2010

[Devika]

Please help me with Q22, and Q18

Question 22 has been discussed earlier, but I din't grasp much.

http://economicsentrance.weebly.com/uploads/1/1/0/5/1105777/option-a-2010.pdf

Thank you :)

[duck]

Hi Devika.. :)

22) X1,X2..Xn are random variables which are uniformly distributed in [0,1]. Now, we need to find the mean of another random variable(Y) which a function of (X1,..Xn)
i.e. Y=max{X1,X2,....Xn}
(As Y is  a max observed completion time)
In order to find the mean of Y, we must know its disrtibution.

So, lets first find out its distribution.
(Pr(Y ≤ y) = Pr(max{X1,X2,...,Xn} ≤ y)
               = Pr(X1 ≤ y,X2 ≤ y,...,Xn ≤ y)
               = Pr(X1 ≤y)·Pr(X2 ≤y)···Pr(Xn ≤y)...by independence of Xi’s
               = y^n
Therefore, pdf of Y , f(y) = n*y^n-1.


Therefore, mean of Y = integral of [(y)*f(y)]dy from 0 to 1.
                              = integral of [y*(n)*(y^n-1)]dy from 0 to 1.
                              = n/n+1

[Devika]

Heyy Duck! Thank you for getting back :)

I understood most of it. Thank you. Have a question though, why is P(X1<=y), and the remaining probabilities, = y?

[Devika]

Got it! Thanks for the solution duck!
Could you explain 18 as well?