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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, Cesareo May 31 at 10:03

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  • $\begingroup$ You need to assume that $\xi, \eta, \zeta$ are independent... $\endgroup$ – David C. Ullrich May 30 at 14:35
  • $\begingroup$ @DavidC.Ullrich, yes, I have forgotten to add this) $\endgroup$ – Matthew5335 May 30 at 14:41
  • $\begingroup$ This indicates it doesn't math.stackexchange.com/questions/1021455/… Perhaps the variables are not independent after all? $\endgroup$ – Benedict W. J. Irwin May 30 at 14:50
  • 1
    $\begingroup$ @BenedictW.J.Irwin, there is an another type of "equation". It has small differences $\endgroup$ – Matthew5335 May 30 at 14:53
  • $\begingroup$ 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 $\zeta$. $\endgroup$ – 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 $\exp(-t^2/2)$.

  • $\begingroup$ Oh, thank you very much! $\endgroup$ – Matthew5335 May 30 at 16:05

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