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Let $\lambda_{n,1} \leq \lambda_{n,2} \leq \dots \leq \lambda_{n,n} $ be the $n$ real eigenvalues of a random symmetric matrix, $X_n$ of dimension $n$. Then in this case one seems to define the "empirical distribution" as the ``random probability measure",

$L_n = \frac{1}{n} \sum_{i=1}^n \delta_{\lambda_{n,i}} $

  • Can someone kindly explain what exactly is this quantity?

    I am used to thinking of probabilty measures as finitely additive functions on some $\sigma/Borel$ algebra of subsets of a set. This above thing $L_n$ doesn't look like anything like that!

  • I don't understand why this $L_n$ seems capable of acting on functions like $<L_n,f>$. One typically wants to define the probability distribution $\bar{L}_n := \mathbb{E}[L_n] $ by the relation, $<\bar{L}_n,f> := \mathbb{E} [ <L_n, f> ]$

    Can someone kindly explain the above construction?

  • Like a particular statement which makes sense is that if $f= x^k$ in the above then, $<\bar{L}_n, x^k > = \frac{1}{n} \mathbb{E} [ Tr [X_n^k ]]$

    Can someone help derive this?

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user6818
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  • By the way, probability measures are $\sigma$-additive in addition to being finitely additive. – Zoran Loncarevic Oct 07 '15 at 11:38
  • I am not really familiar with the theory of random matrices, but $L_n$ defined above is a random probability measure, i.e. a measurable mapping from probability space to the space of probability measures on $\mathbb{R}$, in the same sense as random matrix is a measurable map from your probability space to the appropriate space of matrices. – Zoran Loncarevic Oct 07 '15 at 11:48
  • @ZoranLoncarevic Can you kindly explain how "random matrix is a measurable map from your probability space to the appropriate space of matrices" ? – user6818 Oct 07 '15 at 21:55
  • Well, every stochastic process $(X_t){t \in T}$ with values in measurable space $(\Xi,\mathcal X)$ and sample paths all in $U \subset \Xi^T$ defined on the probability space $(\Omega, \mathcal A,P)$ can be seen as a measurable map from $(\Omega, \mathcal A,P)$ into $(U,U \cap \Xi^T)$ where $\Xi^T$ is product $\sigma$-algebra $\bigotimes{t \in T} \Xi$. – Zoran Loncarevic Oct 08 '15 at 13:08
  • In that sense, random matrix is measurable map $(\Omega,\mathcal A) \to (M_n,M_n \cap \mathfrak{B}(\mathbb{R}^{n^2}))$ where $M_n$ is space of symmetric matrices, and random probability measure defined on $\sigma$-algebra $\mathcal B$ is, in general, a measurable map $(\Omega,\mathcal A) \to (M,M \cap \mathfrak{B}(\mathbb{R}^+)^{\mathcal B})$ where $M$ is set of all probability measures. – Zoran Loncarevic Oct 08 '15 at 13:18
  • Sorry, I made a typo in the first comment, change the last three occurrences of $\Xi$ to $\mathcal X$. – Zoran Loncarevic Oct 08 '15 at 13:24

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Let $(\Omega, \mathcal A, \mathsf P)$ be a probability space, $(S,\mathcal S)$ a measurable space and $X \colon (\Omega, \mathcal A) \to (S, \mathcal S)$ a (general) random variable. Then, for $X = X(\omega)\in S$ one can define a measure on $S$ by setting for any $A \in \mathcal S$ $$ \delta_{X(\omega)} (A) = \begin{cases} 1, & \mbox{ if }X(\omega) \in A, \\ 0, & \mbox{ if }X(\omega)\notin A. \end{cases}$$ That is, the measure of $A$ is one if and only if the realization of $X$ is in $A$. Note that this measure is random because it depends on the value of $\omega$. Though, the argument of $\omega$ is usually suppressed in the notation.

For any function $f \colon S \to \mathbf R$ we can take integrals with respect to $\delta_X$ $$ \int_S f(x) \, d\, \delta_X(x) = f(X). $$ In other words, by taking integrals (expectations) of $f$ with respect to the measure $\delta_X$ we basically evaluate $f$ at the value of the random variable $X$. This follows from the definitions of $\delta_X$ - the Dirac measure at $X$, see also this post.

