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I’m studying on auction. But I couldn’t understand some questions about this topic. And I cannot solve such type questions. For example,

I know that a set of symmetric and risk neutral bidders participate in a first-price sealed bid auction. For the number of bidders $n\ge 0$,

F(v) = the distribution function from which bidders’ valuations (independently drawn and it has uniform distribution $v\in [0,1]$, similarly f(v) is density function. $b(v)$ is strategies in auction and calculated by

$$b(v)=\int_0^v x \frac{(n-1)F(x)^{n-2}f(x)}{F(v)^{n-1}} dx$$

Firstly I try to show that $b(v) is increasing in v. Secondly, how can I show that b(v) is equilibrium strategy.

As for second price sealed bid, I try to show there exist an equilibrium for all bidders to bid their true valuations. And finally I want to compare the expected selling prices in the two auctions for conditional on a fixed highest valuation on $v_n$ by arranging bidders’ valuations from highest to lowest $v_1\ge … \ge v_n$. Well,how can I decide which auctions would a risk averse sellers prefer?

Question is this. And I cannot produce any proper solution for such type questions.

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now, I mention about my solution trials;

I can only show that the bidding function b(v) is increasing in v.

For that I rewrite the bidding function as follows:

$$b(v)=\frac{1}{F(v)^{n-1}}\int_0^v x dF^{n-1}(x)dx$$

Since each bidders value is uniformly distributed on [0,1], I can say F(v)=v and f(v)=1, so if there are n bidders then each employs the bidding function

$$b(v)={1\over v^{n-1}}\int^v_0 xdx^{n-1}={1\over v^{n-1}}\int^v_0 x(n-1)x^{n-2}dx$$ $$={n-1\over v^{n-1}}\int^v_0 x^{n-1}dx={1\over v^{n-1}}{1\over n} v^n = v-(v/N)=((n-1)v)/n$$

So as v increases, the bidding function will increase.

However, after this part of the question, I cannot solve. I also read some materials (like that https://dvikan.no/ntnu-studentserver/kompendier/14%20jehle%20reny.pdf ) but I could not. Please help me to solve these types questions. Thanks

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In first-price sealed bid auctions it is not a dominant strategy to bid your valuation. Consider the expected payoff function:

$$\Pi(V_{i},b_{i})=(V_{i}-b_{i})*\mathcal{Pr}(b_{i} \quad wins)$$

now what would be your payoff if you bid according to your value? It's 0. However, if you bid $b_{i}<V_{i}$ this actually gives you the chance of winning the object at a price lower than your valuation. Hence, there is an optimal amount of bid shading. So how to compute this? Right, using the first order condition, just as in basic micro:

$$\dfrac{\mathrm{d}\Pi(V_{i},b_{i})}{\mathrm{d}b_{i}}=-\mathcal{Pr}(b_{i} \quad wins) +(V_{i}-b_{i})*\dfrac{\mathrm{d}}{\mathrm{d}b_{i}}\mathcal{Pr}(b_{i} \quad wins)$$

now you set this to zero and solve for $b_{i}$

$$b_{i} = V_{i}-\dfrac{\mathcal{Pr}(b_{i} \quad wins)}{\dfrac{\mathrm{d}}{\mathrm{d}b_{i}}\mathcal{Pr}(b_{i} \quad wins)}$$

Given your assumptions, and the uniform distribution, it should be easy for you to find out what $\mathcal{Pr}(b_{i} \quad wins)$ actually is. For the first question to demonstrate the increasing bid function it might be helpful for you to substitute the pdf and CDF of the uniform for your general expression. If done correctly, you will note that the bid function actually is given by

$$b_{i}(V_{i}) = \dfrac{N-1}{N}V_{i}$$

which is clearly increasing in values.

I hope this helps you to proceed. Feel free to ask questions if you need more help.

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  • $\begingroup$ I understand now how to show this is increasing. Yes I need help. Please show the other parts. I’m really difficult in doing the other parts as well. And you do really good explanation and solutions. Thanks a lot😊 $\endgroup$ – mnm123 May 12 '18 at 2:01

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