From pdf to distribution function
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0
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I have the following probability density function (pdf):
$f(x)={0$ if $x<= 11$
$x-11$ if $11leq x leq 12$
$13-x$ if $12leq x leq 13 }$
My aim is to find the distribution function $F(x)$ where the integral from $-infty$ to $+infty$ of the pdf $f(x)$ equals one. I know I have to integrate the pdf over the various sub-intervals indicated above, however, I get something that is not correct. Could somebody please explain me step by step how to do it?
I apologize for my bad typing of the problem, I hope you can understand.
Thank you in advance
probability-theory
asked Nov 16 at 11:47
Matteo
1
add a comment |
up vote
0
down vote
favorite
I have the following probability density function (pdf):
$f(x)={0$ if $x<= 11$
$x-11$ if $11leq x leq 12$
$13-x$ if $12leq x leq 13 }$
My aim is to find the distribution function $F(x)$ where the integral from $-infty$ to $+infty$ of the pdf $f(x)$ equals one. I know I have to integrate the pdf over the various sub-intervals indicated above, however, I get something that is not correct. Could somebody please explain me step by step how to do it?
I apologize for my bad typing of the problem, I hope you can understand.
Thank you in advance
probability-theory
asked Nov 16 at 11:47
Matteo
1
1
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
Yes it is also a cdf
– Matteo
Nov 16 at 13:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following probability density function (pdf):
$f(x)={0$ if $x<= 11$
$x-11$ if $11leq x leq 12$
$13-x$ if $12leq x leq 13 }$
My aim is to find the distribution function $F(x)$ where the integral from $-infty$ to $+infty$ of the pdf $f(x)$ equals one. I know I have to integrate the pdf over the various sub-intervals indicated above, however, I get something that is not correct. Could somebody please explain me step by step how to do it?
I apologize for my bad typing of the problem, I hope you can understand.
Thank you in advance
probability-theory
asked Nov 16 at 11:47
Matteo
1
I have the following probability density function (pdf):
$f(x)={0$ if $x<= 11$
$x-11$ if $11leq x leq 12$
$13-x$ if $12leq x leq 13 }$
My aim is to find the distribution function $F(x)$ where the integral from $-infty$ to $+infty$ of the pdf $f(x)$ equals one. I know I have to integrate the pdf over the various sub-intervals indicated above, however, I get something that is not correct. Could somebody please explain me step by step how to do it?
I apologize for my bad typing of the problem, I hope you can understand.
Thank you in advance
probability-theory
probability-theory
asked Nov 16 at 11:47
Matteo
1
asked Nov 16 at 11:47
Matteo
1
asked Nov 16 at 11:47
Matteo
1
asked Nov 16 at 11:47
Matteo
1
asked Nov 16 at 11:47
Matteo
1
1
1
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
Yes it is also a cdf
– Matteo
Nov 16 at 13:04
add a comment |
1
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
Yes it is also a cdf
– Matteo
Nov 16 at 13:04
1
1
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
Yes it is also a cdf
– Matteo
Nov 16 at 13:04
Yes it is also a cdf
– Matteo
Nov 16 at 13:04
add a comment |
1 Answer
1
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up vote
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Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.
From $11$ to $12$ ($11leq yleq 12$)
$$F(y) = int_{11}^{y}(x - 11)dx = frac{y^2}{2} - 11y + frac{121}{2} $$
For $12$ to $13$ ($12leq yleq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$
Now for $12$ to $13$ ($12leq yleq 13$)
$$F(y) = 0.5 + int_{12}^{y}(13 - x)dx = 13y - frac{y^2}{2} - 83.5$$
After $13$, i.e. $ygeq 13$
$$F(y) = 1$$
as can be seen by putting $y = 13$ in the second equation.
answered Nov 16 at 12:17
Sauhard Sharma
3108
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.
