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Imagine I have $x\stackrel{d}{=}\vert\xi A_i U\vert$,

where $\xi$ is 1-dimensional, non negative random variable with finite 2nd moment and independent of $U$, $A_i$ is a fixed $n$-dimensional vector and $U$ is uniformly distributed on the $n$-dimensional unit sphere $\mathbb{S}_{n-1}$ (see for a definition here: https://en.wikipedia.org/wiki/Unit_sphere#Unit_spheres_and_balls_in_Euclidean_space - the first equation)

Now I want to calculate the median of it; Or more generally I don't need to calculate it explicitly, but I want to show the following:

Let $x\stackrel{d}{=}\vert\xi A U\vert$, where $\xi$ is defined the same way, U as well but $A$ denotes a symmetric positive definite $n \times n$ matrix;

Then $x$ is also $n$-dimensional; What I want to prove now is:

$$\frac{\text{med}(\vert\xi A_i U\vert)}{dA_iA_i^\top }=\frac{\text{med}(\vert\xi A_j U\vert)}{dA_jA_j^\top }$$ for all $i,j \in \{1, ...,n\}$; d denotes here: $\sqrt{\mathbb E(\xi^2)}$;

But I really have no clue on how to do it - calculating a median is tricky because even if two random variables are independent, the median of the product can't be separated generally;

If you have any ideas - even if they don't lead to the solution immediately, even if they later turn out to be wrong - let me know; I really appreciate any help.

Thank you


For Marcus I try to extend my question so you can see where it is coming from:

I have an elliptical distributed random variable- so I have:

$x\stackrel{d}{=}\mu+\xi A U$,

where $\xi$ is 1-dimensional, non negative random variable with finite 2nd moment and independent of $U$, $A$ is a fixed $n \times n$, symmetric positive definite matrix and $U$ is uniformly distributed on the $n$-dimensional unit sphere $\mathbb{S}_{n-1}$ and is a column vector;

Then $x$ is an $n$-dimensional vector and we have:

$x_i\stackrel{d}{=}\mu+\xi A_i U$,

where $A_i$ denotes the ith row; Now I want the median absolute deviation and the standard error; more precisely I want to prove that:

$$\frac{\text{med}(\vert x_i-\text{med}(x_i)\vert)}{sd(x_i)}=\frac{\text{med}(\vert x_j-\text{med}(x_j)\vert)}{sd(x_j)}$$

Here sd means the square root of the variance and the upper term is just the definition of the median absolute deviation; According to google the median of $x$ is $\mu$, that's why the first term reduces to the median of $\vert\xi A_i U\vert$ - for the standard deviation we use the definition of the variance:

Var$(x_i)=\mathbb E((x_i-\mathbb E(x_i))(x_i-\mathbb E(x_i))^\top)=E((\xi A_i U)(\xi A_i U)^\top)=\mathbb E(\xi^2)E(A_i UU^\top A_i^\top)=\mathbb E(\xi)^2A_iE(UU^\top )A_i^\top=\mathbb E(\xi)^2A_iA_i^\top$

where we used the fact that 1. $\mathbb E(x_i)=\mu_i$ ($\mu$ is median and expectation) 2.independence of $\xi$ and $U$ and that $U*U^\top=1$ by definition of being on the unit sphere;

Then I really need to correct myself; I want to show that:

$$\frac{\text{med}(\vert\xi A_i U\vert)}{d\sqrt{A_iA_i^\top} }=\frac{\text{med}(\vert\xi A_j U\vert)}{d\sqrt{A_jA_j^\top} }$$ for all $i,j \in \{1, ...,n\}$; d denotes here: $\sqrt{\mathbb E(\xi^2)}$;

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    $\begingroup$ A couple of questions: By $A_i U$ do you mean the dot product? Also, $A_i$ and $A_j$ are columns (or rows) of $A$ here? Finally, do you perhaps want $\sqrt{A_i A_i^T}$ in the denominator here? $\endgroup$
    – Marcus M
    Jul 23, 2017 at 23:10
  • $\begingroup$ Sorry; A_i is a row vector, U is a column vector - so it's a regular vector multiplication and we get as a result a 1-dimensional number; Therefore $A_i$ and $A_j$ are rows; And I think they way I wrote it is correct - I extended my question, then you maybe understand where it is coming from $\endgroup$
    – user299124
    Jul 24, 2017 at 0:12
  • $\begingroup$ Okay I was wrong, there is a square root missing - sorry totally forgot to add it; But the $d$ should still be in the denominator $\endgroup$
    – user299124
    Jul 24, 2017 at 0:35

1 Answer 1

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First notice that if you have a real-valued random variable $X$ then $c\cdot\mathrm{med}(X) = \mathrm{med}(cX)$. So what you're looking at is showing $$\mathrm{med}\left(\left|\frac{\xi}{d}\cdot \frac{A_i}{\| A_i \|}U \right|\right) = \mathrm{med}\left(\left|\frac{\xi}{d}\cdot \frac{A_j}{\| A_j \|}U \right|\right). $$

But since both $A_i/\|A_i\|$ and $A_j / \|A_j\|$ are unit vectors, there exists an orthogonal matrix $Q$ s.t. $(A_i/\|A_i\|) Q = A_j / \|A_j\|$. We can then rewrite $$\left|\frac{\xi}{d}\cdot \frac{A_i}{\| A_i \|}U \right| = \left|\frac{\xi}{d}\cdot \frac{A_i}{\| A_i \|}QQ^TU \right| = \left|\frac{\xi}{d}\cdot \frac{A_j}{\| A_j \|}(Q^TU) \right|.$$

But since $U$ is uniform on the unit sphere, $Q^T U \stackrel{d}{=} U$ thereby completing the proof.

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  • $\begingroup$ Genius - so simple after you wrote it...Thank you so much..!!!! May I ask why the median of $x_i=\mu_i+\xi A_iU$ is just $\mu_i$? I mean we have $\text{med}(x_i)=\mu_i+\text{med}(\xi A_i*U)=\mu_i+A_i \text{med}(\xi*U)$ but how can I continue? I mean it seems reasonable that the median of the latter term is 0 but I don't know how to properly prove it $\endgroup$
    – user299124
    Jul 24, 2017 at 16:56
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    $\begingroup$ For the latter median note that $U$ and $-U$ have the same distribution. $\endgroup$
    – Marcus M
    Jul 24, 2017 at 17:08
  • $\begingroup$ Great - thank you; You really helped a lot - can't thank you enough $\endgroup$
    – user299124
    Jul 24, 2017 at 17:10
  • $\begingroup$ @J.Doe Happy to help! $\endgroup$
    – Marcus M
    Jul 24, 2017 at 17:15

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