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  1. Let $X$ be a uniform random vector. Is any linear transformation $AX$ of $X$ still uniformly distributed? I know it is yes when $A$ is square and invertible, by using the change of variable formula. But not sure when $A$ is square and not invertible, or $A$ is not square.
  2. If $X$ and $Y$ are both uniform random variables (or vectors), will $(X^T,Y^T)^T$ be a random vector?

Thanks!

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Since you say "I know it is yes when..." could you include a proof, or tell us where you read, that if $X$ and $Y$ are uniformly distributed, then for any invertible $2\times 2$ matrix $A = [a_{i,j}]$, the random variables $a_{1,1}X + a_{1,2}Y$ and $a_{2,1}X + a_{2,2}Y$ are uniformly distributed? –  Dilip Sarwate Oct 25 '12 at 1:54
    
$AX$ is a uniform random vector and its components are uniform random variables are not equivalent. @DilipSarwate –  Tim Oct 25 '12 at 2:04
    
So could you enlighten me as to what the difference is? In the top hit on Google for "uniform random vector" it says "Hi all, I'm trying to generate uniform random vectors with $n$ dimensions. To be more precise, each of the elements of the vector must be a uniform distributed variable in $[0,1].$" –  Dilip Sarwate Oct 25 '12 at 2:17
    
Sure. a random vector that takes values within a circle, and whose density function is constant within the circle. –  Tim Oct 25 '12 at 2:19
    
$A = \begin{bmatrix}2&0\\0&1\end{bmatrix}$ is a square invertible matrix. So are you saying that if the random point $(X,Y)$ is uniformly distributed on the interior of a circle, then the random point $(2X,Y)$ also is uniformly distributed on the interior of a circle? What is the radius of this circle? –  Dilip Sarwate Oct 25 '12 at 2:55

1 Answer 1

up vote 1 down vote accepted

Try the case $A = \pmatrix{1 & 1\cr 0 & 0\cr}$ or the $1 \times 2$ matrix $A = (1,1)$.

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Thanks! (1) The examples explain the answer to the first part is no. So I guess the correct statement should be any invertible linear transformation of a uniform random vector is still uniformly distributed? (2) How about the second part? I guess it is no too, because it seems possible to construct a distribution in a 2D shape, such that the projections to both axes will be uniform distributions. –  Tim Oct 25 '12 at 0:37

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