Probability of lying in the intersection of two subspaces over a finite field
$begingroup$
Let $A$ be a subspace spanned by $d_Aleq M$ linearly independent $M$-dimensional vectors whose elements are uniformly randomly drawn from a finite field $mathbb{F}_q$ of size $q$.
Let $mathcal{B}$ be another subspace spanned by $d_Bleq M$ linearly independent $M$-dimensional vectors, whose elements are also uniformly randomly drawn from $mathbb{F}_q$. The drawing is independent with that of $A$.
Now suppose that a new $M$-dimensional vector $mathbf{x}$ is uniformly randomly drawn from $mathbb{F}_q^{Mtimes 1}$. I would like to write $mathrm{Pr}{mathbf{x}in(Acap B)}$, which denotes the probability that $mathbf{x}$ is in the intersection of $A$ and $B$, as an expression of $d_A$, $d_B$, and $k=mathrm{dim}{Acap B}$. Similarly, I also want to determine $mathrm{Pr}{mathbf{x}in(bar{A}cap B)}$, $mathrm{Pr}{mathbf{x}in(Acap bar{B})}$, and $mathrm{Pr}{mathbf{x}in(bar{A}cap bar{B})}$, which are the probabilities that $mathbf{x}$ lies in $B$ but not in $A$, in $A$ but not in $B$, and neither in $A$ nor $B$, respectively.
linear-algebra probability combinatorics finite-fields
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
$endgroup$
add a comment |
$begingroup$
Let $A$ be a subspace spanned by $d_Aleq M$ linearly independent $M$-dimensional vectors whose elements are uniformly randomly drawn from a finite field $mathbb{F}_q$ of size $q$.
Let $mathcal{B}$ be another subspace spanned by $d_Bleq M$ linearly independent $M$-dimensional vectors, whose elements are also uniformly randomly drawn from $mathbb{F}_q$. The drawing is independent with that of $A$.
Now suppose that a new $M$-dimensional vector $mathbf{x}$ is uniformly randomly drawn from $mathbb{F}_q^{Mtimes 1}$. I would like to write $mathrm{Pr}{mathbf{x}in(Acap B)}$, which denotes the probability that $mathbf{x}$ is in the intersection of $A$ and $B$, as an expression of $d_A$, $d_B$, and $k=mathrm{dim}{Acap B}$. Similarly, I also want to determine $mathrm{Pr}{mathbf{x}in(bar{A}cap B)}$, $mathrm{Pr}{mathbf{x}in(Acap bar{B})}$, and $mathrm{Pr}{mathbf{x}in(bar{A}cap bar{B})}$, which are the probabilities that $mathbf{x}$ lies in $B$ but not in $A$, in $A$ but not in $B$, and neither in $A$ nor $B$, respectively.
linear-algebra probability combinatorics finite-fields
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
$endgroup$
1
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19
add a comment |
$begingroup$
Let $A$ be a subspace spanned by $d_Aleq M$ linearly independent $M$-dimensional vectors whose elements are uniformly randomly drawn from a finite field $mathbb{F}_q$ of size $q$.
Let $mathcal{B}$ be another subspace spanned by $d_Bleq M$ linearly independent $M$-dimensional vectors, whose elements are also uniformly randomly drawn from $mathbb{F}_q$. The drawing is independent with that of $A$.
Now suppose that a new $M$-dimensional vector $mathbf{x}$ is uniformly randomly drawn from $mathbb{F}_q^{Mtimes 1}$. I would like to write $mathrm{Pr}{mathbf{x}in(Acap B)}$, which denotes the probability that $mathbf{x}$ is in the intersection of $A$ and $B$, as an expression of $d_A$, $d_B$, and $k=mathrm{dim}{Acap B}$. Similarly, I also want to determine $mathrm{Pr}{mathbf{x}in(bar{A}cap B)}$, $mathrm{Pr}{mathbf{x}in(Acap bar{B})}$, and $mathrm{Pr}{mathbf{x}in(bar{A}cap bar{B})}$, which are the probabilities that $mathbf{x}$ lies in $B$ but not in $A$, in $A$ but not in $B$, and neither in $A$ nor $B$, respectively.
linear-algebra probability combinatorics finite-fields
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
$endgroup$
Let $A$ be a subspace spanned by $d_Aleq M$ linearly independent $M$-dimensional vectors whose elements are uniformly randomly drawn from a finite field $mathbb{F}_q$ of size $q$.
Let $mathcal{B}$ be another subspace spanned by $d_Bleq M$ linearly independent $M$-dimensional vectors, whose elements are also uniformly randomly drawn from $mathbb{F}_q$. The drawing is independent with that of $A$.
Now suppose that a new $M$-dimensional vector $mathbf{x}$ is uniformly randomly drawn from $mathbb{F}_q^{Mtimes 1}$. I would like to write $mathrm{Pr}{mathbf{x}in(Acap B)}$, which denotes the probability that $mathbf{x}$ is in the intersection of $A$ and $B$, as an expression of $d_A$, $d_B$, and $k=mathrm{dim}{Acap B}$. Similarly, I also want to determine $mathrm{Pr}{mathbf{x}in(bar{A}cap B)}$, $mathrm{Pr}{mathbf{x}in(Acap bar{B})}$, and $mathrm{Pr}{mathbf{x}in(bar{A}cap bar{B})}$, which are the probabilities that $mathbf{x}$ lies in $B$ but not in $A$, in $A$ but not in $B$, and neither in $A$ nor $B$, respectively.
linear-algebra probability combinatorics finite-fields
linear-algebra probability combinatorics finite-fields
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
asked Dec 10 '18 at 8:45
leeyeeleeyee
656
656
1
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19
add a comment |
1
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19
1
1
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033652%2fprobability-of-lying-in-the-intersection-of-two-subspaces-over-a-finite-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033652%2fprobability-of-lying-in-the-intersection-of-two-subspaces-over-a-finite-field%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Correct me if I am wrong, but isn't $P(xin Acap B)$ simply $q^k/q^M$? There are $q^k$ vectors in $Acap B$, and $q^M$ vectors total. Did you get that far?
$endgroup$
– Mike Earnest
Dec 11 '18 at 22:19