Thread: Independent Uniform Random Variables

If X1,X2, . . . ,Xn are independent U(0, 1) random variables and Y=min (X1,X2, . . . ,Xn) has PDF given by

fY(x) = { , 0<=x<=1}, 0, otherwise

Identify this distribution, and hence state E[Y] and Var(Y).

Im thinkin it looks similar to a binomial distribution? Am i on the right lines?

Originally Posted by sirellwood

If X1,X2, . . . ,Xn are independent U(0, 1) random variables and Y=min (X1,X2, . . . ,Xn) has PDF given by

fY(x) = { , 0<=x<=1}, 0, otherwise Mr F says: What you have posted here is wrong. It's not even a pdf ....

Identify this distribution, and hence state E[Y] and Var(Y).

Im thinkin it looks similar to a binomial distribution? Am i on the right lines?

There is a (simple) mistake in your pdf which I'll leave you to figure out and fix.

The correct pdf has a beta distribution (and you will need to determine what value each parameters has).

Ah thank you mr fantastic! its edited now!

So i first need to find the sample mean and variance right? Because it is uniformally distributed, is the sample mean just (x1+xn)/2?

Im getting a little confused with the word "min"(x1,x2....) in the Y term.

minimum of X1,...,Xn is the smallest of these rvs
It's also known as the smallest order stat.
It couldn't be a binomial since the underlying distribution here is continuous (uniform)

for any 0<a<1





So

THIS is a Beta with parameters 1 and n.
Look that up for your mean and variance.

Originally Posted by sirellwood

Ah thank you mr fantastic! its edited now!

So i first need to find the sample mean and variance right? Because it is uniformally distributed, is the sample mean just (x1+xn)/2?

Im getting a little confused with the word "min"(x1,x2....) in the Y term.

The min business is irrelevant (unless you want to derive the given pdf). The bottom line is that you're given a pdf and asked to identify it and hence get it's mean and variance. Note that you can check your answers by calculating the mean and variance directly from the pdf.