From $11$ to $12$ ($11leq yleq 12$)
$$F(y) = int_{11}^{y}(x - 11)dx = frac{y^2}{2} - 11y + frac{121}{2} $$
For $12$ to $13$ ($12leq yleq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$
Now for $12$ to $13$ ($12leq yleq 13$)
$$F(y) = 0.5 + int_{12}^{y}(13 - x)dx = 13y - frac{y^2}{2} - 83.5$$
After $13$, i.e. $ygeq 13$
$$F(y) = 1$$
as can be seen by putting $y = 13$ in the second equation.
answered Nov 16 at 12:17
Sauhard Sharma
3108
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
add a comment |
up vote
0
down vote
Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.
From $11$ to $12$ ($11leq yleq 12$)
$$F(y) = int_{11}^{y}(x - 11)dx = frac{y^2}{2} - 11y + frac{121}{2} $$
For $12$ to $13$ ($12leq yleq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$
Now for $12$ to $13$ ($12leq yleq 13$)
$$F(y) = 0.5 + int_{12}^{y}(13 - x)dx = 13y - frac{y^2}{2} - 83.5$$
After $13$, i.e. $ygeq 13$
$$F(y) = 1$$
as can be seen by putting $y = 13$ in the second equation.
answered Nov 16 at 12:17
Sauhard Sharma
3108
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.
From $11$ to $12$ ($11leq yleq 12$)
$$F(y) = int_{11}^{y}(x - 11)dx = frac{y^2}{2} - 11y + frac{121}{2} $$
For $12$ to $13$ ($12leq yleq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$
Now for $12$ to $13$ ($12leq yleq 13$)
$$F(y) = 0.5 + int_{12}^{y}(13 - x)dx = 13y - frac{y^2}{2} - 83.5$$
After $13$, i.e. $ygeq 13$
$$F(y) = 1$$
as can be seen by putting $y = 13$ in the second equation.
answered Nov 16 at 12:17
Sauhard Sharma
3108
Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.
From $11$ to $12$ ($11leq yleq 12$)
$$F(y) = int_{11}^{y}(x - 11)dx = frac{y^2}{2} - 11y + frac{121}{2} $$
For $12$ to $13$ ($12leq yleq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$
Now for $12$ to $13$ ($12leq yleq 13$)
$$F(y) = 0.5 + int_{12}^{y}(13 - x)dx = 13y - frac{y^2}{2} - 83.5$$
After $13$, i.e. $ygeq 13$
$$F(y) = 1$$
as can be seen by putting $y = 13$ in the second equation.
answered Nov 16 at 12:17
Sauhard Sharma
3108
answered Nov 16 at 12:17
Sauhard Sharma
3108
answered Nov 16 at 12:17
Sauhard Sharma
3108
answered Nov 16 at 12:17
Sauhard Sharma
3108
3108
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
add a comment |
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
Thank you Sauhard! However, the solution that I have reads: $0$ for $xleq 11$; $frac{1}{2}(x-11)^2$ for $11 leq x leq 12$; $1-frac{1}{2}(13-x)^2$ for $12 leq x leq 13$ and $1$ for $x geq 12$
– Matteo
Nov 16 at 13:05
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
You can factorise my equations so that they fit your way of solution. For $11leq y leq 12$ , rewrite my equation of $F(y) = frac{y^2}{2} - 11y + frac{121}{2} = frac{1}{2}(y^2 - 22y + 121) = frac{1}{2}(y-11)^2$. For $12 leq y leq 13$, rewrite $F(y) = 13y - frac{y^2}{2} - 83.5 = 1 +13y - frac{y^2}{2} - 84.5 = 1 - frac{1}{2}(y^2 - 26y + 169) = 1 - frac{1}{2}(13 - y)^2$
– Sauhard Sharma
Nov 16 at 13:21
Thank you very much!
– Matteo
Nov 16 at 14:18
Thank you very much!
– Matteo
Nov 16 at 14:18
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
If you don't mind, could you please accept the answer
– Sauhard Sharma
20 hours ago
add a comment |
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1
The $F(x)$ in your 5th line, is it a cumulative distribution function ?
– Sauhard Sharma
Nov 16 at 11:55
Yes it is also a cdf
– Matteo
Nov 16 at 13:04