0
$\begingroup$

I know that the sum of $k$ independent exponentially distributed random variables each with density function: $$\displaystyle \lambda\,{{\rm e}^{-\lambda\,x}}$$ has an Erlang distribution: $$\displaystyle {\frac {{\lambda}^{k}{x}^{k-1}{{\rm e}^{-\lambda\,x}}}{ \left( k-1 \right) !}} $$ What would the sum density function look like for a sum of k independent exponentially distributed random variables, each with a density function: $$\displaystyle \lambda\,{{\rm e}^{-\lambda\,(x-a)}},$$ where $a$ is a shift parameter.

$\endgroup$
2
  • 2
    $\begingroup$ I am convinced that a small amount of thinking can lead you to the answer. $\endgroup$
    – Xi'an
    May 31, 2021 at 9:30
  • 2
    $\begingroup$ Perhaps ask yourself, what is the distribution of $X-a$ here? $\endgroup$
    – Ben
    May 31, 2021 at 11:17

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.