Bizarre behaviour of a probability density function [closed]
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It is said in my lecture notes that if $f$ is a probability density function, then it must fulfill the condition:
$$int_{-infty}^{infty} f(x)dx = 1$$
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
Then it can be observed when integrating over $[0,40]$, the definite integral is 1. But consider also:
$$12.5int_{-infty}^{infty} f(x)dx = 0$$
Does this disqualify $f$ from being a probability density function? What am I misunderstanding here?
probability-distributions random-variables
closed as unclear what you're asking by Martin R, NCh, José Carlos Santos, Cesareo, Lord_Farin Nov 28 at 22:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
1
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It is said in my lecture notes that if $f$ is a probability density function, then it must fulfill the condition:
$$int_{-infty}^{infty} f(x)dx = 1$$
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
Then it can be observed when integrating over $[0,40]$, the definite integral is 1. But consider also:
$$12.5int_{-infty}^{infty} f(x)dx = 0$$
Does this disqualify $f$ from being a probability density function? What am I misunderstanding here?
probability-distributions random-variables
closed as unclear what you're asking by Martin R, NCh, José Carlos Santos, Cesareo, Lord_Farin Nov 28 at 22:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It is said in my lecture notes that if $f$ is a probability density function, then it must fulfill the condition:
$$int_{-infty}^{infty} f(x)dx = 1$$
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
Then it can be observed when integrating over $[0,40]$, the definite integral is 1. But consider also:
$$12.5int_{-infty}^{infty} f(x)dx = 0$$
Does this disqualify $f$ from being a probability density function? What am I misunderstanding here?
probability-distributions random-variables
It is said in my lecture notes that if $f$ is a probability density function, then it must fulfill the condition:
$$int_{-infty}^{infty} f(x)dx = 1$$
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
Then it can be observed when integrating over $[0,40]$, the definite integral is 1. But consider also:
$$12.5int_{-infty}^{infty} f(x)dx = 0$$
Does this disqualify $f$ from being a probability density function? What am I misunderstanding here?
probability-distributions random-variables
probability-distributions random-variables
asked Nov 25 at 14:07
hephaes
1627
1627
closed as unclear what you're asking by Martin R, NCh, José Carlos Santos, Cesareo, Lord_Farin Nov 28 at 22:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Martin R, NCh, José Carlos Santos, Cesareo, Lord_Farin Nov 28 at 22:20
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13
add a comment |
3
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13
3
3
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13
add a comment |
2 Answers
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2
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This is a case where good notation might be really helpful. You write
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit
$$ f(x) = begin{cases}
frac{12.5}{(10+x)^2} & text{if $xin[0,40]$, and} \
0 & text{if $x in (-infty,0)cup(40,infty)$}. end{cases} $$
This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-infty,0)cup(40,infty)$ on which it is zero.
Integrating this over $mathbb{R}$, we obtain
begin{align}
int_{-infty}^{infty} f(x) ,mathrm{d}x
&= int_{-infty}^{0} f(x) ,mathrm{d}x + int_{0}^{40} f(x) ,mathrm{d}x + int_{40}^{infty} f(x) ,mathrm{d}x tag{1}\
&= int_{-infty}^{0} 0 ,mathrm{d}x + int_{0}^{40} frac{12.5}{(10+x)^2} ,mathrm{d}x + int_{40}^{infty} 0 ,mathrm{d}x tag{2}\
&= 0 + 1 + 0 tag{3} \
&= 1.
end{align}
In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $int_a^b 0 ,mathrm{d}x= 0$ for any $a le b$, where either could be infinite.
add a comment |
up vote
1
down vote
The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)
There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This is a case where good notation might be really helpful. You write
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit
$$ f(x) = begin{cases}
frac{12.5}{(10+x)^2} & text{if $xin[0,40]$, and} \
0 & text{if $x in (-infty,0)cup(40,infty)$}. end{cases} $$
This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-infty,0)cup(40,infty)$ on which it is zero.
