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I have a vector of experiment outcomes, $Q$, and I assume that $Q_i$ are generated by a Gaussian distribution, i.i.d., such that the likelihood is the standard $$\mathcal{L}(q_1, ..., q_n) = \prod_{i=1}^{n} f_{Q_i}(q_i)= \frac{1}{(2\pi\sigma)^{n/2}} \exp{\left\{-\sum_{i=1}^{n} \frac{(q_i - \mu_i)^2}{2\sigma^2} \right\}} $$ In my application, we normalize $Q$ to be percentage of initial as such: $$q = \left[ \frac{Q_{1}}{Q_1}, \frac{Q_{2}}{Q_1}, ..., \frac{Q_n}{Q_1} \right]$$ so that a typical sample usually looks something like $q = [1.00, 0.98, 0.96,...]$. I should note that $Q$ is almost always decreasing, so the Gaussian assumption is not the best, but that's a separate issue.

QUESTION: How does this process affect the likelihood?

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There are two cases here:

  1. You still have the value of Q1 salted away somewhere. In that case, reconstitute the original data set and analyse it as the nice, independent set of normal variates that it is. The values of $\mu$ and $\sigma$ that maximize the original data will also maximize the likelihood of the transformed data.
  2. You have discarded the value of Q1 and are stuck with the "normalized" observations. You should stop doing this because it really doesn't help.

Under case 2, your likelihood will be based on the joint distribution of $[q_2, \ldots, q_n]$, which is going to be ugly. The $q_i$ are not independent. Marginally, each one of them is a Gaussian ratio distribution , but collectively, it's going to be more complicated than that.

People normally transform data so that the transformed numbers are more normal than the raw. This is the reverse.

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  • $\begingroup$ Thank you for the response. Thankfully, I am in situation 1: I figured that it would be the case that the maximum is still attained at the MLEs. Now, I just have to convince my boss that the Gaussian assumption is simply unfounded. $\endgroup$ – call-in-co Aug 13 '15 at 1:59
  • $\begingroup$ Good luck with your boss. How about plotting a histogram? With overlayed density? $\endgroup$ – Placidia Aug 13 '15 at 2:01
  • $\begingroup$ He doesn't even know what a density is. I'm slowly teaching, but I want to make sure that my statistical intuition is correct. $\endgroup$ – call-in-co Aug 13 '15 at 2:05

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