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Mi question is the following.

I have two independent 2-dimensional normal distributions with the same mean vector and different covariance matrixes, lets say $X_1 \sim N_2( \mu, C)$ and $X_2 \sim N_2(\mu, C')$.

How can I prove that the random vector $X_1-X_2$ is itself normal too, with the zero vector as the mean (that's so easy) and with its covariance matrix given by $C+C'$ ?.

Thanks people!!!.

I suppose using the convolution Mathew has mentioned is similar to treat with moment-generating functions (show how the moment-generating function of the difference of 2-dimension normal distribution is the moment-generating function of the variable wanted ) and i have got it through this approach.

However, I think it's easier to handle the issue with Dilip approach (now I realise what was the key of the problem hehehe) , since it does not require using further functions, only some basic algebra results, making it more intuitive.

So, thank you to both :).

An additional question is what do you think is the more convenient approach to solve the question in my end-of-degree project, the first or the second?

gung - Reinstate Monica
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Angel Lopez Oriona
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    This is a sit down and calculate situation! The outline is that that the PDF of a difference of two random variables is the convolution (https://stats.stackexchange.com/questions/331973/why-is-the-sum-of-two-random-variables-a-convolution) of the individual pdfs. You would have to write down this convolution, and fuss with the integrals until it looks like the PDF of N(0, C+C'). – Matthew Drury Jan 04 '19 at 16:52
  • Hint: $(X_1,X_2)$ is a random vector whose distribution is a 4-dimensional normal distribution with mean vector $(\mu,\mu)$ and covariance matrix $$C = \left[\begin{matrix}\mathbf C & \mathbf 0\\ \mathbf 0 & \mathbf C^\prime\end{matrix} \right]$$ and $(X_1,X_2) \to X_1-X_2$ is a _linear_ map from 4-dimensional space to 2-dimensional space. – Dilip Sarwate Jan 04 '19 at 19:06

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