What does the value of a PDF mean?
$begingroup$
I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?
probability
asked 1 hour ago
Lea EnglandLea England
161
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add a comment |
$begingroup$
I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?
probability
asked 1 hour ago
Lea EnglandLea England
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago
add a comment |
$begingroup$
I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?
probability
asked 1 hour ago
Lea EnglandLea England
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I understand that the integral of a PDF provides tangible value --i.e., the integral of a PDF allows one to see the probability of a value or less than that value, under a particular distribution, occurring. But, what does the value of just the output of the PDF provide? In other words, what does the PDF of the standard normal distribution at x=0.5 mean?
probability
probability
asked 1 hour ago
Lea EnglandLea England
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Lea EnglandLea England
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Lea EnglandLea England
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Lea EnglandLea England
161
asked 1 hour ago
Lea EnglandLea England
161
161
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lea England is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago
add a comment |
2
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago
2
2
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.
But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.
answered 1 hour ago
cspruncsprun
21414
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csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
The value of the PDF is sort of like the "weighting" that each event gets for the CDF.
answered 1 hour ago
Jacob SmithJacob Smith
111
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$begingroup$
Definition (Gut, 2005): A distribution function F is
- discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.
- continuous iff it is continuous for all x.
- absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
- singular iff F $neq$ 0, F' exists and equals 0 a.e.
By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.
However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.
answered 26 mins ago
Victoria MVictoria M
9113
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
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$begingroup$
I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.
But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.
answered 1 hour ago
cspruncsprun
21414
New contributor
csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.
But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.
answered 1 hour ago
cspruncsprun
21414
New contributor
csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.
But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.
answered 1 hour ago
cspruncsprun
21414
New contributor
csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I would first say that you should try to be OK with it not having a super down-to-earth meaning. It can sometimes help to let yourself be abstract and use the mathematical tools you have to get somewhere you might not have gotten by purely reasoning through with statements you might use to describe other things in your daily life. On the other hand, it is often helpful to have somewhat concrete ways of thinking about things that are abstract, so it's good that you're investigating this.
But as the PDF is the derivative of the CDF, the value of PDF $p(x)$ at $x$ is the (slope of the) best linear approximation of the CDF near $x$, i.e., that if $epsilon$ is super small, then if you want to approximate the probability that an event between $x$ and $x+epsilon$ happens linearly in $epsilon$, your best bet is to say $epsilon cdot p(x)$.
answered 1 hour ago
cspruncsprun
21414
New contributor
csprun is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 20 mins ago
answered 1 hour ago
cspruncsprun
21414
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answered 1 hour ago
cspruncsprun
21414
answered 1 hour ago
cspruncsprun
21414
21414
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add a comment |
add a comment |
$begingroup$
The value of the PDF is sort of like the "weighting" that each event gets for the CDF.
answered 1 hour ago
Jacob SmithJacob Smith
111
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Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
The value of the PDF is sort of like the "weighting" that each event gets for the CDF.
answered 1 hour ago
Jacob SmithJacob Smith
111
New contributor
Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The value of the PDF is sort of like the "weighting" that each event gets for the CDF.
answered 1 hour ago
Jacob SmithJacob Smith
111
New contributor
Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The value of the PDF is sort of like the "weighting" that each event gets for the CDF.
answered 1 hour ago
Jacob SmithJacob Smith
111
New contributor
Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Jacob SmithJacob Smith
111
New contributor
Jacob Smith is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Jacob SmithJacob Smith
111
answered 1 hour ago
Jacob SmithJacob Smith
111
111
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add a comment |
add a comment |
$begingroup$
Definition (Gut, 2005): A distribution function F is
- discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.
- continuous iff it is continuous for all x.
- absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
- singular iff F $neq$ 0, F' exists and equals 0 a.e.
By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.
However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.
answered 26 mins ago
Victoria MVictoria M
9113
$endgroup$
add a comment |
$begingroup$
Definition (Gut, 2005): A distribution function F is
- discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.
- continuous iff it is continuous for all x.
- absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
- singular iff F $neq$ 0, F' exists and equals 0 a.e.
By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.
However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.
answered 26 mins ago
Victoria MVictoria M
9113
$endgroup$
add a comment |
$begingroup$
Definition (Gut, 2005): A distribution function F is
- discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.
- continuous iff it is continuous for all x.
- absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
- singular iff F $neq$ 0, F' exists and equals 0 a.e.
By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.
However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.
answered 26 mins ago
Victoria MVictoria M
9113
$endgroup$
Definition (Gut, 2005): A distribution function F is
- discrete iff for some countable set of numbers {$x_j$} and point masses {$p_j$}, $$F(x) = sum_{x_jleq x} p_j, quad text{for all x} in Bbb R.$$ The function p is called the probability function.
- continuous iff it is continuous for all x.
- absolutely continuous iff there exists a non-negative, Lebesgue integrable function f, such that $$F(b) - F(a) = int_a^b f(x)text{ dx} quad {for space all space x < b.}$$ The function f is called the density of F.
- singular iff F $neq$ 0, F' exists and equals 0 a.e.
By the definition, evaluating a probability distribution function F is simply an integration of the density of said distribution. As mentioned, in the case for a $<$ b, this gives us the probability for a given range of values.
However, if evaluating the distribution function F at a single value, the integral clearly vanishes. To find the probability of a single value, F must be discrete thereby allowing us to compute $$F(a) = sum_{a in X} p_j(a).$$ By taking the limit on $int_a^bf(x)text{dx}$ to F(a + $epsilon$) - F(a - $epsilon$) = $int_{a-epsilon}^{a+epsilon}f(x)text{dx}$, we then thus find the relative probability of F(a) occuring.
answered 26 mins ago
Victoria MVictoria M
9113
answered 26 mins ago
Victoria MVictoria M
9113
answered 26 mins ago
Victoria MVictoria M
9113
answered 26 mins ago
Victoria MVictoria M
9113
9113
add a comment |
add a comment |
Lea England is a new contributor. Be nice, and check out our Code of Conduct.
Lea England is a new contributor. Be nice, and check out our Code of Conduct.
Lea England is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
The PDF is the rate of change of the CDF. No more, no less.
$endgroup$
– JimB
1 hour ago
$begingroup$
Possible duplicate of What does the value of a probability density function (PDF) at some x indicate?
$endgroup$
– Victoria M
1 hour ago