Is it true that

If $a$ is random vector independent of $x$ for which $P(a'\Sigma a=0)=0$, then $$f\sim N(0,1)$$ and $f$ is independent of $a$ ?

Here $x\sim N_p(\mu,\Sigma)$ and $f={a'(x-\mu)\over\sqrt{a'\Sigma a}}$

  • $\begingroup$ Since $f$ is a scalar, don't you mean to write "$N(0,1)$" instead of "$N(0,I)$"? $\endgroup$ – whuber Feb 8 '17 at 20:52
  • $\begingroup$ @whuber Okay done $\endgroup$ – Qwerty Feb 8 '17 at 22:58
  • 4
    $\begingroup$ could you make question title informative? $\endgroup$ – hxd1011 Feb 8 '17 at 23:00

Yes, $f$ is indeed standard normal and independent of $a$.

To simplify notation, define $y=x-\mu$. Then $y\sim N_p(0,\Sigma)$ with multivariate characteristic function $\varphi_y (t)=\exp(-\frac{1}{2} t' \Sigma t)$. The (univariate!) characteristic function of $f$ can be computed by conditioning on $a$... $$\varphi_f (t)=\mathrm E[\exp(itf)]=\mathrm E[\mathrm E[\exp(itf)|a]] $$ Now $\mathrm E[\exp(itf)|a]=\mathrm E[\exp(ita'y/\sqrt{a'\Sigma a})|a]=\varphi_y (ta/\sqrt{a'\Sigma a})=\exp(-t^2/2)$, where $\varphi_y$ is the multivariate characteristic function of $y$, because we can think of $a$ as if it were a constant (see Did's nice answer here for a detailed justification of this). Therefore we find that $$\varphi_f (t)=\mathrm E[\exp(-t^2/2)]=\exp(-t^2/2)$$ Hence $f$ must be standard normal. It is also clear that, conditional on $a$, $f|a\sim N(0,1)$, which implies that $f$ is independent of $a$.


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