Integrating this over $mathbb{R}$, we obtain
begin{align}
int_{-infty}^{infty} f(x) ,mathrm{d}x
&= int_{-infty}^{0} f(x) ,mathrm{d}x + int_{0}^{40} f(x) ,mathrm{d}x + int_{40}^{infty} f(x) ,mathrm{d}x tag{1}\
&= int_{-infty}^{0} 0 ,mathrm{d}x + int_{0}^{40} frac{12.5}{(10+x)^2} ,mathrm{d}x + int_{40}^{infty} 0 ,mathrm{d}x tag{2}\
&= 0 + 1 + 0 tag{3} \
&= 1.
end{align}
In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $int_a^b 0 ,mathrm{d}x= 0$ for any $a le b$, where either could be infinite.
add a comment |
up vote
2
down vote
accepted
This is a case where good notation might be really helpful. You write
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit
$$ f(x) = begin{cases}
frac{12.5}{(10+x)^2} & text{if $xin[0,40]$, and} \
0 & text{if $x in (-infty,0)cup(40,infty)$}. end{cases} $$
This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-infty,0)cup(40,infty)$ on which it is zero.
Integrating this over $mathbb{R}$, we obtain
begin{align}
int_{-infty}^{infty} f(x) ,mathrm{d}x
&= int_{-infty}^{0} f(x) ,mathrm{d}x + int_{0}^{40} f(x) ,mathrm{d}x + int_{40}^{infty} f(x) ,mathrm{d}x tag{1}\
&= int_{-infty}^{0} 0 ,mathrm{d}x + int_{0}^{40} frac{12.5}{(10+x)^2} ,mathrm{d}x + int_{40}^{infty} 0 ,mathrm{d}x tag{2}\
&= 0 + 1 + 0 tag{3} \
&= 1.
end{align}
In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $int_a^b 0 ,mathrm{d}x= 0$ for any $a le b$, where either could be infinite.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This is a case where good notation might be really helpful. You write
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit
$$ f(x) = begin{cases}
frac{12.5}{(10+x)^2} & text{if $xin[0,40]$, and} \
0 & text{if $x in (-infty,0)cup(40,infty)$}. end{cases} $$
This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-infty,0)cup(40,infty)$ on which it is zero.
Integrating this over $mathbb{R}$, we obtain
begin{align}
int_{-infty}^{infty} f(x) ,mathrm{d}x
&= int_{-infty}^{0} f(x) ,mathrm{d}x + int_{0}^{40} f(x) ,mathrm{d}x + int_{40}^{infty} f(x) ,mathrm{d}x tag{1}\
&= int_{-infty}^{0} 0 ,mathrm{d}x + int_{0}^{40} frac{12.5}{(10+x)^2} ,mathrm{d}x + int_{40}^{infty} 0 ,mathrm{d}x tag{2}\
&= 0 + 1 + 0 tag{3} \
&= 1.
end{align}
In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $int_a^b 0 ,mathrm{d}x= 0$ for any $a le b$, where either could be infinite.
This is a case where good notation might be really helpful. You write
But then consider this function
$$f(x) = frac{12.5}{(10+x)^2}$$ when $x in [0,40]$, and it is evaluated at 0 otherwise.
It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit
$$ f(x) = begin{cases}
frac{12.5}{(10+x)^2} & text{if $xin[0,40]$, and} \
0 & text{if $x in (-infty,0)cup(40,infty)$}. end{cases} $$
This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-infty,0)cup(40,infty)$ on which it is zero.
Integrating this over $mathbb{R}$, we obtain
begin{align}
int_{-infty}^{infty} f(x) ,mathrm{d}x
&= int_{-infty}^{0} f(x) ,mathrm{d}x + int_{0}^{40} f(x) ,mathrm{d}x + int_{40}^{infty} f(x) ,mathrm{d}x tag{1}\
&= int_{-infty}^{0} 0 ,mathrm{d}x + int_{0}^{40} frac{12.5}{(10+x)^2} ,mathrm{d}x + int_{40}^{infty} 0 ,mathrm{d}x tag{2}\
&= 0 + 1 + 0 tag{3} \
&= 1.
end{align}
In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $int_a^b 0 ,mathrm{d}x= 0$ for any $a le b$, where either could be infinite.
answered Nov 25 at 14:18
Xander Henderson
14k103553
14k103553
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add a comment |
up vote
1
down vote
The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)
There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.
add a comment |
up vote
1
down vote
The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)
There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.
add a comment |
up vote
1
down vote
up vote
1
down vote
The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)
There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.
The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)
There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.
answered Nov 25 at 14:11
Ethan Bolker
40.2k545106
40.2k545106
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3
I don't understand why you believe that the last integral is zero. I also don't know where the factor of 12.5 came from in that integral (unless you are duplicating the factor in $f$ for some reason?).
– Xander Henderson
Nov 25 at 14:09
If the integral in first display is $1$ [I didn't check] then the second display should just be $12.5cdot 1=12.5,$
– coffeemath
Nov 25 at 14:13