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From Introduction to Probability here (PDF page 348, book page 337) we are told:

Definition 7.5.1

A $k$-dimensional random vector $X=(X_1, ..., X_k)$ is Multivariate Normal if every linear combination of the $X_j$ has a Normal distribution. That is, we require

$$t_1X_1 + \cdots + t_kX_k$$

to have a Normal distribution for any constants $t_1, ..., t_k$.

Question set-up

Logically speaking this is of the form: $Q$ if $P$, where:

$P$: We have a $k$-dimensional random vector $X=(X_1, ..., X_k)$ where every linear combination of the $X_j$ has a Normal distribution

$Q$: $X=(X_1, ..., X_k)$ is Multivariate Normal.

So, if $P$ then $Q$ i.e. $P \implies Q$.

Question

If we are told that we have a MVN (i.e. a Bivariate Normal, $k=2$) with random vector $(X, Y)$ can we conclude that every linear combination of the $X, Y$ are Normal? It seems to me we cannot as we are given $Q$ is true, not $P$.

If Definition 7.5.1 above said "if and only if" then I think we could, but not just with "if"?

Note: I know that it does happen to be the case that every linear combination of the $X, Y$ is Normal if $(X, Y)$ are MVN...I'm asking if we can logically deduce it from the information given.

Background

This is from homework question 3 b (page 20 of PDF). Provided below for context:

Let $(X, Y)$ be Bivariate Normal, with $X$ and $Y$ marginally $\mathcal{N}(0, 1)$ and with correlation $\rho$ between $X$ and $Y$.

Show that $(X + Y,X − Y )$ is also Bivariate Normal.

Solution given: the linear combination $s(X + Y ) + t(X − Y ) = (s + t)X + (s − t)Y$ is also a linear combination of $X$ and $Y$, so it is Normal, which shows that $(X +Y,X −Y )$ is MVN.

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    $\begingroup$ Definition 7.5.1 is if and only if. Definitions always imply only if, otherwise they wouldn't be definitions. $\endgroup$ Jan 4, 2022 at 11:02

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Assume not.

Then there is some linear combination that doesn’t satisfy the criteria for being MVN. Thus, the original bivariate distribution is not MVN.

Consequently, it has to be the case that y have the “iff” logic that you want.

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