Let $X_1$ and $X_2$ be independent normal random variables with

$EX_1 = EX_2 = 0$

$Var(X_1) =\sigma_1^2$ and $Var(X_2) =\sigma_2^2$

Let $Y_1 = X_1 + X_2, Y_2 = X_1 - X_2$

Find the joint distribution of $Y_1$ and $Y2$

I know that the sum ($Y_1$) and the difference ($Y_2$) are both normally distributed ~ $N(0, \sigma_1^2+\sigma_2^2)$

However, I am not sure if $Y_1$ and $Y_2$ are independent and how their joint distribution would be if they are not independent. I'd appreciate if anyone can help me.

  • $\begingroup$ You can use a transformation to compute the joint pdf. $\endgroup$ Sep 27 '20 at 0:04
  • $\begingroup$ iid and different variance does not make sense. I think you mean independent and different variances. $\endgroup$ Sep 27 '20 at 0:42
  • $\begingroup$ thanks I just fixed it $\endgroup$
    – H. Ameber
    Sep 27 '20 at 0:46

One way is to use characteristic functions: If $X$ and $Y$ are normal with mean $0$ and variances $A$ and $B$ then $Ee^{it(X+Y)+is(X-Y)}=Ee^{i(t+s)X}Ee^{i(t-s)Y}=e^{-A^{2}(t+s)^{2}/2}e^{-B^{2}(t-s)^{2}/2}$. I will let you simplify this and show that $(X+Y, X-Y)$ is jointly normal.

[The variances are $A^{2}+B^{2}$ and the covariance is $A^{2}-B^{2}$].

  • $\begingroup$ Wouldn’t you need the inverse Fourier transform to obtain the pdf from characteristic functions? And what happened to $i$ after you take the expected value? $\endgroup$
    – Sigma
    Sep 27 '20 at 1:05
  • 1
    $\begingroup$ Do you know how the characteristic functio of a normal distribution looks like? There won't be $i$ in it. And charcateristic functions uniquely determine the distribution. So once you get the characterisitic function in the right form you can conclude that he distribution is normal, No need for any inversion formula. @VictorS. $\endgroup$ Sep 27 '20 at 1:52
  • $\begingroup$ Is having the same variance the only case makes $Y_1$ and $Y_2$ independent? $\endgroup$
    – H. Ameber
    Sep 30 '20 at 0:07
  • $\begingroup$ @H.Ameber Yes, that is true. $\endgroup$ Sep 30 '20 at 0:09
  • $\begingroup$ Is it the only case though? $\endgroup$
    – H. Ameber
    Sep 30 '20 at 0:11

This is not an answer, it just felt too big for a comment:

You are correct in that $Y_1$ and $Y_2$ are not independent, as they are functions of the same random variables. Since $Y_1$ and $Y_2$ are dependent, their joint distribution is not the product of their individual pdfs.

I don't know if I'm correct here but what I would try is

$$ f_{X_1 + X_2, X_1 - X_2}(x_1, x_2) = f_{X_1, X_2}(x_1 + x_2, x_1 - x_2)|J|. $$

Where $J$ is the Jacobian of the bivariate transformation.

  • $\begingroup$ I think you should remove the “correct answer” from my answer because I myself am not sure if I’m correct. You can upvote the answer if you just want to thank the help. $\endgroup$
    – Sigma
    Sep 27 '20 at 1:07

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