>0).$$ Here $E(X)=\tfrac1\lambda$ and $var(X)=\tfrac1{\lambda^2}$. If you want the parameter to be the expectation then is usual to say that $X\sim \mathcal E (\beta)$ if it has a density $$f_X(x)=\frac1\beta e^{-\tfrac x \beta}\quad (x>0).$$

Depending on the convention you follow for the scale/rate parameter you would say that $X\sim \operatorname{Exp}(\theta, \lambda)$ if a density is given by $$f_X(x)=\lambda e^{-\lambda (x-\theta)}\quad (x>\theta),$$ or that $X\sim \operatorname{Exp}(\theta, \beta)$ if a density is given by $$f_X(x)=\frac1\beta e^{-\tfrac {x-\theta} \beta}\quad (x>\theta).$$

For sure there is ambiguity, since $\operatorname{Exp}(3,2)$ could mean both that the density is $$f_X(x)=2 e^{-2 (x-3)}\quad (x>3),$$ or that $$f_X(x)=\frac12 e^{-\tfrac {x-3} 2}\quad (x>3).$$

If $\lambda$ is explictly mentioned, then the first convention is almost for sure in use.

Answer 2

In general "$\text{Exp}(\lambda)$" has PDF $\lambda e^{-\lambda x}$ so I suspect "$\text{Exp}(\theta, \lambda)$" has PDF $\lambda e^{-\lambda (x-\theta)}$.

Answer 3

By $\text{Exp}(\theta,c)$ I think you mean an exponential random variable $X_{\theta,c}$ with intensity $c>0$ and location parameter $\theta\in\mathbb R$. To construct it you can define its cdf by $$ \mathbf P[X_{\theta,c}\le x]:=1-e^{-c(x-\theta)},\quad x>\theta. $$ Then you can check that that it is indeed a cdf, as it is increasing and right-continuous in $x$, and you can check the limits $\mathbf P[X_{\theta,c}=\theta]=0$ and $\mathbf P[X_{\theta,c}< \infty]=1$. The pdf can then be computed as follows $$ \frac{d}{dx}\mathbf P[X_{\theta,c}\le x]=ce^{-c(x-\theta)}. $$