What is the PDF of $Pr(X>x)$ for a uniform distribution? [on hold]
I understand that the integral of CDF, $Pr(Xleq x)$, is pdf, and that the pdf of a unifrom distribution is $1/(b-a)$ for $[a,b]$.
But if I have to find the PDF of $Pr(X>x)$ then what would the limits and values of the pdf be?
Thanks.
probability-distributions
edited Dec 26 at 3:12
Rócherz
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
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Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by StubbornAtom, NCh, Andrew, Eevee Trainer, KReiser 1 hour ago
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I understand that the integral of CDF, $Pr(Xleq x)$, is pdf, and that the pdf of a unifrom distribution is $1/(b-a)$ for $[a,b]$.
But if I have to find the PDF of $Pr(X>x)$ then what would the limits and values of the pdf be?
Thanks.
probability-distributions
edited Dec 26 at 3:12
Rócherz
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by StubbornAtom, NCh, Andrew, Eevee Trainer, KReiser 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24
add a comment |
I understand that the integral of CDF, $Pr(Xleq x)$, is pdf, and that the pdf of a unifrom distribution is $1/(b-a)$ for $[a,b]$.
But if I have to find the PDF of $Pr(X>x)$ then what would the limits and values of the pdf be?
Thanks.
probability-distributions
edited Dec 26 at 3:12
Rócherz
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I understand that the integral of CDF, $Pr(Xleq x)$, is pdf, and that the pdf of a unifrom distribution is $1/(b-a)$ for $[a,b]$.
But if I have to find the PDF of $Pr(X>x)$ then what would the limits and values of the pdf be?
Thanks.
probability-distributions
probability-distributions
edited Dec 26 at 3:12
Rócherz
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 at 3:12
Rócherz
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 26 at 3:12
Rócherz
2,7612721
edited Dec 26 at 3:12
Rócherz
2,7612721
edited Dec 26 at 3:12
Rócherz
2,7612721
2,7612721
asked Dec 26 at 1:15
Yasir Khan
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 26 at 1:15
Yasir Khan
1
asked Dec 26 at 1:15
Yasir Khan
1
1
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yasir Khan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by StubbornAtom, NCh, Andrew, Eevee Trainer, KReiser 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by StubbornAtom, NCh, Andrew, Eevee Trainer, KReiser 1 hour ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – StubbornAtom, Eevee Trainer, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24
add a comment |
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24
add a comment |
1 Answer
1
active
oldest
votes
Welcome to the Math.SE!
In fact, the probability density function of the random variable $X$, which is distributed uniformly on the interval $[a, b]$ is
$$f(x) = frac{1}{b-a}$$
To find the cumulative distribution function $F(x)$ one should integrate the density function, i.e.
$$F(x) = mathbb{P}(X leq x) = int_{a}^{x}{frac{dx}{b - a}} = frac{x-a}{b-a}$$
Now note that $$mathbb{P}(X > x) = 1 - mathbb{P}(X leq x) = 1 - frac{x - a}{b-a} = frac{b-x}{b-a}$$
On the other hand, we obtain the same result by integration
$$mathbb{P}(X > x) = int_{x}^{b} {frac{dx} { b - a }} = frac{b-x}{b-a}$$
answered Dec 26 at 1:25
hyperkahler
1,501714
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Welcome to the Math.SE!
In fact, the probability density function of the random variable $X$, which is distributed uniformly on the interval $[a, b]$ is
$$f(x) = frac{1}{b-a}$$
To find the cumulative distribution function $F(x)$ one should integrate the density function, i.e.
$$F(x) = mathbb{P}(X leq x) = int_{a}^{x}{frac{dx}{b - a}} = frac{x-a}{b-a}$$
Now note that $$mathbb{P}(X > x) = 1 - mathbb{P}(X leq x) = 1 - frac{x - a}{b-a} = frac{b-x}{b-a}$$
On the other hand, we obtain the same result by integration
$$mathbb{P}(X > x) = int_{x}^{b} {frac{dx} { b - a }} = frac{b-x}{b-a}$$
answered Dec 26 at 1:25
hyperkahler
1,501714
add a comment |
Welcome to the Math.SE!
In fact, the probability density function of the random variable $X$, which is distributed uniformly on the interval $[a, b]$ is
$$f(x) = frac{1}{b-a}$$
To find the cumulative distribution function $F(x)$ one should integrate the density function, i.e.
$$F(x) = mathbb{P}(X leq x) = int_{a}^{x}{frac{dx}{b - a}} = frac{x-a}{b-a}$$
Now note that $$mathbb{P}(X > x) = 1 - mathbb{P}(X leq x) = 1 - frac{x - a}{b-a} = frac{b-x}{b-a}$$
On the other hand, we obtain the same result by integration
$$mathbb{P}(X > x) = int_{x}^{b} {frac{dx} { b - a }} = frac{b-x}{b-a}$$
answered Dec 26 at 1:25
hyperkahler
1,501714
add a comment |
Welcome to the Math.SE!
In fact, the probability density function of the random variable $X$, which is distributed uniformly on the interval $[a, b]$ is
$$f(x) = frac{1}{b-a}$$
To find the cumulative distribution function $F(x)$ one should integrate the density function, i.e.
$$F(x) = mathbb{P}(X leq x) = int_{a}^{x}{frac{dx}{b - a}} = frac{x-a}{b-a}$$
Now note that $$mathbb{P}(X > x) = 1 - mathbb{P}(X leq x) = 1 - frac{x - a}{b-a} = frac{b-x}{b-a}$$
On the other hand, we obtain the same result by integration
$$mathbb{P}(X > x) = int_{x}^{b} {frac{dx} { b - a }} = frac{b-x}{b-a}$$
answered Dec 26 at 1:25
hyperkahler
1,501714
Welcome to the Math.SE!
In fact, the probability density function of the random variable $X$, which is distributed uniformly on the interval $[a, b]$ is
$$f(x) = frac{1}{b-a}$$
To find the cumulative distribution function $F(x)$ one should integrate the density function, i.e.
$$F(x) = mathbb{P}(X leq x) = int_{a}^{x}{frac{dx}{b - a}} = frac{x-a}{b-a}$$
Now note that $$mathbb{P}(X > x) = 1 - mathbb{P}(X leq x) = 1 - frac{x - a}{b-a} = frac{b-x}{b-a}$$
On the other hand, we obtain the same result by integration
$$mathbb{P}(X > x) = int_{x}^{b} {frac{dx} { b - a }} = frac{b-x}{b-a}$$
answered Dec 26 at 1:25
hyperkahler
1,501714
answered Dec 26 at 1:25
hyperkahler
1,501714
answered Dec 26 at 1:25
hyperkahler
1,501714
answered Dec 26 at 1:25
hyperkahler
1,501714
1,501714
add a comment |
add a comment |
On what probability space? I guess that what you ask for, is to find the probability density function of the random variable $P(X>x)$. But the problem is, we don't know what the probability space is, so we don't have enough information for characterization of this random variable
– Jakobian
Dec 26 at 1:33
If you know the probability $P(A)$ of an event $A$, then the complement event $overline{A}$ has probability $PBig(,overline{A},Big) = 1 - P(A)$. This rule applies to probability functions like the pdf's and cdf's in a discrete or continuous random variable.
– Benedict Voltaire
Dec 26 at 1:56
“Integral of CDF, [...], is pdf” no. Integral of PDF is the CDF. $mathbb{P}(X>c)$ is the complementary CDF, or sometimes survival distribution. It is associated to the same PDF as the underlying RV never changes. Review the definitions.
– LoveTooNap29
Dec 26 at 3:24