1
$\begingroup$

Question Let $(X,Y)$ be a random point drawn from a two-dimensional distribution. Suppose that $X\cosβ+Y\sinβ\sim N(0,1)$ for any $β∈ \mathbb{R}$. Show that $X$ and $Y$ are independent $N(0,1)$ random variables.


Attempt to Solution Let $Z=X\cosβ+Y\sinβ\sim N(0,1)$, so using mgf,

\begin{align} M_Z(t) & = \exp(1/2(x^2\cos^2\beta+Y^2\sin^2\beta)) \\[8pt] & = \exp\left(\frac{x^2\cos^2\beta}{2}\right) \exp \left( \frac{y^2 \sin^2 \beta}{2}\right) \\[8pt] &=M_X(s)M_Y(t), \\[8pt] \text{and } X & \sim N(0,\cos^2\beta), \quad Y\sim N(0,\sin^2\beta). \end{align}

My confusion

  1. Is my method correct?
  2. Can I conclude $X\sim N(0,1), Y\sim N(0,1)$, since $\beta \in \mathbb{R}$?
$\endgroup$
9
  • 2
    $\begingroup$ How about simply considering what happens for $\beta=0$ and for $\beta=\frac{\pi}2$? $\endgroup$
    – dfnu
    Dec 2, 2019 at 12:23
  • $\begingroup$ @dfnu I get what you mean, but I feel special cases like $0,\frac{\pi}{2}$ may not be strict since $\beta$ may vary. $\endgroup$
    – Chris Tang
    Dec 2, 2019 at 12:47
  • 2
    $\begingroup$ In this way you guarantee that $X$ and $Y$ are "special cases" of $Z$, therefore they have the same distribution. Then as in the answer below, show incorrelation (wich is equivalent to independence, for gaussian r.v.. Why?) $\endgroup$
    – dfnu
    Dec 2, 2019 at 12:52
  • $\begingroup$ @christ your method is not correct since $M_z(t)$ is a function of $t$ and not of $x$ and $y$. Second, if (in some way) you arrive at $X\sim N(0,\cos^2\beta), Y\sim N(0,\sin^2\beta).$, you can not say that $X\sim N(0,1), Y\sim N(0,1)$. $\endgroup$ Dec 2, 2019 at 13:01
  • $\begingroup$ @dfnu I see. First find X and Y's distribution and second prove they are independent. However, I still cannot figure out why we can use covariance to determine the independence. $\endgroup$
    – Chris Tang
    Dec 2, 2019 at 13:11

2 Answers 2

1
$\begingroup$

I add little to Jethro's answer (which I do not understand exactly why was downvoted), but I want to put some order to the discussion in comments.

From the hypothesis we can conclude the following.

  1. Taking $\beta = 0$ and $\beta = \frac{\pi}2$ leads to $X\sim N(0,1)$, and $Y\sim N(0,1)$, respectively.
  2. For any $a,b\in \Bbb R$, $aX+bY \sim N(0,a^2+b^2)$. In fact we have $$aX+ bY=\sqrt{a^2+b^2}(X\cos\beta + Y\sin \beta)=\sqrt{a^2+b^2}Z,$$where $$\beta=\arctan \left(\frac ba\right).$$Therefore $(X,Y)$ have a bivariate normal distribution.
  3. In particular $W=X+Y\sim N(0,2)$, implying $\mbox{E}\left[W^2\right]=2$. Hence the result, already shown, that $$\mbox{Cov}(X,Y)=\mbox{E}[XY]=\frac12\mbox{E}\left[W^2-X^2-Y^2\right]=0.$$Thus $X$ and $Y$ are uncorrelated.
  4. Uncorrelation and 2. guarantee independence. $\blacksquare$
$\endgroup$
4
  • $\begingroup$ Should it be that $\beta = \arctan(b/a)$? It looks like using $a/b$ instead gives $aY + bX$ but I'm unsure. $\endgroup$
    – Slade
    Dec 2, 2019 at 21:46
  • $\begingroup$ Correct, I'll amend it. Thanks @Slade $\endgroup$
    – dfnu
    Dec 2, 2019 at 21:51
  • $\begingroup$ Or $\beta=\operatorname{atan2}(b,\,a)$, if you want to be a stickler. $\endgroup$
    – J.G.
    Dec 2, 2019 at 21:54
  • $\begingroup$ @J.G. yes, definitely that is even better... $\endgroup$
    – dfnu
    Dec 2, 2019 at 21:57
1
$\begingroup$

We just need to show that $\text{Cov} (X,Y)=0$. To do this, note that when $\beta = \pi/4$, we have $\frac 1{\sqrt 2} (X+Y)\sim N(0,1)$, so $E[(X+Y)^2/2]=1, E(X+Y)^2=2$.

$$ \text{Cov} (X,Y)=E(XY)-E(X)E(Y)\\ =\frac12 E[(X+Y)^2-X^2-Y^2]-0\times 0\\ =\frac{1}{2}\times (2-1-1)=0. $$ So, they are independent.

Note I have used the fact that $E(X)=E(Y)=0, E(X^2)=E(Y^2)=1$.

$\endgroup$
2
  • 1
    $\begingroup$ Your way of calculating $E(XY)$ inspires me. However, based on my knowledge, $Cov(X,Y)=0$ cannot conclude $X,Y$ are independent. $\endgroup$
    – Chris Tang
    Dec 2, 2019 at 12:49
  • 2
    $\begingroup$ @christ : It is true that this answer falls somewhat short of proving independence, but perhaps not as far short as you seem to suggest. There is a theorem stating that if the pair $(X,Y)$ is so distributed that for every pair $(a,b)$ of constants $\text{(“constant''} = \text{not random})$ the linear combination $aX+bY$ is normally distributed, then $X,Y$ are independent if their covariance is $0. \qquad$ $\endgroup$ Dec 2, 2019 at 20:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .