Asked by Lex
                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 .
            
            
        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 .
Answers
                    Answered by
            Lex
            
    5.) As  k  (the number of samples) becomes very large, what is  E[M]  in terms of  n ?
As k→∞ , E[M]→
    
As k→∞ , E[M]→
                    Answered by
            Agawata
            
    Please help us!!!
    
                    Answered by
            melo
            
    1. (m/n)^k
2. (1/n)^k
3. 1-(1/n)^k
    
2. (1/n)^k
3. 1-(1/n)^k
                    Answered by
            melo
            
    4. E[M] = 2−1/2^k, Var[M] = 1/2^k - 1/2^(2*k)
5. E[M] = n
 
    
5. E[M] = n
                    Answered by
            Lo
            
    I think 1. is:
P(M<=m)=(1/m)^k.
    
P(M<=m)=(1/m)^k.
                    Answered by
            Fred
            
    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???
    
                    Answered by
            Anonymous
            
    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
?
    
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
?
                    Answered by
            Anonymous
            
    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
    
Anon is right. P(M=m) = P(M<=m) - P(M<=m-1) = (m/n)^k - [(m-1)/n)]^k
                    Answered by
            Anonymous
            
    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
    
P(M=m e {2,...,n})=P(M<=m) - P(M=1) = (m/n)^k - (1/n)^k
                    Answered by
            BiggusDickus
            
    3. (1/n)*(m/n)^(k-1)
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.