Mixing
Bounded function with finitely many discontinuities is integrable $overset{?}{Rightarrow}$ density of...
$begingroup$
The density function of the distribution function of a continuous random variable is not uniquely defined.
A new density function can be obtained by changing the value of the function at finite number of points to some non-negative value, without changing the integral of the function. We then get a new density function for the same continuous distribution.
Does this follow from the theorem-
A bounded function with finite number of discontinuities over an interval is Riemann integrable.
or is there a different theorem supporting the above claim? Is the theorem a sufficient justification?
probability-theory probability-distributions random-variables riemann-integration density-function
asked Jan 20 at 4:49
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
The density function of the distribution function of a continuous random variable is not uniquely defined.
A new density function can be obtained by changing the value of the function at finite number of points to some non-negative value, without changing the integral of the function. We then get a new density function for the same continuous distribution.
Does this follow from the theorem-
A bounded function with finite number of discontinuities over an interval is Riemann integrable.
or is there a different theorem supporting the above claim? Is the theorem a sufficient justification?
probability-theory probability-distributions random-variables riemann-integration density-function
asked Jan 20 at 4:49
Za IraZa Ira
161115
$endgroup$
2
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
2
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23
add a comment |
$begingroup$
The density function of the distribution function of a continuous random variable is not uniquely defined.
A new density function can be obtained by changing the value of the function at finite number of points to some non-negative value, without changing the integral of the function. We then get a new density function for the same continuous distribution.
Does this follow from the theorem-
A bounded function with finite number of discontinuities over an interval is Riemann integrable.
or is there a different theorem supporting the above claim? Is the theorem a sufficient justification?
probability-theory probability-distributions random-variables riemann-integration density-function
asked Jan 20 at 4:49
Za IraZa Ira
161115
$endgroup$
The density function of the distribution function of a continuous random variable is not uniquely defined.
A new density function can be obtained by changing the value of the function at finite number of points to some non-negative value, without changing the integral of the function. We then get a new density function for the same continuous distribution.
Does this follow from the theorem-
A bounded function with finite number of discontinuities over an interval is Riemann integrable.
or is there a different theorem supporting the above claim? Is the theorem a sufficient justification?
probability-theory probability-distributions random-variables riemann-integration density-function
probability-theory probability-distributions random-variables riemann-integration density-function
asked Jan 20 at 4:49
Za IraZa Ira
161115
asked Jan 20 at 4:49
Za IraZa Ira
161115
asked Jan 20 at 4:49
Za IraZa Ira
161115
asked Jan 20 at 4:49
Za IraZa Ira
161115
asked Jan 20 at 4:49
Za IraZa Ira
161115
161115
2
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
2
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23
add a comment |
2
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
2
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23
2
2
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
2
2
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The theorem you state certainly suffices to justify that that the new density function is integrable, but this is not enough to justify the full claim, which is much stronger. The Reimann integral is not quite enough to fully formalize all of the concepts at play here. In particular, we need to be very careful about what exactly we mean by "density". What you really want is the Lebesgue integral, which in turn requires some understanding of measure theory. Based on your question, I am unsure how familiar you are with these concepts (correct me if I'm wrong), so I will try to at least informally hint at some of the concepts at stake here (so there will be some handwaving).
Let's say we have some distribution over the real line in the sense that we have specified $P(xin A)$ for all "sufficiently well behaved" $A$ and these probabilities satisfies a few axioms. We will say that $P$ is a continuous distribution if there is some $f:mathbb R to mathbb R^+$ such that for any $A$ where the probability above is well defined, we have
$$P(x in A) = int_A f(x),mathrm dx$$
where the integral is the Lebesgue integral with respect to the Lebesgue measure on $mathbb R$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $f$ in the above definition will count as a "density". Now consider modifying $f$ at a finite set of points $S$. Formally, we can write this new density as $g = f + h$ where $h = 0$ everywhere except on $S$. Let's first integrate $h$. Even in the Reimann theory of integration, it is true that $int_A h(x),mathrm dx = 0$. This carries over to Lebesgue integration. Then for any $A$ that is "well behaved", we must have
$$int_A g(x),mathrm dx = int_A f(x) + g(x),mathrm dx = int_A f(x),mathrm dx + int g(x),mathrm dx = int_A f(x),mathrm dx = P(xin A)$$
But that means that $g$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $f$ on any $S$ that satisfies
$$int 1{S},mathrm dx = 0$$
will still give us a valid density (where $1{S}$ is the indicator function for the set $S$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.
