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Freeness for operator valued random variables
The Next CEO of Stack OverflowInteger-valued random variables must converge in distribution to a integer-valued random variable?Criterion for independency of random variablesKolmogorov continuity theorem for Banach space valued random processesWhat is a real-valued random variable?The limit of integer valued random variables must be integer valued?Two random variables from the same probability density function: how can they be different?Independence of two non-negative integer valued random variablesIs a collection of random variables always a random vector?When are Hilbert space valued random variables independent?Defining a “dependence structure” for random variables
$begingroup$
Threre is a definition of freeness of random variables clasically. For operator valued random variables is there any analogue of free ness depending on the base space?
probability-theory operator-algebras
asked Mar 28 at 7:00
mathlovermathlover
166110
$endgroup$
add a comment |
$begingroup$
Threre is a definition of freeness of random variables clasically. For operator valued random variables is there any analogue of free ness depending on the base space?
probability-theory operator-algebras
asked Mar 28 at 7:00
mathlovermathlover
166110
$endgroup$
add a comment |
$begingroup$
Threre is a definition of freeness of random variables clasically. For operator valued random variables is there any analogue of free ness depending on the base space?
probability-theory operator-algebras
asked Mar 28 at 7:00
mathlovermathlover
166110
$endgroup$
Threre is a definition of freeness of random variables clasically. For operator valued random variables is there any analogue of free ness depending on the base space?
probability-theory operator-algebras
probability-theory operator-algebras
asked Mar 28 at 7:00
mathlovermathlover
166110
asked Mar 28 at 7:00
mathlovermathlover
166110
edited Mar 28 at 9:32
mathlover
asked Mar 28 at 7:00
mathlovermathlover
166110
asked Mar 28 at 7:00
mathlovermathlover
166110
asked Mar 28 at 7:00
mathlovermathlover
166110
166110
add a comment |
add a comment |
1 Answer
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$begingroup$
The analogue is known as freeness with amalgamation.
The definition is made in the same way: given an expectation $E: M rightarrow A$, we say that ($*$-)subalgebras $M_1, ..., M_n$ are ($*$-)free with amalgamation over $A$ if $E(X_i_1...X_i_m) = 0$ whenever $E(X_i_j) = 0$, $X_i_j in M_i_j$, and $i_j neq i_j+1$ for all $j$. We say that a family $X_1, ... X_n$ of $A$-valued random variables are free with amalgamation over $A$ if the $*$-algebras they generate with $A$ are free with amalgamation over $A$. Freeness for classical random variables is then just freeness with amalgamation over $mathbbC$.
You can check out Free Random Variables by Voiculescu, Dykema, and Nica for more, but there are other good references depending on what you're looking for. Speicher has a book called Combinatorial Theory of the Free Product with Amalgamation and Operator-Valued Free Probability Theory that covers the combinatorics in a lot of detail.
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
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$begingroup$
The analogue is known as freeness with amalgamation.
The definition is made in the same way: given an expectation $E: M rightarrow A$, we say that ($*$-)subalgebras $M_1, ..., M_n$ are ($*$-)free with amalgamation over $A$ if $E(X_i_1...X_i_m) = 0$ whenever $E(X_i_j) = 0$, $X_i_j in M_i_j$, and $i_j neq i_j+1$ for all $j$. We say that a family $X_1, ... X_n$ of $A$-valued random variables are free with amalgamation over $A$ if the $*$-algebras they generate with $A$ are free with amalgamation over $A$. Freeness for classical random variables is then just freeness with amalgamation over $mathbbC$.
You can check out Free Random Variables by Voiculescu, Dykema, and Nica for more, but there are other good references depending on what you're looking for. Speicher has a book called Combinatorial Theory of the Free Product with Amalgamation and Operator-Valued Free Probability Theory that covers the combinatorics in a lot of detail.
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
$endgroup$
add a comment |
$begingroup$
The analogue is known as freeness with amalgamation.
The definition is made in the same way: given an expectation $E: M rightarrow A$, we say that ($*$-)subalgebras $M_1, ..., M_n$ are ($*$-)free with amalgamation over $A$ if $E(X_i_1...X_i_m) = 0$ whenever $E(X_i_j) = 0$, $X_i_j in M_i_j$, and $i_j neq i_j+1$ for all $j$. We say that a family $X_1, ... X_n$ of $A$-valued random variables are free with amalgamation over $A$ if the $*$-algebras they generate with $A$ are free with amalgamation over $A$. Freeness for classical random variables is then just freeness with amalgamation over $mathbbC$.
You can check out Free Random Variables by Voiculescu, Dykema, and Nica for more, but there are other good references depending on what you're looking for. Speicher has a book called Combinatorial Theory of the Free Product with Amalgamation and Operator-Valued Free Probability Theory that covers the combinatorics in a lot of detail.
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
$endgroup$
add a comment |
$begingroup$
The analogue is known as freeness with amalgamation.
The definition is made in the same way: given an expectation $E: M rightarrow A$, we say that ($*$-)subalgebras $M_1, ..., M_n$ are ($*$-)free with amalgamation over $A$ if $E(X_i_1...X_i_m) = 0$ whenever $E(X_i_j) = 0$, $X_i_j in M_i_j$, and $i_j neq i_j+1$ for all $j$. We say that a family $X_1, ... X_n$ of $A$-valued random variables are free with amalgamation over $A$ if the $*$-algebras they generate with $A$ are free with amalgamation over $A$. Freeness for classical random variables is then just freeness with amalgamation over $mathbbC$.
You can check out Free Random Variables by Voiculescu, Dykema, and Nica for more, but there are other good references depending on what you're looking for. Speicher has a book called Combinatorial Theory of the Free Product with Amalgamation and Operator-Valued Free Probability Theory that covers the combinatorics in a lot of detail.
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
$endgroup$
The analogue is known as freeness with amalgamation.
The definition is made in the same way: given an expectation $E: M rightarrow A$, we say that ($*$-)subalgebras $M_1, ..., M_n$ are ($*$-)free with amalgamation over $A$ if $E(X_i_1...X_i_m) = 0$ whenever $E(X_i_j) = 0$, $X_i_j in M_i_j$, and $i_j neq i_j+1$ for all $j$. We say that a family $X_1, ... X_n$ of $A$-valued random variables are free with amalgamation over $A$ if the $*$-algebras they generate with $A$ are free with amalgamation over $A$. Freeness for classical random variables is then just freeness with amalgamation over $mathbbC$.
You can check out Free Random Variables by Voiculescu, Dykema, and Nica for more, but there are other good references depending on what you're looking for. Speicher has a book called Combinatorial Theory of the Free Product with Amalgamation and Operator-Valued Free Probability Theory that covers the combinatorics in a lot of detail.
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
answered 4 hours ago
Josh KenedaJosh Keneda
1,783717
1,783717
add a comment |
add a comment |
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