Convergence in Distribution

statmajor · #1 $Old$ November 5th, 2009, 02:08 PM

Let Xn be a sequence of p dimensional random vectors. Show that

Xn converges in distribution to $N_p(\mu,\Sigma)$ iff $a'X_n$ converges in distribution to $N_1(a' \mu, a' \Sigma a).$

So I start by finding the MGF to a'Xn:

$E(e^{(a'X_n)t} = E(e^{(a't)X_n}) = e^{a't \mu + 0.5t^2(a' \Sigma a)}$

Hence, {a'Xn} converges $N(a' \mu, a' \Sigma a).$ in distribution.

Is that it, cause I'm not to sure about what I'm supposed to prove.

Moo · #2 $Old$ November 7th, 2009, 01:52 PM

Hello,

What you did was finding the MGF of a normal distribution. But all we know is that there is a convergence in distribution. Plus, you can't just change the order of t, a, Xn, because they're matrices/vectors !

' denotes the transpose

Let's suppose $X_n \xrightarrow[n\to\infty]{distribution} N_p(\mu,\Sigma)$

This means that there is convergence of the MGF's :

$\forall t\in\mathbb{R}^p ~,~ \mathcal{M}_1(t)=E\left[\exp\left(\langle t,X_n\rangle\right)\right]\xrightarrow[n\to\infty]{} \exp\left(\mu't+\tfrac 12 \cdot t'\Sigma t\right) \quad \quad (*)$ ( $\langle,\rangle$ denotes the scalar product)

Now we have the MGF of $a'X_n$ which is $\mathcal{M}_2(t)=E\left[\exp\left(\langle t,a'X_n\rangle\right)\right]$

By remembering that $\langle x,y \rangle=x'y$ , it's very easy to show that $\langle t,a'X_n\rangle=\langle at,X_n\rangle$

So we have $\mathcal{M}_2(t)=\mathcal{M}_1(at)$ . By (*), it follows that :

$\mathcal{M}_2(t)\xrightarrow[n\to\infty]{}\exp\left(\mu'at+\tfrac 12 \cdot t'a'\Sigma at\right) = \exp\left((a'\mu)'t+\tfrac 12 \cdot t' (a'\Sigma a)t \right)$

which is exactly the MGF of $Z\sim \mathcal{N}_p(a'\mu,a'\Sigma a)$

Now for the other way of the equivalence, just apply this property to $b(a'X_n)$ , where b is such that $ba'=I_p$ (the identity matrix)

statmajor · #3 $Old$ November 7th, 2009, 02:48 PM

Thanks a lot for that explanation, which cleared up a lot of things.

Just have a question, the MGF of $M_2(t)$ is $N_p(a' \mu, a' \Sigma a)$ ?

Moo · #4 $Old$ November 7th, 2009, 02:53 PM

Quote:

Originally Posted by statmajor $View Post$

Thanks a lot for that explanation, which cleared up a lot of things.

Just have a question, the MGF of $M_2(t)$ is $N_p(a' \mu, a' \Sigma a)$ ?

No, it tends to the MGF of $N_p(a'\mu,a'\Sigma a)$

Sorry I'll modify a bit my post above so that there is no confusion ^^'

you have to remember that the MGF's of Xn or a'Xn have their limit equal to the MGF's of $N_p(\mu,\Sigma)$ or $N_p(a'\mu,a'\Sigma a)$

statmajor · #5 $Old$ November 7th, 2009, 03:13 PM

Quote:

Originally Posted by Moo $View Post$

you have to remember that the MGF's of Xn or a'Xn have their limit equal to the MGF's of $N_p(\mu,\Sigma)$ or $N_p(a'\mu,a'\Sigma a)$

That would be a limiting distribution (as my textbook/prof refer to it)?

Moo · #6 $Old$ November 7th, 2009, 03:40 PM

Yes, the limiting distribution of $X_n$ is $N_p(\mu,\Sigma)$ iff the limiting distribution of $a'X_n$ is $N_p(a'\mu,a'\Sigma a)$ , as proved above.

And if D is the limiting distribution of a rv Xn, then the mgf associated to D is the limit of the mgf of Xn.

statmajor · #7 $Old$ November 7th, 2009, 04:33 PM

Again, thanks for all of your help.