answered Jan 20 at 5:36
stats_modelstats_model
612412
$endgroup$
add a comment |
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%2f3080195%2fbounded-function-with-finitely-many-discontinuities-is-integrable-overset%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The theorem you state certainly suffices to justify that that the new density function is integrable, but this is not enough to justify the full claim, which is much stronger. The Reimann integral is not quite enough to fully formalize all of the concepts at play here. In particular, we need to be very careful about what exactly we mean by "density". What you really want is the Lebesgue integral, which in turn requires some understanding of measure theory. Based on your question, I am unsure how familiar you are with these concepts (correct me if I'm wrong), so I will try to at least informally hint at some of the concepts at stake here (so there will be some handwaving).
Let's say we have some distribution over the real line in the sense that we have specified $P(xin A)$ for all "sufficiently well behaved" $A$ and these probabilities satisfies a few axioms. We will say that $P$ is a continuous distribution if there is some $f:mathbb R to mathbb R^+$ such that for any $A$ where the probability above is well defined, we have
$$P(x in A) = int_A f(x),mathrm dx$$
where the integral is the Lebesgue integral with respect to the Lebesgue measure on $mathbb R$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $f$ in the above definition will count as a "density". Now consider modifying $f$ at a finite set of points $S$. Formally, we can write this new density as $g = f + h$ where $h = 0$ everywhere except on $S$. Let's first integrate $h$. Even in the Reimann theory of integration, it is true that $int_A h(x),mathrm dx = 0$. This carries over to Lebesgue integration. Then for any $A$ that is "well behaved", we must have
$$int_A g(x),mathrm dx = int_A f(x) + g(x),mathrm dx = int_A f(x),mathrm dx + int g(x),mathrm dx = int_A f(x),mathrm dx = P(xin A)$$
But that means that $g$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $f$ on any $S$ that satisfies
$$int 1{S},mathrm dx = 0$$
will still give us a valid density (where $1{S}$ is the indicator function for the set $S$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.
answered Jan 20 at 5:36
stats_modelstats_model
612412
$endgroup$
add a comment |
$begingroup$
The theorem you state certainly suffices to justify that that the new density function is integrable, but this is not enough to justify the full claim, which is much stronger. The Reimann integral is not quite enough to fully formalize all of the concepts at play here. In particular, we need to be very careful about what exactly we mean by "density". What you really want is the Lebesgue integral, which in turn requires some understanding of measure theory. Based on your question, I am unsure how familiar you are with these concepts (correct me if I'm wrong), so I will try to at least informally hint at some of the concepts at stake here (so there will be some handwaving).
Let's say we have some distribution over the real line in the sense that we have specified $P(xin A)$ for all "sufficiently well behaved" $A$ and these probabilities satisfies a few axioms. We will say that $P$ is a continuous distribution if there is some $f:mathbb R to mathbb R^+$ such that for any $A$ where the probability above is well defined, we have
$$P(x in A) = int_A f(x),mathrm dx$$
where the integral is the Lebesgue integral with respect to the Lebesgue measure on $mathbb R$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $f$ in the above definition will count as a "density". Now consider modifying $f$ at a finite set of points $S$. Formally, we can write this new density as $g = f + h$ where $h = 0$ everywhere except on $S$. Let's first integrate $h$. Even in the Reimann theory of integration, it is true that $int_A h(x),mathrm dx = 0$. This carries over to Lebesgue integration. Then for any $A$ that is "well behaved", we must have
$$int_A g(x),mathrm dx = int_A f(x) + g(x),mathrm dx = int_A f(x),mathrm dx + int g(x),mathrm dx = int_A f(x),mathrm dx = P(xin A)$$
But that means that $g$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $f$ on any $S$ that satisfies
$$int 1{S},mathrm dx = 0$$
will still give us a valid density (where $1{S}$ is the indicator function for the set $S$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.
answered Jan 20 at 5:36
stats_modelstats_model
612412
$endgroup$
add a comment |
$begingroup$
The theorem you state certainly suffices to justify that that the new density function is integrable, but this is not enough to justify the full claim, which is much stronger. The Reimann integral is not quite enough to fully formalize all of the concepts at play here. In particular, we need to be very careful about what exactly we mean by "density". What you really want is the Lebesgue integral, which in turn requires some understanding of measure theory. Based on your question, I am unsure how familiar you are with these concepts (correct me if I'm wrong), so I will try to at least informally hint at some of the concepts at stake here (so there will be some handwaving).
