You observe k i.i.d. copies of the discrete uniform random variable Xi , which takes values 1 through n with equal probability.

Question

Define the random variable M as the maximum of these random variables, M=maxi(Xi) .

1.) Find the probability that M≤m , as a function of m , for m∈{1,2,…,n} .

2.) Find the probability that M=1 .

3.) Find the probability that M=m for m∈{2,3,…n} .

4.) For n=2 , find E[M] and Var(M) as a function of k .

Answer 1

5.) As k (the number of samples) becomes very large, what is E[M] in terms of n ?
As k→∞ , E[M]→

Answer 2

Please help us!!!

Answer 3

1. (m/n)^k
2. (1/n)^k
3. 1-(1/n)^k

Answer 4

4. E[M] = 2−1/2^k, Var[M] = 1/2^k - 1/2^(2*k)
5. E[M] = n

Answer 5

I think 1. is:
P(M<=m)=(1/m)^k.

Answer 6

I think 1 is: (n-m+1)/n . There are always n numbers to choose from but only a subset of those will be >= the current max???

Answer 7

3.) Find the probability that M=m for m∈{2,3,…n} .
3. 1-(1/n)^k
Shouldn't this probability depend on m somehow?
P(M=m)=P(M<=m+1)-P(M<=m)=(m+1/n)^k-(m/n)^k
?

Answer 8

Yeah melo, your number 3 is clearly wrong. It is dependent on m, first of all

Anon is right. P(M=m) = P(M<=m) - P(M<=m-1) = (m/n)^k - [(m-1)/n)]^k

Answer 9

Isn't this the probability of m=all possible values minus the prob of m=1?

P(M=m e {2,...,n})=P(M<=m) - P(M=1) = (m/n)^k - (1/n)^k

Answer 10

3. (1/n)*(m/n)^(k-1)