Similarly, for any random variables $X_1, \dots, X_n$ we can define the linear combination of the measures $\delta_{X_1}, \dots, \delta_{X_n}$ by setting for $A\in\mathcal S$ $$ \left(\frac{1}{n} \sum_{i=1}^n \delta_{X_i}\right) (A) := \frac{1}{n} \sum_{i=1}^n \delta_{X_i}(A).$$ $\frac{1}{n} \sum_{i=1}^n \delta_{X_i}$ is then called the empirical measure of the variables $X_1, \dots, X_n$, and is again a random measure on $(S,\mathcal S)$. We can take the expectation of $f$ with respect to the empirical measure $$ \int_S f(x) \, d \, \left(\frac{1}{n} \sum_{i=1}^n \delta_{X_i}\right)(x) = \frac{1}{n} \sum_{i=1}^n f(X_i).$$ If we now take the expectation of the previous formula (with respect to $X_1, \dots, X_n$), we get a real number $$\mathbf E \frac{1}{n} \sum_{i=1}^n f(X_i) = \frac{1}{n} \sum_{i=1}^n \mathbf E f(X_i).$$

In your setting, $S = \mathbf R$ and $\lambda_{n,i}$, $i=1,\dots,n$ are the random variables used for the construction of the empirical measure $L_n$ as above. For $f \colon \mathbf R \to \mathbf R$ you have $$ < L_n, f > = \int_\mathbf R f(x) \, d \, L_n(x) = \frac{1}{n} \sum_{i=1}^n f(\lambda_{n,i}),$$ which is a random variable, and $$ \mathbf E < L_n, f > = \frac{1}{n} \sum_{i=1}^n \mathbf E f(\lambda_{n,i}),$$ which is a real number. Therefore, $\overline L_n$ is a (non-random) measure on $\mathbf R$ such that for any $f$ $$ <\overline L_n,f> = \int_\mathbf R f(x) \, d \, \overline L_n (x) = \frac{1}{n} \sum_{i=1}^n \mathbf E f(\lambda_{n,i}). $$ In particular, for $f = x^k$ you get $$ <\overline L_n,x^k> = \mathbf E < L_n, x^k > = \frac{1}{n} \sum_{i=1}^n \mathbf E \lambda_{n,i}^k,$$ where the last expression is equal to $$ \frac{1}{n} \mathbf E\,\, Tr(X_n^k).$$

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Jano Kakara
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  • Thanks for the effort. But there are quite a few things that I am not understanding. (1) If $\omega$ an element of a space of outcomes? Then shouldn't $X(\omega)$ be a real number? (2) When you say $(S,\cal{S})$ is a measure space I guess you mean that $\cal{S}$ is a Borel algebra on $S$. Then $A \in \cal{S}$ would mean that $A$ is a set, a subset of $S$. Then why should $X(\omega)$ which is a real number be an element of $A$? – user6818 Oct 08 '15 at 18:17
  • (3) $\int _S f(x) d(\delta_X (x) ) = f(X) $ a definition? Why is this true? – user6818 Oct 08 '15 at 18:18
  • (1) Yes, $\omega$ is an outcome; $X(\omega)$ is an element of $S$ (yes, it is real number when $(S,\mathcal S)=(\mathbb R,\mathfrak B(\mathbb R))$ and $X$ is a real random variable -- an answer just assumes, more generally, that random variable $X$ takes values in arbitrary measurable space $(S,\mathcal S)$ – Zoran Loncarevic Oct 08 '15 at 20:03
  • (2) You are right, $\mathcal S$ is $\sigma$-algebra on $S$, so $A \in \mathcal S$ is subset of $S$. What is the problem then, for $X(\omega) \in S$ to belong to $A \subset S$? – Zoran Loncarevic Oct 08 '15 at 20:08
  • (3) For a fixed $\omega \in \Omega$, the $\delta_{X(\omega)}$ defined above is Dirac measure concentrated on ${X(\omega)}$. Integral of any function $f: S \to \mathbb R$ with respect to this measure is $f(X(\omega))$. This is the consequence of definition of integral (and theory of integration). – Zoran Loncarevic Oct 08 '15 at 20:23
  • I edited the answer a bit, providing the definition of the random variable, and a link to integrating wrt Dirac measures. I hope it's clearer now. Thanks, @ZoranLoncarevic for comments. – Jano Kakara Oct 09 '15 at 07:26
  • (1) Well $(S,\cal{S})$ could be an arbitrary measure space but $X(\omega)$ would always be a real number. Then what does it mean to say $X(\omega) \in A \subset \cal{S})$? (2) Is there a good reference to see some exposition about this particular Dirac measure? (I am familiar with this construction as used hugely in physics literature but I wonder what is the mathematical understaning of it that you are using here) – user6818 Oct 09 '15 at 07:34
  • @user6818 Forget about $(S,\mathcal S)$ and read instead $(\mathbb R, \mathfrak B(\mathbb R))$. Then you have $X(\omega) \in A \subset \mathbb R$, and everything is OK. – Zoran Loncarevic Oct 09 '15 at 07:50
  • (1) No, if $X \colon \Omega \to S$, then $X(\omega)$ is an element of $S$, not necessarily a real number. Take, for instance $S$ to be the space of all symmetric $n\times n$ matrices as is your question. Then $X(\omega)$ is a matrix, not a number, and we can ask if it belongs to $A\subset S$, $A$ being the set of matrices with unit trace. (2) Dirac measure, or point mass, is the simplest example of measure you can think of, the reference is any good book on measure theory. – Jano Kakara Oct 09 '15 at 07:52