# How does normalizing the response affect likelihood?

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?

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.
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.