Question

Question 3 Jo and Jim are bidding in a second-price, independent, private value auction. Their values for the object are independently and uniformly distributed on the interval (1,2). (1) What is the expected value of the winning bidder? (2) What is the expected value of the price?

Answer 1

Hi

The answer of the following question is given below as follows :

Ans.1) first of all we will assume the value as X be the random variables denoting the Jo's bidding in second price during the auction and Y be the random variable that denotes the Jim's bidding in a second price during the auction.

Now the equation will be as follows

X~U(1,2) & Y~U(1,2).

Now coming to the part 1 of the answer.

Z= Max.(X,Y).

P(Z<=z) = PX(Max.(X,Y)<=z)

P(X<=z) P(Y<=z)

Note : Here X & Y are the independent distributed random variable.

(z-1/2-1)(z-1/2-1)=(z-1)²

Then the expected value of the running table as follows :

=E(Z)

{^2₁ z fz(z)dz

By solving the above equation of limit.

We get the ans. As 1.667

Ans.2) Now we'll concentrate on the second and the last part of the question .

So the minimum price at which the second selling price is U.

Then the U is the random variable .

U = Min.{X,Y}

U+Z=X+Y ------1

So the price will lie in between the U & Z

So the Expected value of the price will be given as

E(U+Z/2)=E (X+Y/2) - BY PUTTING THE EQ.1

=1/2 {E(X)+E(Y)}

=1/2{3/2+3/2}

= 1.5 Ans.

I hope I have served the purpose well.

Thanks.