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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!

Suppose $Xsim N(mu,sigma^2)$.

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

beginalign
Sigma=sigma^2beginpmatrixb^2 & b \ b & 1 endpmatrix
endalign

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.

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

Let $(Omega,mathcalF,P)$ 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 $Z_jsim N(0,1)$ for each $j=1,ldots,n$ and for each
$i=1,ldots,m$, $X_iinmboxspan1, Z_1,Z_2,ldots,Z_n$
(i.e., $X_i=sum_j=1^nalpha_ijZ_j+beta_i$ for some $alpha_ij,beta_iinmathbbR$).
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 $Z=fracX-mu_Xsigma_X$,
then $Zsim N(0,1)$. Now $a+bXinmboxspan(Z)$ and $Xinmboxspan(Z)$,
so $(a+bX,X)$ is jointly normal.

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{
beginarraycc
x & p=1 \
o.w & p=0
endarray
right.
$ $hspace1cm$ (1)

$F_(X,Y)(x,y)=P(Xleq x, Yleq y)=P(A)=E(I_A)=E(E(I_A)|X)
=int E(I_A|X=t) f_X(t) dt=int E(I_(Xleq x, Yleq y)|X=t) f_X(t) dtoverset(1)=
int E(I_(Xleq x, a+bXleq y)|X=t) f_X(t) dt=
int E(I_(tleq x, a+btleq y)|X=t) f_X(t) dt=
int E(I_(tleq x, tleq fracy-ab)|X=t) f_X(t) dt=
int_-infty^min(x,fracy-ab) E(I_(tleq x, tleq fracy-ab)|X=t) f_X(t) dt+0=int_-infty^min(x,fracy-ab) E(1|X=t) f_X(t) dt=
int_-infty^min(x,fracy-ab) f_X(t) dt=F_X(min(x,fracy-ab))
$

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

$F_(X,Y)(x,y)=F_X(min(x,fracy-ab))$

$f_(X,Y)(x,y)=fracd^2dx dyF_(X,Y)(x,y)=0$

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

to say yes you should check this:

$X_ktimes 1 sim N(mu_ktimes 1, Sigma) iff $ there exist

$mu in R^k $ and $A_Ktimes L in R^ktimes L$ such that $X_Ktimes 1=A_Ktimes LZ_Ltimes 1+mu_Ktimes 1 $ for $Z_noverseti.i.dsim N(0,1)$

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

$left[ beginarrayc X \ a+bX endarray right]
=A_2times L Z_Ltimes 1 +mu $

go and find $A$ and check this:

( $left[ beginarrayc X \ a+bX endarray right]$
and $A_2times L Z_Ltimes 1 $ have a same family and $Z_noverseti.i.dsim N(0,1)$ "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
$left[ beginarrayc X \ a+bX endarray right]$ is joint normal so
conditional distribution should be normal. note that conditional distribution exists and not normal!