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    (Refer to image) I don't understand how to get part a of this questionName:  stats 2.jpg
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    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?
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    (Original post by zomgleh)
    (Refer to image) I don't understand how to get part a of this questionName:  stats 2.jpg
Views: 48
Size:  24.5 KB

    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

    a(b+1)^2 = 0 \Rightarrow a=0

    but you have a second constraint. Namely

    a(b-1)^2 = 1

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

    Indeed, under normal circumstances

    a(b+1)^2 = 0 \Rightarrow a=0

    but you have a second constraint. Namely

    a(b-1)^2 = 1

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