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The standard exponential density is only defined for $x>0$ and scale parameter $\lambda > 0$ where the density is given by $f(x) = 1/ \lambda * exp(-x/ \lambda)$.

My question is, does a similar density where $\lambda < 0$ and $x < 0$ make any sense and have any interpretation? Similarly, what about the case of $\lambda = 0$?

Thanks!

felix
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    $\lambda=0$ is not allowed because division by zero is not allowed. $\lambda < 0$ would give $f(x)<0$ for all $x$, which is not a probability density function. I believe it would be ok to take $f(x) = \exp(-x/\lambda)/|\lambda|$ for $x<0$, with $\lambda<0$, and I think this would be the density of -1 times an exponential random variable. – mark999 Oct 12 '12 at 20:09
  • Thank you, yes, the $\lambda = 0$ of course makes sense. I think my actual problem is this, say we have a variable x, that is exponentially distributed with $\lambda = 1$. What is the density function of $Y= \lambda X$ when $\lambda \neq 1$? It is obvious when $y > 0$ and $\lambda > 0$, $f(y) = 1/ \lambda * exp(-y/ \lambda)$ . But is the case of $\lambda < 0$ and $y < 0$ defined and sensible?

    In particular a suggested answer is that then the density of y is: $f(y) = 1/(- \lambda) * exp(-y/ \lambda)$ when $\lambda < 0$ and $y < 0$ but I am not sure this is correct.

     – felix Oct 12 '12 at 20:43
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    Yes, that is correct. You can get that by writing the cumulative distribution function $F_Y (y) = P(Y \leq y)$ in terms of $X$ and then taking the derivative to get the density function. – mark999 Oct 12 '12 at 21:08

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Yes, it makes sense if you restrict the support, in which case it is called a truncated exponential distribution, which is a useful exponential family. As mark999 points out, if you don't restrict the support, you won't be able to normalize the distribution.

Neil G
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