# Maximum likelihood observation with a constraint

I have a vector $$\mathbf{x}$$ of $$n$$ real random variables. The random variables can be assumed to follow a multivariate normal or log-normal distribution (assuming $$\mathbf{x}>0$$ is ok). I would like to analytically obtain the probability that the sum of the random variables decreases below a certain threshold, i.e. $${\mathbf{1}^T\mathbf{x}}\leq\sqrt{\mathbf{x}^T\mathbf{Q}\mathbf{x}}$$, where $$\mathbf{1}$$ is a vector $$n$$ ones and $$\mathbf{Q}$$ is a positive definite $$n$$ x $$n$$ real matrix.

I cannot find the distribution of $${\mathbf{1}^T\mathbf{x}}-\sqrt{\mathbf{x}^T\mathbf{Q}\mathbf{x}}$$ or $${\mathbf{1}^T\mathbf{x}}/\sqrt{\mathbf{x}^T\mathbf{Q}\mathbf{x}}$$. Therefore I have started to look for alternative ways to find the probability: Is there any sense to find the maximum likelihood observation s.t. the constraint and then obtain the probability from a cdf? For example, could I apply the normal log-likelihood as follows:

$$\arg\max$$ $$f(\mathbf{x} | \mathbf{\mu},\mathbf{\Sigma}) =\frac{1}{2} [\text{ln}(|\mathbf{\Sigma}|) + \text{n ln}\left (2\pi \right ) + (\mathbf{x}-\mathbf{\mu})^T\mathbf{\Sigma}^{-1}(\mathbf{x}-\mathbf{\mu})]$$

subject to $${\mathbf{1}^T\mathbf{x}}-\sqrt{\mathbf{x}^T\mathbf{Q}\mathbf{x}} \leq 0$$.

If this is nonsense please elaborate why.

• The maximum likelihood is a maximisation in $\theta$ while the constraint is over $\mathbf x$. And a maximisation in $\mathbf x$ returns a curve of lesser dimension than $n$, hence with zero probability. Commented Jan 17, 2021 at 14:05
• Here is a paper on the moments of a ratio of two Normal quadratic forms. Commented Jan 17, 2021 at 14:08

Lets assume $$Q$$ is one of the parameters and you can go on with this optimization. You already constrain your function such that $$1^Tx - \sqrt{x^TQx} \leq 0$$ since the distrb. you end up will satisfy the condition for all values, you will have a empirical prob. of 1. But this will be wrong since you are already forcing it to have prob. of 1.
Here I assuming you obtain the parameters $$\Sigma$$ and $$\mu$$ through MLE then use the CDF of estimated distrb. to find the probabilities.
You need to find the distrb. of $$1^Tx - \sqrt{x^TQx}$$ directly and that is tricky, even in the case you obtain the distrb. there is a good chance there is no close form solution for CDF.
One way I can think of, calculate $$1^Tx - \sqrt{x^TQx}$$ with the data you have, fit a kernel density estimator to values you obtain, then calculate the prob. through CDF of kernel density estimator.