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Let's say one has a univariate Gaussian distribution with mean $\mu = 0$ and standard deviation $\sigma$.

It is easy to see that the distance from $\mu$ to +1$\sigma$ is...well...$\sigma$.

Let's now say we have a bivariate Gaussian distribution with

$\vec{\mu} = {[ 0 \space 0]^T}$ and covariance matrix $\Sigma = \begin{bmatrix} \sigma_{x,x}=var(X) & \sigma_{y,x}=cov(Y,X) \\ \sigma_{x,y}=cov(X,Y) & \sigma_{y,y}=var(Y) \end{bmatrix} $

where cov(X,Y) $\ne 0 $, and because cov. matrices are always symmetric, cov(X,Y)=cov(Y,X)

I am wanting to calculate the distance d (shown in picture)enter image description here

I understand that this distance is precisely the largest eigenvalue of the covariance matrix. Is there another method that one can use to determine d?

Update 1:
One way of doing this -I believe- is by finding the largest eigenvalue of the covariance matrix. I am looking for a method that does not use the eigenvalues of the covariance matrix.

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    $\begingroup$ You refer to "another" method. Which method are you currently using to find $d$ and why do you need another? $\endgroup$ – whuber Mar 24 '17 at 19:51
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    $\begingroup$ I don't follow: your question asks for an eigenvalue of a matrix, but you would like a method that doesn't use eigenvalues?? $\endgroup$ – whuber Mar 24 '17 at 20:53
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    $\begingroup$ are you missing a correlation coefficient in $\Sigma$? What you've got there is rank 1 so your problem reduces to the univariate case $\endgroup$ – jld Mar 24 '17 at 22:28
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    $\begingroup$ @anr what whuber is explaining is that since the solution to this problem is finding the first eigenvalue, then any solution to this problem is a way of finding eigenvalues. It is like asking "how can I calculate $1+1$ without using addition?" $\endgroup$ – Neil G Mar 25 '17 at 4:14
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    $\begingroup$ This question would be a lot better if you explained why you don't want to calculate eigenvalues in the usual way so that we can propose solutions that fit your constraints. $\endgroup$ – Neil G Mar 25 '17 at 4:21

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