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If a sequence of quadratic forms converges in probability $Q_n\xrightarrow{P}Q$ and a random vector converges in distribution $X_n\xrightarrow{d}X$ then $X_n^TQ_nX_n\xrightarrow{d}X^TQX$.

This is a statement from an online source in statistics. It follows by Slutsky's theorem and the continuous mapping theorem. I can also see how intuitively it should be true, but I'm having trouble setting up the argument. No matter what I do, in the end I have a product of two things converging only in distribution.

Rodrigo
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  • This is also a question I am currently interested in. In one dimension, it is just similar question in essence as t-test converges to z-test asymptotically and very easy to prove. But I have trouble with 2 and more dimensional cases. And where is this online source you found, mind sharing this? Intuitively, I feel it is correct, but I never see such formal result in 2 dimensions or vector cases. Probably a good question. – lzstat Dec 30 '15 at 21:57
  • @Izstat look at their solution of exercise 1.2 http://www.statlect.com/convergence_of_transformations_exercise_set_1.htm For me, their passage from the second to the third equation is just hand-waving. – Rodrigo Dec 30 '15 at 22:20
  • it looks like correct if it is ok to use countious mapping theorem(CMT). I remember this kind of proof is usually depending on CMT and slutsky theorem. maybe you could state more clear about the problem you have in this post about the step you have question. thx for the information – lzstat Dec 30 '15 at 22:40
  • @Izstat, you're welcome. I find this website very good in general by the way. Oh, wait a second, maybe I got it too: coming up in my next post... – Rodrigo Dec 30 '15 at 22:50

2 Answers2

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A first step is to prove that the sequence $\left(X_n^T(Q_n-Q)X_n\right)_{n\geqslant 1}$ converges in probability to $0$. To see this, fix a positive $\varepsilon$. There is some $R$ for which for each $n$, $\mathbb P\{\lVert X_n\rVert\gt R\} \lt \varepsilon$. Then \begin{align} \mathbb P\{\left|X_n^T(Q_n-Q)X_n\right|\gt \delta\}&\leqslant \mathbb P\{\lVert X_n\rVert \lVert Q_n-Q\rVert \lVert X_,\rVert \gt \delta\} \\ &\leqslant 2\mathbb P\{\lVert X_n\rVert \gt R\}+\mathbb P\{\lVert Q_n-Q\rVert \gt \varepsilon/R^2\}\\ &\leqslant 2\varepsilon+ \mathbb P\{\lVert Q_n-Q\rVert \gt \varepsilon/R^2\}. \end{align} Therefore, the question reduces to the case where $Q_n=Q$ for each $n$.

It is true that the sequence $\left(X_n^TQX_n\right)_{n\geqslant 1}$ is tight, hence it admits a subsequence which converges in distribution.

But if $X$ is a symmetric non-degenerated random variable and $X_n:=e^{(-1)^nX}$, then $X_n$ has the same distribution as $e^{X}$; if $Q=e^{2X}$, then the distribution on $X_nQX_n$ is that of $1$ or $e^{4X}$, hence this does not converge in distribution.

However, we do have $X_n^TQX_n\to X^TQX$ if $Q$ is not random, since we can use the continuous mapping theorem with $x\mapsto x^TQx$.

Rodrigo
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Davide Giraudo
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  • How can saying that it admits a subsequence that converges in distribution imply that it converges in distribution? And why is the sequence $(X_n^TQX)_{n\ge1}$ tight? And do you replace the last $X_n$ by its distributional limit? But if you could look at my argument below, I think I actually proved it. – Rodrigo Dec 30 '15 at 23:45
  • I did not say that "it admits a subsequence that converges in distribution imply that it converges in distribution". For tightness of $(X_n^TQX)_{n\ge1}$, use similar arguments as in the three displayed lines: $P(||X_n^TQX||\gt R^3)\lt P(||X_n||\gt R)+P(||Q||\gt R)+P(||X||\gt R)$. What do you mean by "the last $X_n$"? – Davide Giraudo Dec 30 '15 at 23:49
  • I meant that we should be studying the sequence $X_n^TQX_n$, but instead you wrote everywhere $X_n^TQX$. You see, you replaced the second $X_n$ by $X$. Also, in the end, I'd like to have convergence in distribution, so only a subsequence is not sufficient. – Rodrigo Dec 31 '15 at 12:59
  • There is a typo in the question, you should edit. // I showed that the convergence in distribution may not hold. – Davide Giraudo Dec 31 '15 at 13:48
  • Oh, I'm terribly sorry! It's now fixed. I'll upvote you for your example, but maybe you want to clarify in your answer that you are a little out of the question now. – Rodrigo Dec 31 '15 at 15:11
  • I preferred to update the answer. – Davide Giraudo Dec 31 '15 at 15:26
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Write $Q_n = \Sigma_n^T\Sigma_n \xrightarrow{p}\Sigma^T\Sigma = Q$

By Slutsky theorem $X_n^T\Sigma_n^T \xrightarrow{d} X^T\Sigma^T = (\Sigma X)^T$

Similarly, $\Sigma_n X_n\xrightarrow{d} \Sigma X$.

Now $(\Sigma X)^T\Sigma X$ is a quadratic polynomial in $\dim(X)$ variables, but each of its terms is a monomial in one variable. So the continue mapping theorem applies.

Rodrigo
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