Mixing
How to find 2 constants in a probability distribution function?
$begingroup$
How to find a and c in this function? In the example they are using limits, but I don't exactly understand why and how.
The answer is: $a = -1, c = 0$
probability statistics
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
$endgroup$
add a comment |
$begingroup$
How to find a and c in this function? In the example they are using limits, but I don't exactly understand why and how.
The answer is: $a = -1, c = 0$
probability statistics
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
$endgroup$
add a comment |
$begingroup$
How to find a and c in this function? In the example they are using limits, but I don't exactly understand why and how.
The answer is: $a = -1, c = 0$
probability statistics
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
$endgroup$
How to find a and c in this function? In the example they are using limits, but I don't exactly understand why and how.
The answer is: $a = -1, c = 0$
probability statistics
probability statistics
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
edited Jan 30 at 1:39
Graham Kemp
87.6k43578
87.6k43578
asked Jan 30 at 0:23
qwerfdqwerfd
32
asked Jan 30 at 0:23
qwerfdqwerfd
32
asked Jan 30 at 0:23
qwerfdqwerfd
32
32
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
When $x = 1$, the cumulative distribution must be continuous with value $1$, so $a cdot 1^2 + 2 cdot 1 + c = 1$. Also, at $x = 0$ the distribution must be continuous with value $0$, so $a cdot 0^2 + 2 cdot 0 + c = 0$. From the second equation, $c = 0$. Plugging into the first gives $a = -1$.
Here's the function:
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When $x = 1$, the cumulative distribution must be continuous with value $1$, so $a cdot 1^2 + 2 cdot 1 + c = 1$. Also, at $x = 0$ the distribution must be continuous with value $0$, so $a cdot 0^2 + 2 cdot 0 + c = 0$. From the second equation, $c = 0$. Plugging into the first gives $a = -1$.
Here's the function:
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
add a comment |
$begingroup$
When $x = 1$, the cumulative distribution must be continuous with value $1$, so $a cdot 1^2 + 2 cdot 1 + c = 1$. Also, at $x = 0$ the distribution must be continuous with value $0$, so $a cdot 0^2 + 2 cdot 0 + c = 0$. From the second equation, $c = 0$. Plugging into the first gives $a = -1$.
Here's the function:
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
add a comment |
$begingroup$
When $x = 1$, the cumulative distribution must be continuous with value $1$, so $a cdot 1^2 + 2 cdot 1 + c = 1$. Also, at $x = 0$ the distribution must be continuous with value $0$, so $a cdot 0^2 + 2 cdot 0 + c = 0$. From the second equation, $c = 0$. Plugging into the first gives $a = -1$.
Here's the function:
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
$endgroup$
When $x = 1$, the cumulative distribution must be continuous with value $1$, so $a cdot 1^2 + 2 cdot 1 + c = 1$. Also, at $x = 0$ the distribution must be continuous with value $0$, so $a cdot 0^2 + 2 cdot 0 + c = 0$. From the second equation, $c = 0$. Plugging into the first gives $a = -1$.
Here's the function:
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
edited Jan 30 at 1:54
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
answered Jan 30 at 1:12
David G. StorkDavid G. Stork
11.6k41534
11.6k41534
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
add a comment |
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
So the value of x will always be the value of the function itself?
$endgroup$
– qwerfd
Jan 30 at 1:41
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
Huhhh??? No of course not! That would mean that $f(x) = x$... a straight line.
$endgroup$
– David G. Stork
Jan 30 at 1:54
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
$begingroup$
I just realized what I've said. And I think I understood, what I wanted to ask in the first place, myself. Thanks for helping out
$endgroup$
– qwerfd
Jan 30 at 2:07
add a comment |
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