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Lets suppose that we have a 2D Gaussian with zero mean and one covariance and the equation looks as follows

$$f(x,y) = e^{-(x^2+y^2)}$$

If we want to rotate in by an angle $\theta$, does it mean that we rotate the values $x$ and $y$ and then see how the Gaussian is rotated or do we actually rotate the graph of the function.

How this rotation actually be computed analytically and how the graph would look like. Is there any intuitive way of understanding.

How do we explain rotating a general function analytically and geometrically?

Thanks a lot

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Your function is rotationally symmetric about the origin, so the most immediate interpretation of "rotate it by an angle $\theta$" will leave it completely unchanged. – Henning Makholm Jun 23 '12 at 16:16
@Henning, Thanks a lot, I wanted to know things in general, Gaussian is just an example... What if function is not symmetric? – Shan Jun 23 '12 at 16:25
If the covariance is the identity matrix, then it's $e^{-(x^2+y^2)/2}$ (and the whole thing is then multiplied by a normalizing constant). You're missing the factor of $1/2$. – Michael Hardy Jun 23 '12 at 17:18
@Michael, Thanks, you are absolutely right... I just wanted to put a simple function and the constant does not matter that much... What I want to exactly how to get a closed form expression of a function after the rotation... +1 for pointing it out. – Shan Jun 23 '12 at 17:26
up vote 4 down vote accepted

If the covariance is the $2\times2$ identity matrix, then the density is $$e^{−(x^2+y^2)/2}$$ multiplied by a suitable normalizing constant. If $\begin{bmatrix} X \\ Y \end{bmatrix}$ is a random vector with this distribution, then you rotate that random vector by multiplying on the left by a typical $2\times 2$ orthogonal matrix: $$ G \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} X \\ Y \end{bmatrix}. $$

If the question is how to "rotate" the probability distribution, then asnwer is that it's invariant under rotations about the origin since it depends on $x$ and $y$ only through the distance $\sqrt{x^2+y^2}$ from the origin to $(x,y)$.

If you multiply on the left by a $k\times2$ matrix $G$, you have $$ \mathbb{E}\left(G\begin{bmatrix} X \\ Y \end{bmatrix}\right) = G\mathbb{E}\begin{bmatrix} X \\ Y \end{bmatrix} $$ and $$ \operatorname{var}\left( G \begin{bmatrix} X \\ Y \end{bmatrix} \right) = G\left(\operatorname{var}\begin{bmatrix} X \\ Y \end{bmatrix}\right)G^T, $$ a $k\times k$ matrix. If the variance in the middle is the $2\times2$ identity matrix and $G$ is the $2\times 2$ orthogonal matrix given above, then it's easy to see that the variance is $$ GG^T $$ and that is just the $2\times 2$ identity matrix. The only fact you need after that is that if you multiply a multivariate normal random vector by a matrix, what you get is still multivariate normal. I'll leave the proof of that as an exercise.

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@Micheal, great answer, thanks! – Shan Jun 23 '12 at 18:31

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