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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.