You observe k i.i.d. copies of the discrete uniform random variable Xi , which takes values 1 through n with equal probability.
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 .
5.) As k (the number of samples) becomes very large, what is E[M] in terms of n ?
As k→∞ , E[M]→
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
4. E[M] = 2−1/2^k, Var[M] = 1/2^k - 1/2^(2*k)
5. E[M] = n
Please help us!!!
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
1. (m/n)^k
2. (1/n)^k
3. 1-(1/n)^k
3. (1/n)*(m/n)^(k-1)
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
?
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???
I think 1. is:
P(M<=m)=(1/m)^k.