Is $(a+bX,X)$ jointly normal when $X$ is normal?

Question

Let $X$ be a normal random variable and $Y=a+bX$, where $a,b$ are just some constants.

Then, is it true that $(Y,X)$ are jointly normal? If yes, how can I easily see that?

Thanks!

Answer 1

I have some doubt to say no or yes. but I can say:

$(a+bX,X)$ have not joint density ( singular distribution )

$(a+bX,X)$ has not joint density$

$(Y|X=x) = \left\{ \begin{array}{cc} x & p=1 \\ o.w & p=0 \end{array} \right. $ $\hspace{1cm}$ (1)

$F_{(X,Y)}(x,y)=P(X\leq x, Y\leq y)=P(A)=E(I_A)=E(E(I_A)|X) =\int E(I_A|X=t) f_X(t) dt=\int E(I_{(X\leq x, Y\leq y)}|X=t) f_X(t) dt\overset{(1)}{=} \int E(I_{(X\leq x, a+bX\leq y)}|X=t) f_X(t) dt= \int E(I_{(t\leq x, a+bt\leq y)}|X=t) f_X(t) dt= \int E(I_{(t\leq x, t\leq \frac{y-a}{b})}|X=t) f_X(t) dt= \int_{-\infty}^{\min(x,\frac{y-a}{b})} E(I_{(t\leq x, t\leq \frac{y-a}{b})}|X=t) f_X(t) dt+0=\int_{-\infty}^{\min(x,\frac{y-a}{b})} E(1|X=t) f_X(t) dt= \int_{-\infty}^{\min(x,\frac{y-a}{b})} f_X(t) dt=F_X(\min(x,\frac{y-a}{b})) $

so it easy now to say $(a+bX,X)$ have not joint density.

$F_{(X,Y)}(x,y)=F_X(\min(x,\frac{y-a}{b}))$

$f_{(X,Y)}(x,y)=\frac{d^2}{dx dy}F_{(X,Y)}(x,y)=0$

this is a singular distribution. so , p.d.f of(X,Y) are not normal density.

to say yes you should check this:

$X_{k\times 1} \sim N(\mu_{k\times 1}, \Sigma) \iff $ there exist

$\mu \in R^k $ and $A_{K\times L} \in R^{k\times L}$ such that $X_{K\times 1}=A_{K\times L}Z_{L\times 1}+\mu_{K\times 1} $ for $Z_n\overset{i.i.d}{\sim} N(0,1)$

to say yes ,so you should find $A$ such that

$\left[ \begin{array}{c} X \\ a+bX \end{array} \right] =A_{2\times L} Z_{L\times 1} +\mu $

go and find $A$ and check this:

( $\left[ \begin{array}{c} X \\ a+bX \end{array} \right]$ and $A_{2\times L} Z_{L\times 1} $ have a same family and $Z_n\overset{i.i.d}{\sim} N(0,1)$ "i.i.d" )

$(Y|X=x)$ is not a normal

$(Y|X=x)$ is not a normal . it is a degenerated in point ${x}$. if you say $\left[ \begin{array}{c} X \\ a+bX \end{array} \right]$ is joint normal so conditional distribution should be normal. note that conditional distribution exists and not normal!

Answer 2

The answer is Yes. One of the equivalent definition of jointly normal is as follow:

Let $(\Omega,\mathcal{F},P)$ be a probability space. We say that a random vector $(X_{1},X_{2},\ldots,X_{m})$ is jointly normal if there exist independent random variables $Z_{1},Z_{2},\ldots,Z_{n}$ such that $Z_{j}\sim N(0,1)$ for each $j=1,\ldots,n$ and for each $i=1,\ldots,m$, $X_{i}\in\mbox{span}\{1, Z_{1},Z_{2},\ldots,Z_{n}\}$ (i.e., $X_{i}=\sum_{j=1}^{n}\alpha_{ij}Z_{j}+\beta_i$ for some $\alpha_{ij},\beta_i\in\mathbb{R}$). Beware of degerated cases, for example, $(0,0,\ldots,0)$ is regarded as jointly normal.

We go back to your case. Suppose that the given $X$ is not degenerated (i.e., $\sigma_{X}>0$). Define $Z=\frac{X-\mu_{X}}{\sigma_{X}}$, then $Z\sim N(0,1)$. Now $a+bX\in\mbox{span}(Z)$ and $X\in\mbox{span}(Z)$, so $(a+bX,X)$ is jointly normal.

Answer 3

Suppose $X\sim N(\mu,\sigma^2)$.

Then the dispersion matrix of $(Y,X)$ is

\begin{align} \Sigma=\sigma^2\begin{pmatrix}b^2 & b \\ b & 1 \end{pmatrix} \end{align}

Since $\Sigma$ does not have full rank, the joint distribution of $(Y,X)$ is a degenerate bivariate normal.

The degenerate bivariate normal is not expected to possess all the regular properties of the usual (nonsingular) bivariate normal. Notably, $(Y,X)$ does not enjoy a joint density.

You can see this from the fact that the correlation $\rho$ between $Y$ and $X$ satisfies $\rho^2=1$. In other words there exists a perfect linear relationship between $Y$ and $X$, with the random point $(Y,X)$ falling on a fixed line with probability one.

For details on this degenerate distribution, check out this excellent post on Cross Validated.