1
$\begingroup$

Let $X,Y$ be a random sample from $N(0,1)$. Obtain the probability distributions for the following:
$$(1)\frac{X-Y}{\sqrt{2}}\qquad\qquad(2)\frac{(X+Y)^2}{(X-Y)^2}$$ also show that $(2)$ has an F-distribution with $(1,1)$ df.

$$X,Y\sim N(0,1)\implies f_{XY}(x,y)=\frac{1}{2\pi}e^{-\frac12(x^2+y^2)}\qquad\text{Joint density function}$$ Now I am confused how to proceed. Because using CDF to find PDF and recognize the distribution is lengthy process. However If I found $\mathbb E\left(\frac{X-Y}{2}\right)\text{ and }Var\left(\frac{X-Y}{2}\right)$ then I saw, $$\mathbb E\left(\frac{X-Y}{\sqrt2}\right)=\frac{1}{\sqrt2}(\mathbb E[X]-\mathbb E[Y])=\frac{1}{\sqrt2}(0-0)=0$$ $$Var\left(\frac{X-Y}{\sqrt2}\right)\stackrel{1}{=}\frac12(Var[X]+Var[Y]+2.0)=\frac12(1+1)=1\quad {}^1Cov(X,Y)=0$$ That's mean $$\left(\frac{X-Y}{\sqrt2}\right)\sim N(0,1)$$ Similar way I can prove $(2)$ is ratio of two $\chi_{1}^2$ variate hence it is F-distribution with $(1,1)df$.


$(1)$ Now what make me confused$(\text{Or can't justify})$ is without any further investigate how could I conclude that $\left(\frac{X-Y}{\sqrt2}\right)\sim N(0,1)$ using just $\mathbb E\text{ and }Var?$
$(2)$ One addition question is I know the MGF of the sum of $n$ independent random variable is the product of their MGF. But is there any similar thing found for the product of $n$ independent random variable$?$ Or the ratio of two independent random variable$?$
Any help will be appreciated. Thanks in advance.

$\endgroup$
1
  • 1
    $\begingroup$ (1) In general if $(X,Y)$ have a joint normal distribution then $aX+bY+c$ has normal distribution (where $a,b,c$ are constants and $(a,b)\neq(0,0)$). This can be proved e.g. by means characteristic functions. Actually you could call this a characteristic property of the normal distribution. This justifies your conclusion. $\endgroup$
    – drhab
    Jan 13 '20 at 9:45
0
$\begingroup$

Say we have $Y$ and $X$ to be independent.

Note that $X+Y$ and $X-Y$ are jointly normally distributed. Now as the covariance between these variables is zero, they are independent (since they are normally distributed random variables). Further, $X+Y\sim\mathcal{N}(0,2)$ and that $X-Y\sim\mathcal{N}(0,2)$. Therefore $\frac{(X+Y)^2}{2}\sim\chi_{(1)}^2$ and $\frac{(X-Y)^2}{2}\sim\chi_{(1)}^2$ implying that $$\frac{(X+Y)^2/2}{(X-Y)^2/2}=\frac{(X+Y)^2}{(X-Y)^2}$$ is the ratio of two independent $\chi^2_{(1)}$ which is distributed that $F$ with with $1$ and $1$ degrees of freedom.

$\endgroup$
4
  • $\begingroup$ That's mean I can directly use the additive property of normal distribution here? Can you help me for $(2)?$ $\endgroup$ Jan 13 '20 at 8:13
  • $\begingroup$ You can use the properties of the normal distribution. The answer I wrote is for (2). part (1) you did by yourself :-) $\endgroup$
    – Math-fun
    Jan 13 '20 at 8:15
  • $\begingroup$ $(2)$ means my last question about MGF. @Math-fun $\endgroup$ Jan 13 '20 at 8:15
  • $\begingroup$ there is no closed form recipe for the mgf of product of ratio or random variables. $\endgroup$
    – Math-fun
    Jan 13 '20 at 8:19

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.