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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?2012-10-25
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    $AX$ is a uniform random vector and its components are uniform random variables are not equivalent. @DilipSarwate2012-10-25
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    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 $[0,1].$"2012-10-25
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    Sure. a random vector that takes values within a circle, and whose density function is constant within the circle.2012-10-25
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    $A = \begin{bmatrix}2&0\\0&1\end{bmatrix}$ 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?2012-10-25
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    "then the random point (2X,Y) also is uniformly distributed on the interior of a" ELLIPSE.2012-10-25
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    -1 So let me ask again, since _your_ definition of _uniform random vector_ is a random point $(X,Y)$ uniformly distributed on the interior of a _circle_, not any other closed curve, would you _please_ **either** provide a proof of your assertion in the question that if $A$ is a $2\times 2$ square invertible matrix, that $(X,Y)A$, or $A\begin{bmatrix}X\\Y\end{bmatrix}$ if you like that notation better, is a uniform random vector as per _your_ definition, **or** edit your question to state clearly exactly what you mean by a uniform random vector and how you "know" that a certain answer is "yes"?2012-10-25
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    @DilipSarwate: by "[using the change of variable formula](http://en.wikipedia.org/wiki/Probability_density_function#Dependent_variables_and_change_of_variables)" mentioned in my post. Because the original random vector has a constant density function and the Jacobian is constant, the transformed random vector has a constant density function.2012-10-25
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    Oh dear! Whatever happened to "Sure, a random vector that takes values within a **circle**, and whose density function is constant within that **circle**"?? (emphasis added)2012-10-26
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    @DilipSarwate: Do you have a real question?2012-10-26
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    Yes, I am asking what your definition of uniform random vector is. You gave one definition and then a different one. Which one is it?2012-10-26

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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.2012-10-25