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Here is the definition of joint normality in my textbook.

Def: Two random variables $X$ and $Y$ are said to be jointly normal if they have the joint density $$f_{X,Y}(x,y)\\=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left\{-\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_1)^2}{\sigma_1^2} - \frac{2\rho(x-\mu_1)(y-\mu_2)}{\sigma_1 \sigma_2} +\frac{(y-\mu_2)^2}{\sigma_2^2} \right] \right\},$$

where $\sigma_1 >0, \sigma_2>0,|\rho|<1,$ and $\mu_1,\mu_2$ are real numbers.

My textbook states without proof that this definition is equivalent to the statement that linear combinations of jointly normal random variables are still jointly normal.

I tried to google a proof for this equivalence but I didn't manage to find one. Can anyone help me with proving the $(\Longrightarrow)$ direction? Thanks.


Textbook page:

Related textbook page

Stochastic calculus for finance II Continuous time models, Steven E. Shreve.

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  • $\begingroup$ It is easier to use characteristic functions instead of density functions. $\endgroup$ Feb 8 at 10:02
  • $\begingroup$ @KaviRamaMurthy Can you explain a bit more? Thanks. But unfortunately I cannot choose the definition. $\endgroup$
    – Sam Wong
    Feb 8 at 10:08
  • $\begingroup$ the pdf above can be rewritten as $f_{X,Y}(v)=\frac1{2\pi \sqrt{\det \Sigma }}e^{-\frac1{2}(v-\mu )^\top \Sigma ^{-1}(v-\mu )}$ where $\mu:=(\mathrm{E}X,\mathrm{E}Y)$ is the mean of $(X,Y)$ and $\Sigma $ its covariance matrix $\endgroup$
    – Masacroso
    Feb 8 at 10:26
  • $\begingroup$ @Masacroso Thanks for the comment. Yes, I know this vector form of the pdf. But I still don't know how to prove, say $X+Y$ and $Y$ have a joint pdf as $f_{X+Y,Y}(v)=\frac1{2\pi \sqrt{\det \Sigma }}e^{-\frac1{2}(v-\mu )^\top \Sigma ^{-1}(v-\mu )}$, where $\mu=(EX+EY,EY)$ and $\Sigma$ is the covariance matrix of $X+Y$ and $Y$. Can you give me some hints? Thanks. $\endgroup$
    – Sam Wong
    Feb 8 at 10:59
  • $\begingroup$ @SamWong I guess you understood wrongly the statement: $(X+Y,X)$ is not a linear combination of jointly normals. What the statement in reality says that $(X,Y)$ is jointly normal if and only if $aX+bY$ is normal for any chosen $a,b\in \mathbb{R}$, see here $\endgroup$
    – Masacroso
    Feb 8 at 11:28

1 Answer 1

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Sketch for a proof, too long for a comment: I understand that the statement of the book marked in red says that if $X$ and $Y$ are jointly normal then $aX+bY$ and $cX+dY$ are jointly normal also, for arbitrary $a,b,c,d\in \mathbb{R}$. This means that the random variable $(a,c)X+(b,d)Y$ must be multivariate normal, if we follow the definition of jointly normal of wikipedia.

Setting $J:=(a,c)X+(b,d)Y$ this amount to compute it density, given by

$$ \frac{\partial}{\partial s}\frac{\partial}{\partial t}\Pr [J\in (-\infty ,s]\times (-\infty ,t]]=\frac{\partial}{\partial s}\frac{\partial}{\partial t}\int_{\{(x,y)\in \mathbb{R}^2:ax+by\leqslant s, cx+dy\leqslant t\}}f_{X,Y}(x,y)\,d (x,y)\tag1 $$

and show that it is of the desired form. As said in the comments an equivalent condition is easily proved using characteristic functions.

For a direct proof using (1) probably you will need to use some linear algebra, specially knowledge about positive definite matrices, and the theorem of change of variables for the integral.

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  • $\begingroup$ Thanks for the post. What is $(a,c)$? Is it the greatest common divisor of $a$ and $c$? $\endgroup$
    – Sam Wong
    Feb 8 at 13:21
  • $\begingroup$ @SamWong its a vector $(a,c)\in \mathbb{R}^2$ and $(a,c)X$ is scalar multiplication, as $X$ is real valued. That is $(a,c)X+(b,d)Y=(aX+bY,cX+dY)$ $\endgroup$
    – Masacroso
    Feb 8 at 14:02

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