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# valid cumulative distribution functions Watch

1. (Refer to image) I don't understand how to get part a of this question

I realize to be a valid CDF, it must satisfy: F(a)=0 and F(b)=1 and I evaluated the functions in (a) at the endpoints of its domain and ended up with a pair of simultaneous equations : a(b-1)^2 = 1 and a(b+1)^2 = 0

to which I got this result when I expanded and solved simultaneously:
2ab^2 + 2a -1=0. Not sure how to solve that for a and b though?

If we solve a(b+1)^2 = 0 we get b=-1 (which is right), which when substituted into the 1st equation a(b-1)^2 = 1 gives the correct value (according to the answer) for a= 0.25.

But wouldn't a(b+1)^2 = 0 also imply that a=0?
2. (Original post by zomgleh)
(Refer to image) I don't understand how to get part a of this question

I realize to be a valid CDF, it must satisfy: F(a)=0 and F(b)=1 and I evaluated the functions in (a) at the endpoints of its domain and ended up with a pair of simultaneous equations : a(b-1)^2 = 1 and a(b+1)^2 = 0

to which I got this result when I expanded and solved simultaneously:
2ab^2 + 2a -1=0. Not sure how to solve that for a and b though?

If we solve a(b+1)^2 = 0 we get b=-1 (which is right), which when substituted into the 1st equation a(b-1)^2 = 1 gives the correct value (according to the answer) for a= 0.25.

But wouldn't a(b+1)^2 = 0 also imply that a=0?
No. The answer is in your question.

Indeed, under normal circumstances

but you have a second constraint. Namely

This equation has to be true at the same time.
3. (Original post by Indeterminate)
No. The answer is in your question.

Indeed, under normal circumstances

but you have a second constraint. Namely

This equation has to be true at the same time.
just seen this, thank you!

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