Finding PDF of Y from Uniform Distribution of X1-Xn

[jetoso]

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.

 

[EnumaElish]

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.

 

[jetoso]

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.