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S2 Helpp ):

Im doing cumulative distribution functions and there's something really confusing me. See image.

The blue bit in the square is what i don't get. Why are we adding the integrated part for 0 ≤ x ≤ 1 to the integrated part of 1 ≤ x ≤ 2??

Thanks in advance.
Maths.jpg54.2KB
Reply 1
Avatar for hajs
hajs
OP
11
anyonnee?
Original post by hajs
...


I'll use "s" to avoid confusion: F(s)=∫−∞sf(x)  dxF(s)=\int_{-\infty}^{s}f(x)\; dx

Because f(x) is zero for x < 0 in this case this reduces to

F(s)=∫0sf(x)  dxF(s)=\int_{0}^{s}f(x)\; dx


For s greater than 1, because the formula for f(x) changes at x=1 in this case, we have to split our integral into two parts, to get:

F(s)=∫0sf(x)  dx=∫01f(x)  dx+∫1sf(x)  dxF(s)=\int_{0}^{s}f(x)\; dx=\int_{0}^{1}f(x)\; dx+\int_{1}^{s}f(x)\; dx

and substitute for f(x) as appropriate.
Reply 3
Avatar for hajs
hajs
OP
11
Original post by ghostwalker
I'll use "s" to avoid confusion: F(s)=∫−∞sf(x)  dxF(s)=\int_{-\infty}^{s}f(x)\; dx

Because f(x) is zero for x < 0 in this case this reduces to

F(s)=∫0sf(x)  dxF(s)=\int_{0}^{s}f(x)\; dx


For s greater than 1, because the formula for f(x) changes at x=1 in this case, we have to split our integral into two parts, to get:

F(s)=∫0sf(x)  dx=∫01f(x)  dx+∫1sf(x)  dxF(s)=\int_{0}^{s}f(x)\; dx=\int_{0}^{1}f(x)\; dx+\int_{1}^{s}f(x)\; dx

and substitute for f(x) as appropriate.


is it alwayys 1 or does it vary dependent on the equations given?
Original post by hajs
is it alwayys 1 or does it vary dependent on the equations given?


Depends on the format of f(x). In this case it changed at x=1, but it could be anywhere.
Notice the intervals on which f(x) is defined.

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