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Say I have the following : $$ (X, Y) \sim N_2(\mu, \Sigma) $$

Then what would be the distribution of $(2X,2Y)$ ?

Let $\Sigma = \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2\\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix}$

I think $\rho$ shouldn't change since : $$ corr(๐‘Ž๐‘‹+๐‘,๐‘๐‘Œ+๐‘‘)=corr(๐‘‹,๐‘Œ) $$ and that I should use the univariate transformation for the sigmas, so it would be : $$ (2X, 2Y) \sim N_2(2\mu, 4\Sigma) $$ but can't prove it, and my simulations don't seem to back it. Does anybody have a clue about this ? I can't find anything on the internet about multiplying a bivariate gaussian random variable with a constant

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  • $\begingroup$ The random variable setting is a red herring: this is just a change-of-units problem, no different than converting pecks to gallons. When you view it this way you should immediately be able to write down the answer. $\endgroup$
    – whuber
    Feb 12, 2020 at 15:21

1 Answer 1

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First, they'll be still jointly normal because this is a linear transformation. The rest is pretty straightforward. Each entry in the covariance matrix and the mean vector can be calculated individually, e.g. $$\operatorname{var}(2X)=4\operatorname{var}(X), \operatorname{cov}(2X,2Y)=4\operatorname{cov}(X,Y),E[2X]=2E[X]$$ So, your result is correct.

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