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Suppose $X,Y$ are probability spaces with appropriate measures. Suppose $f_n : X \rightarrow \mathbb R$ and $g_n : Y \rightarrow \mathbb R$ be sequences of random variables satisfying the conditions of central limit theorem. Then, does the sequence $(f_n, g_n) : X \times Y \rightarrow \mathbb{R}^2$ satisfy some kind of central limit theorem?

If so, what is the precise statement? And what is the relation of the mean, variance etc of the production function, to the mean and variance of $f_n$ and $g_n$ ?

What if $X=Y$, and we are considering the functions $(f_n, g_n) : X \rightarrow \mathbb{R}^2$ ?

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This is a theorem on page 28 of Serfling's Approximation Theorems of Mathematical Statistics. Besides variances, one also takes into account covariances, and the inverse of the matrix of covariances is in the role that $1/\sigma^2$ plays in the scalar-valued case.

Here's a formulation with more standard notations and hypotheses. Let $\Omega$ be the probability space and $X_n : \Omega \longrightarrow \mathbb{R}^p$, $n=1,2,3,\ldots$ be an i.i.d. sequence of random vectors with finite second moments. (And normally $\Omega$ isn't even mentioned in the statement of the theorem.)

Suppose $\mu = \mathbb{E}(X) \in \mathbb{R}^p$ is the mean and $$ \Sigma = \mathbb{E}((X-\mu)(X-\mu)^\top) \in \mathbb{R}^{p\times p} $$ is the variance, or as it is sometimes called (because its entries are covariances), the covariance matrix.

Then $$ \frac{1}{\sqrt{n}}\left(\left(\frac1n\sum_{i=1}^nX_i\right) - \mu\right) $$ converges in distribution to the $p$-dimensional normal distribution with expectation $0\in\mathbb{R}^p$ and variance $\Sigma \in \mathbb{R}^{p\times p}$.

If $\det\Sigma\ne0$, then there is a density function: $$ f(x) = \frac{1}{\sqrt{2\pi}^p\sqrt{|\det\Sigma|}} \exp\left(\frac{-1}{2} (x-\mu)^\top \Sigma^{-1}(x-\mu) \right). $$

However, you've formulated the question as being on product spaces, which for present purposes means the components are independent. That's a narrow special case, and the conclusion from it is that all of the off-diagonal entries in $\Sigma$ are zero.

You've also in effect specified that $p=2$.

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I initially had $\sqrt{2\pi}$ where I needed $\sqrt{2\pi}^p$. I've fixed it. –  Michael Hardy Oct 23 '11 at 20:36

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