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I have a random variable with an exponential distribution and have solved an inequality to determine the maximum a posteriori rule (where if $x > \alpha$, I will choose hypothesis 1 over hypothesis 2). I doubt my finding, however, because the value I found for alpha is negative (between $-1$ and $0$).

Per my limited understanding of the exponential distribution, $x$ must be greater than $0$. Is this correct, or is it possible that $x$ can be negative?

I found what appears to be a similar question. I don't follow the entire conversation, but it seems to suggest it is possible that $x$ can be negative (link). I'd appreciate any thoughts on this.

The random variable $x$ that I refer to can be seen in the exponential PDF below.

$f(x) = \lambda * e^{-\lambda*x}$

Edit: The original question incorrectly referred to the random variable $x$ as a "parameter."

vintagedeek
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  • $\lambda$ is parameter, and $x$ is random variable. – user158565 Nov 08 '18 at 02:52
  • Note that the only way the posterior probability of $\alpha < 0$ is itself $>0$ is if your prior places nonzero probability on that occurring. In that case, you don't have to have any $x < 0$ for your MAP estimate of $\alpha$ to be $<0$; consider the trivial case of a prior with $p(\alpha = -1) = 0.99$ and $p(\alpha = -2) = 0.01$; your MAP estimate of $\alpha$ is assured to be $<0$ regardless of the data. This is a consequence of your prior. You should post your prior to clarify this issue... – jbowman Nov 08 '18 at 04:20
  • Also, how did you calculate a posterior w/o observing $x$? You should know whether there were any $x<0$ or not! – jbowman Nov 08 '18 at 04:20
  • Thanks, a_statistician, I have noted the error and updated my question. jbowman, I calculated the MAP rule for alpha without a specific observed value for $x$. Under hypothesis 1, I have a parameter $\mu$ that is greater than $\lambda$, which is the parameter under hypothesis 2. I plugged in different options for these two parameters, where in each case $\mu$ was greater than $\lambda$. Some of these resulted in a negative $\alpha$ and some resulted in a positive $\alpha$. – vintagedeek Nov 08 '18 at 12:37
  • I think that you should better explain your problem how you got to values $x < 0$ (or $\alpha < 0$) and make this the point of your question. Because $x$ should (without doubt) be greater then 0. You can use a similar expressions that allow $x < 0$ but it will be a different type of distribution (ie, truncated or shifted distribution or distribution for $x<0$). – Sextus Empiricus Nov 08 '18 at 14:02
  • I agree with @MartijnWeterings - and please provide your prior distributions, because the problem may well lie there. – jbowman Nov 09 '18 at 04:56

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