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I have two independent 2-dimensional normal distributions with the same mean vector and different covariance matrices, 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 $C+C'$ ?.

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.

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?

kjetil b halvorsen
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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
  • Applying the right definition of a Normal distribution makes short work of this: just compute the sum of the cumulant generating functions. You don't even need to do the computation, because it suffices to observe the original cgfs are purely quadratic (that's what it means to be Normal!) and therefore so is their sum and then it's immediately obvious what the coefficient of the quadratic term is. – whuber Mar 20 '23 at 20:36

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