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.

  • 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

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.

  • $\begingroup$ 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

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