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S1 Help - Cumulative distribution function

Hi there,

Just a quick question, I am slightly confused as to the mark scheme for this question:

(b) The continuous random variable Y takes values between 1 and 2 and its cumulative distribution function F is given, for 1 < y < 2, by
F(y ) = ay + by^2.
Find the values of the constants a and b.

Now I understand to input F(1) into it so I get a+b=0 (like what the mark scheme says) and then F(2) so 2a+4b.. but then it equals this second equation to 1 for some reason?

I know how to find a and b after getting the two equations but I just wanted to know why that second equation equals 1?
Reply 1
Original post by Ryan44
Hi there,

Just a quick question, I am slightly confused as to the mark scheme for this question:

(b) The continuous random variable Y takes values between 1 and 2 and its cumulative distribution function F is given, for 1 < y < 2, by
F(y ) = ay + by^2.
Find the values of the constants a and b.

Now I understand to input F(1) into it so I get a+b=0 (like what the mark scheme says) and then F(2) so 2a+4b.. but then it equals this second equation to 1 for some reason?

I know how to find a and b after getting the two equations but I just wanted to know why that second equation equals 1?


F(y) F(y) represents the cumulative distribution function, its range is given as 0<y<1 0 < y < 1

So to solve we know that at the lowest point of range, F(1)=0 F(1) = 0 , and at the highest point, F(2)=1 F(2) = 1

The 1 1 represents the total probability(the total area under the p.d.f graph)
Reply 2
Original post by raheem94
F(y) F(y) represents the cumulative distribution function, its range is given as 0<y<1 0 < y < 1

So to solve we know that at the lowest point of range, F(1)=0 F(1) = 0 , and at the highest point, F(2)=1 F(2) = 1

The 1 1 represents the total probability(the total area under the p.d.f graph)


So does F(y) always have a range of 0 < y < 1?

Thanks for your answer!
Reply 3
Original post by Ryan44
So does F(y) always have a range of 0 < y < 1?

Thanks for your answer!


No, it depends upon the question.

Example:
It F(y) F(y) has a range of 2<y<5 2 < y < 5 . Then F(2)=0 and F(5)=1 F(2)=0 \ and \ F(5)=1
Reply 4
Original post by raheem94
No, it depends upon the question.

Example:
It F(y) F(y) has a range of 2<y<5 2 < y < 5 . Then F(2)=0 and F(5)=1 F(2)=0 \ and \ F(5)=1


I see what you mean as in my question it is between 1 < y < 2 and that F(1)=0 and F(2)=1

But does the F(higher boundary) always equal to 1 and the F(lower boundary) always = to 0?
Reply 5
Original post by Ryan44
I see what you mean as in my question it is between 1 < y < 2 and that F(1)=0 and F(2)=1

But does the F(higher boundary) always equal to 1 and the F(lower boundary) always = to 0?


Yes.

Do you understand what does F(y) actually represent?

Do you know the difference between f(y) and F(y) ? f(y) \ and \ F(y) \ ?
Reply 6
Original post by raheem94
Yes.

Do you understand what does F(y) actually represent?

Do you know the difference between f(y) and F(y) ? f(y) \ and \ F(y) \ ?


Couldn't give you the definition between the two although I know how to convert f(y) to F(y) and vice versa correctly.

f(y) to F(y) you integrate? and F(y) to f(y) you differentiate the function?
Reply 7
Original post by Ryan44
Couldn't give you the definition between the two although I know how to convert f(y) to F(y) and vice versa correctly.

f(y) to F(y) you integrate? and F(y) to f(y) you differentiate the function?


Yes, you are correct. I thought that you might be misunderstanding them.

We know the total area under the f(y) f(y) curve is 1, so if its range is 0<y<3 0 < y < 3
Then the total area is 03f(y) dy=1=F(3) \displaystyle \int ^3 _0 f(y) \ dy = 1 = F(3)

Hope it makes sense.

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