Thanks for your answer @Aleksei. So you used the fact that for O orthonormal we have ON(0,I) = N(0,OIOT) =  N(0,I). This says that y=Ox has the same distribution as x. While E[yyT /(yTy)] = E[OxxTOT/(xTx)] = OE[xxT /(xT x)] OT. But as y has the same distribution as x, we have OE[xxT /(xTx)] OT=E[xxT/(xTx)].

Hi Remi, thanks for answering, but I don't what you mean by "a one dimensional random variable with n entries". It seems like a oxymoron.  A one dimensional r.v. has only one entry, in which case, the desired identity is trivially E[1] = 1.

Sorry to come in so late, but my recent retirement caused a change of email address such that I have been missing notifications from ResearchGate.

Anyway, I agree with Remi, including his use of the phrase “one dimensional array”, since in all the programming languages I know, that means a vector (as opposed to a scalar or a higher-dimensional array). Still, just calling it a “vector” would have been my preference. On the other hand, the phrasing of the original question casts a random n-vector into the form of an nxn matrix, apparently to facilitate the definition of an n-vector whose elements are all the square of a zero-mean unit-variance Gaussian random variable. The matrix defined is diagonal, so it is equal to its own transpose.

I see a potential problem with the definition of y, however. It is a function of x, not x^2, and it is not clear what is meant by ||x||. If that means the determinant of x or its expectation value, then since the elements of x are zero-mean, the expectation value of the determinant is zero, making y singular. The determinant of x is the product of n zero-mean unit-variance Gaussian random variables and will have a highly nontrivial density function. For example, for n=2, the determinant is the product of 2 zero-mean unit-variance Gaussian random variables, for which the density function is a modified Bessel function of the second kind and order zero to within a scale factor, sharply peaked at zero, the expectation value.

On the other hand, if ||x|| is some innocent nonzero constant, then y is just a rescaled random n-vector with elements of N(0,1/||x||^2), so its variance is just an n-vector with all elements equal to 1/||x||^2.

The question however refers to the covariance of y, not the variance, suggesting that the elements of x might be correlated. Unless they are uncorrelated, the complete covariance matrix for y cannot be computed without a specification of the x covariance matrix. But usually the lack of such a specification is taken to mean that the elements of x are uncorrelated.

Hope I didn’t completely misinterpret the question!