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Let $x_1 , x_2 , x_3 ~ N(0,1)$ iid. $Y = \frac{x_1 + x_2 * x_3}{\sqrt{1+ x_3^2}}$

Does Y and $X_3$ jointly gaussian ?

I already showed that $E[Y|X3]=0 $ and $Var[Y|X3]=0$ And Y|X3 is gaussian... But how can I proceed?

Maybe there is a mistake in the question that make it better?

Yar Sha
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2 Answers2

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$ Z=aX_1+bX_2\sim N(0,a^2+b^2).$ Therefore $Y|X_3\sim N(0,1)$ is independent of $X_3$, and $Y, X_3$ are Gaussian and independent.

Let $U,V$ be two rv with joint distribution $\pi(du)K(u,dv)$ where $U\sim \pi$ and $V|U\sim K(u,dv).$ Then $U$ and $V$ are independent if and only if $U$ and $V|U$ are independent.

Proof. $\Rightarrow$ Let $V\sim K(dv)$. We get $\pi(du)K(u,dv)=\pi(du)K(dv)$ and $K(u,dv)=K(dv),$ at least $\pi(du)$ almost everywhere.

$\Leftarrow$ $u\mapsto K(u,dv)$ is a constant.

NN2: True, up to the subtilities of conditioning, I do not quite understand your question.

Letac Gérard
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    How do you do this step?$$$$ "$Y|X_3\sim N(0,1)$ is independent of $X_3 \implies $ $Y, X_3$ are Gaussian and independent" – NN2 Feb 21 '24 at 17:25
  • It seems to me obvious. What about the formal following proof $$E[E(e^{sY+tX_3}|X_3)]=E{E(e^{sY}|X_3)\times e^{tX_3}]=e^{s^2/2}\times e^{t^2/2}.?$ – Letac Gérard Feb 21 '24 at 22:19
  • But why is $Y|X_3$ independent of $X_3$? – NN2 Feb 22 '24 at 00:08
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It seems to me obvious. What about the formal following proof $$E(e^{sY+tX_3})=E[E(e^{sY+tX_3}|X_3)]=E[E(e^{sY}|X_3)\times e^{tX_3}]=e^{s^2/2}\times e^{t^2/2}?$$

Letac Gérard
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