# Proving that the $\frac {\xi +\zeta\eta}{\sqrt {1+\zeta^2}}$ has normal distribution (0,1) [duplicate]

The task: $$\xi, \eta, \zeta \sim N(0,1)$$ and independent. Prove, that $$\frac {\xi +\zeta\eta}{\sqrt {1+\zeta^2}} \sim N(0,1).$$ (1)

It is clear, that with fixed $$\zeta$$ we get, that (1) has expected value = 0 (as the sum of normal distributed values) and variance = 1 (as the sum of $$(\frac {1}{\sqrt {1+\zeta^2}})^2$$ and $$(\frac {\zeta}{\sqrt {1+\zeta^2}})^2$$). And what to do with un-fixed value I don't know. There was a small tip -imagine, that $$\zeta$$ is discrete value (for example getting 3 different values) and use the full probability formula $$(P(B)=\sum P(B|A_{j})P(A_{j}))$$.

## marked as duplicate by StubbornAtom, YuiTo Cheng, Lord Shark the Unknown, Lee David Chung Lin, CesareoMay 31 at 10:03

• You need to assume that $\xi, \eta, \zeta$ are independent... – David C. Ullrich May 30 at 14:35
• Calculate the characteristic function $E\exp(itR)$ of your ratio $R$ by conditioning on $\zeta$ and noticing that the conditional expectation does not depend on $\zeta$. – kimchi lover May 30 at 15:01
Let $$R=\frac{\xi+\zeta\eta}{\sqrt{1+\zeta^2}}$$. You want to show $$E\exp(itR)=\exp(-t^2/2)$$. Write $$E\exp(itR)=E(E[\exp(itR)|\zeta])$$. The inner, or conditional expectation is \begin{align*}E[\exp(itR)|\zeta]&=\tag{*} E\exp(it\xi/\sqrt{1+\zeta})|\zeta) \times E\exp(it\eta/\sqrt{1+\zeta})|\zeta)\\ &= \exp(-\frac{t^2}{2(1+\zeta^2)}) \exp (-\frac{t^2\zeta^2}{2(1+\zeta^2)})\\ &= \exp(-\frac{t^2}2). \end{align*} The first step, at (*), is because $$\xi$$ and $$\eta$$ are conditionally independent given $$\zeta$$. So the outer expectation is also $$\exp(-t^2/2)$$.