Let X1,X2,...,Xn be independent random variables, each having a uniform distribution over (0, 1). Let M=maximum (X1,X2,...,Xn). Show that the distribution
function of M,FM(·), is given by
FM(x)=x^n.
Solution:
As the random variables are independent, probability that all the random variables are less than or equal to a value say "a" would be (a)^n i.e P(X1<=a,X2<=a.........Xn<=a)=P(X1<=a)*P(X2<=a)........P(Xn<=a) where 0<a<1( due to independence and uniform distribution).
Now we are given that M=max(X1,X2....Xn). Thus M too lies in the interval (0,1).
Now lets consider FM(a) which is equal to P(M<=a). As the value of M is a maximum function over (X1.....Xn), the values M<=a can be obtained in many ways. If I am correct, how do I proceed from here?
function of M,FM(·), is given by
FM(x)=x^n.
Solution:
As the random variables are independent, probability that all the random variables are less than or equal to a value say "a" would be (a)^n i.e P(X1<=a,X2<=a.........Xn<=a)=P(X1<=a)*P(X2<=a)........P(Xn<=a) where 0<a<1( due to independence and uniform distribution).
Now we are given that M=max(X1,X2....Xn). Thus M too lies in the interval (0,1).
Now lets consider FM(a) which is equal to P(M<=a). As the value of M is a maximum function over (X1.....Xn), the values M<=a can be obtained in many ways. If I am correct, how do I proceed from here?