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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.

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    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Commented Jan 17, 2021 at 14:05
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    $\begingroup$ Here is a paper on the moments of a ratio of two Normal quadratic forms. $\endgroup$
    – Xi'an
    Commented Jan 17, 2021 at 14:08

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you can only constrain the parameters or functions of parameters you are trying to optimize, you cannot really constrain something that is observed.

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.

Note that kernel density estimator has its own parameters to optimize.

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