Let's say we have some distribution over the real line in the sense that we have specified $P(xin A)$ for all "sufficiently well behaved" $A$ and these probabilities satisfies a few axioms. We will say that $P$ is a continuous distribution if there is some $f:mathbb R to mathbb R^+$ such that for any $A$ where the probability above is well defined, we have
$$P(x in A) = int_A f(x),mathrm dx$$
where the integral is the Lebesgue integral with respect to the Lebesgue measure on $mathbb R$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $f$ in the above definition will count as a "density". Now consider modifying $f$ at a finite set of points $S$. Formally, we can write this new density as $g = f + h$ where $h = 0$ everywhere except on $S$. Let's first integrate $h$. Even in the Reimann theory of integration, it is true that $int_A h(x),mathrm dx = 0$. This carries over to Lebesgue integration. Then for any $A$ that is "well behaved", we must have
$$int_A g(x),mathrm dx = int_A f(x) + g(x),mathrm dx = int_A f(x),mathrm dx + int g(x),mathrm dx = int_A f(x),mathrm dx = P(xin A)$$
But that means that $g$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $f$ on any $S$ that satisfies
$$int 1{S},mathrm dx = 0$$
will still give us a valid density (where $1{S}$ is the indicator function for the set $S$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.
answered Jan 20 at 5:36
stats_modelstats_model
612412
$endgroup$
The theorem you state certainly suffices to justify that that the new density function is integrable, but this is not enough to justify the full claim, which is much stronger. The Reimann integral is not quite enough to fully formalize all of the concepts at play here. In particular, we need to be very careful about what exactly we mean by "density". What you really want is the Lebesgue integral, which in turn requires some understanding of measure theory. Based on your question, I am unsure how familiar you are with these concepts (correct me if I'm wrong), so I will try to at least informally hint at some of the concepts at stake here (so there will be some handwaving).
Let's say we have some distribution over the real line in the sense that we have specified $P(xin A)$ for all "sufficiently well behaved" $A$ and these probabilities satisfies a few axioms. We will say that $P$ is a continuous distribution if there is some $f:mathbb R to mathbb R^+$ such that for any $A$ where the probability above is well defined, we have
$$P(x in A) = int_A f(x),mathrm dx$$
where the integral is the Lebesgue integral with respect to the Lebesgue measure on $mathbb R$ (which you can think of as behaving exactly like the Reimann integral for most normal cases, but with some modifications to allow integration of a considerably larger class of functions). Any $f$ in the above definition will count as a "density". Now consider modifying $f$ at a finite set of points $S$. Formally, we can write this new density as $g = f + h$ where $h = 0$ everywhere except on $S$. Let's first integrate $h$. Even in the Reimann theory of integration, it is true that $int_A h(x),mathrm dx = 0$. This carries over to Lebesgue integration. Then for any $A$ that is "well behaved", we must have
$$int_A g(x),mathrm dx = int_A f(x) + g(x),mathrm dx = int_A f(x),mathrm dx + int g(x),mathrm dx = int_A f(x),mathrm dx = P(xin A)$$
But that means that $g$ also satisfies our definition of what it means to be a density. As mentioned by the comments before me, all of this can be generalized considerably. One way to put it is that modifying $f$ on any $S$ that satisfies
$$int 1{S},mathrm dx = 0$$
will still give us a valid density (where $1{S}$ is the indicator function for the set $S$). Apologies if this explanation is not particularly satisfying. I am not sure how much measure theory knowledge to assume, so I am sweeping a ton under the rug. If you're interested in learning more about the formalization of probability, Billingsley's Probability and Measure is an excellent exposition.
answered Jan 20 at 5:36
stats_modelstats_model
612412
answered Jan 20 at 5:36
stats_modelstats_model
612412
answered Jan 20 at 5:36
stats_modelstats_model
612412
answered Jan 20 at 5:36
stats_modelstats_model
612412
612412
add a comment |
add a comment |
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%2f3080195%2fbounded-function-with-finitely-many-discontinuities-is-integrable-overset%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
2
$begingroup$
You can even change an infinite number of values. For example, $1_{[0,1]setminus mathbb{Q}}$ is the density of a uniform r.v. on $[0,1]$.
$endgroup$
– d.k.o.
Jan 20 at 5:19
2
$begingroup$
Definition of density functions involves Lebesgue integration. If $f$ is a density function so is $g$ whenever $g$ is measurable and $f=g$ almost everywhere w.r.t. Lebesgue measure. Riemann integrability is not required for a function to be a density function.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 5:23