# Obtain the probability distributions for $\frac{X-Y}{\sqrt2}$ and $\frac{(X+Y)^2}{(X-Y)^2}$

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) 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. 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.
• That's mean I can directly use the additive property of normal distribution here? Can you help me for $(2)?$ Jan 13 '20 at 8:13
• $(2)$ means my last question about MGF. @Math-fun Jan 13 '20 at 8:15