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Finding pdf

  • Thread starter jetoso
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  • #1
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Suppose X1, . . . ,Xn are independently and identically from the uniform distribution on [0, 1]. Find the probability density function of Y = min[X1, X2, ... , Xn].
I do not know how to formulate this problem. I know that the pdf has to be some integral, but no clue so far.
 

Answers and Replies

  • #2
EnumaElish
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First find the CDF of Y: G(y) = Prob(Y<y) = Prob(min{X1,...,Xn} < y) = Prob(at least one X is < y) = Prob(not all X are > y) = 1 - Prob(all X are > y) = 1 - [1 - F(y)]n = 1 - (1 - y)n. Now find the PDF by differentiating with respect to y.
 
  • #3
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Thanks

That's true; since we have that X1,...,Xn are iid, and does not matter if we have <= or < because is a continuous function, then pdf = f(y) = F'(y) = -n(1-y)^(n-1)(-1) = n(1-y)^(n-1).

Thanks.
 

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