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Disclaimer, this is my first question/interaction in this forum.

Let's assume I have random variables that are normally distributed. Then, say I take the observations that are greater than the mean, i.e the right half of the observations. So $X \sim \mathcal{N}(\mu,\sigma^2)$. We take all $X > \mu$. Assume $\mu \ge 0$. Let's call these RVs $Z$.

Now imagine a new random variable $Y$. My goal is to take the $Z$ observations as the $\Omega$ of $Y$, and fit its values using MLE assuming an exponential distribution although they are not exponentially distributed as random variables, I have a strong argument that the measure of $Y$ of these sample points is well approximated by an exponential distribution. Assuming this is correct, is this probability space well-defined?

The whole issue is that there is no good way for me to empirically find the measure (let's say using some logistic regression), yet, if we assume an exponential distribution of these sample points it answers a question practically very well. The whole weight of the argument then falls on the assumption of the measure but again I have a strong argument for that to be the case.

If I have mistakes in notation or reasoning please advise.

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  • $\begingroup$ So $Z = \sigma|R| + \mu$ where $R \sim N(0,1)$ is the standard normal? $\endgroup$ Commented Nov 27, 2023 at 17:28
  • $\begingroup$ I think that sounds right. Also, I thought of a potential problem to my approach and then refuted it. Basically MLE is specifically designed to be used for RVs where they are already following some distribution, so me using MLE on RVs not distributed exponentially will not guarantee that the created measure distributes these sample points as I would hope. However, empirically it seems the lowest observations of $Z$, in the cdf function have a cumulative probability of zero, and the biggest observations have a probability approaching 1. This is perfect for what I intend to accomplish. $\endgroup$ Commented Nov 28, 2023 at 8:18

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