How are they making such a statement?

[vernal oak]

Franklin244

[heady steppe]

The characteristic function of a random variable completely determines the distribution. What they have shown is that the characteristic function of the joint distribution has the form of a characteristic function of a normal distribution.

[tidal plaza]

By characteristic function do you mean moment generating function which is defined by $\phi(t)=E[e^{Xt}]$ for a random variable $X$ ?

[vernal oak]

Franklin244

[heady steppe]

Oh, you're using MGFs. But yeah, it still holds.

Characteristic functions just exists more generally than MGFs, but they have a 1-1 relation. Characteristic functions are defined as φ(u) = E(exp(iuX)) where i² = -1.

[tidal plaza]

Ok, so they basically showed that the joint MGF of the random variables $X_1,...,X_n$ is equal to the MGF for a normal random variable. And since there's a definition which states that :

If the joint MGF of the random variables $X_1,...,X_n$ is similar to the MGF of a particular type of random variable X (be it discrete or continuous) then, we say that the random variables $X_1,...,X_n$ have the same distribution as that of $X.$

So they say that $X_1,...,X_n$ have a joint normal distribution, right?

Did I get it?

[vernal oak]

Franklin244

[heady steppe]

Well, usually stated as a theorem and not definition. But yeah.

[tidal plaza]

Did you see this edited message? I mean while u were replying I was editing my previous message. So, do you still agree with this?

[tidal plaza]

@heady steppe Are you there?!?

If u could lend me a helping hand, I'll be grateful 🥲

[tidal plaza]

.solved

[grim ice]

Solved

Post marked as solved by @tidal plaza.

Use .unsolved if this was a mistake.

[tidal plaza]

.unsolved