User talk:Malin84

Monotone Utility and Junk Food
I've checked the past exams and this is not a prelim question. Must be a homework question then. The question reads like this:

Assume your preference over food is described by the following function U(t,h), where "t" stands for the tastiness of the food and "h" stands for the healthiness of the food. Your utility is strictly increasing in both dimensions of food: you like the food better if it is more tasty(t higher) or if it is more healthy(h higher).

Prove: If you want to sustain a constant level of utility, then among the food that you eat, junk food must be tasty.

Well.... this is embarrassingly easy to prove. We want U(t,h) to be a constant and it is strictly increasing in both arguments, therefore:

$$ \forall  $$ such that $$ U(t,h) = \bar U$$,

we must have

$$h_1 \le h_2 \implies t_1 \ge t_2 $$.

This is really just a complicated way to say, 鱼与熊掌不可得兼.

The implication is actually quite interesting. It is not that we don't enjoy something both tasty and healthy suppose such food exists, then we can always twist it so that it becomes a little bit more a tastier and a little bit less healthier  yet we enjoy it the same. Repeat the twisting many times, then at the end of the day, among the food that we choose to eat, the really tasty ones must be the least healthy ones as well.

Average Utility and Marginal Utility
只有在边际效用是常数的时候，平均效用才会和边际效用相等. 如果你要衡量"单位物品在价格为P的时候的每一美元的平均效用",那么公式应该是：


 * $$\frac{\int_0^q MU(s)ds}{P\cdot q}$$

这里面分子是消费量为q的时候商品带给你的总效用，分母是为q单位商品支付的总财富.

上面这个式子只有在

$$\int_0^q MU(s)ds=MU(q)\cdot q$$

的时候才会等于$$\frac{MU(q)}{P}$$. 而上式说明边际效用为常数？

Weak Axiom Implies x(p,w) Homogenous of Degree Zero
假设x(p,w)满足弱公理，则有：

if $$p\cdot x(p',w')\leq w \cap x(p,w)\ne x(p',w')$$ then $$p'\cdot x(p,w)>w'$$

设 $$p=p_0,w=w_0,p'=\alpha p_0, w'=\alpha w_0$$,直接代如弱公理，我们有：

if $$p_0\cdot x(\alpha p_0,\alpha w_0)\leq w_0 \cap x(p_0,w_0)\ne x(\alpha p_0,\alpha w_0)$$ then $$\alpha p_0\cdot x(p_0,w_0)>\alpha w_0$$

这个then后面的推论显然是不成立的，因为对于任何的p,w我们必须有$$p\cdot x(p,w)\leq w$$. 因此，if 部分也必然不成立，我们因此有if部分的否命题成立：

$$p_0\cdot x(\alpha p_0,\alpha w_0) > w_0 \cup x(p_0,w_0)= x(\alpha p_0,\alpha w_0)$$

到这一步我们还是没法推论$$x(p_0,w_0)= x(\alpha p_0,\alpha w_0)$$，因为有前面那个“或”的命题. 现在考虑“或”前面的那个命题. 这个命题也是不可能成立的，因为如果在不等式两边同时乘$$\alpha$$，我们就有：

$$\alpha p_0\cdot x(\alpha p_0,\alpha w_0) >\alpha w_0$$ 跟then 命题必然不成立的道理一样. “或”前面的命题必然不成立，而整个命题成立，那么我们就必然有

$$x(p_0,w_0)= x(\alpha p_0,\alpha w_0)$$

成立了. x(p,w)零阶齐次证毕.

Is It Possible to Have a Decreasing Contract Curve in Exchange Economy?
考虑这个问题，假设每个人的效用函数分别是 $$ u_1(x,y),u_2(x,y)$$,两种商品的禀赋分别是$$w_x,w_y$$,那么契约曲线在内点一定满足


 * $$\frac{u_{1x}(x,y)}{u_{1y}(x,y)}=\frac{u_{2x}(w_x-x,w_y-y)}{u_{2y}(w_x-x,w_y-x)}$$

重新整理得：


 * $$ F(x,y)=u_{1x}(x,y)u_{2y}(w_x-x,w_y-x)-u_{1y}(x,y)u_{2x}(w_x-x,w_y-y)$$

其中$$u_{ix}$$是第i个人对商品x的边际效用. 因为我们考虑的是扭曲的问题，所以应该只需要讨论内点解. 上式定义了一个$$y$$对$$x$$的隐含数，如果这个曲线左上到右下扭曲的话，那么也就是说这个隐含数是递减的. 假定这个函数可导（纯粹为了省事，当然可能不可导），那么用隐含数定理可知:


 * $$\frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}$$
 * $$=\frac{u_{1xx}u_{2y}+u_{2xx}u_{1y}-u_{2yx}u_{1x}-u_{1yx}u_{2x}}{u_{2yy}u_{1x}+u_{1yy}u_{2x}-u_{1xy}u_{2y}-u_{2xy}u_{1y}}$$

现在我们就可以讨论了：

如果我们要求偏好是单调的且凸的话，那么我们就有$$u_{ix},u_{iy}\geq 0,\;u_{ixx},u_{iyy}\leq 0$$. 但我们不知道$$u_{ixy}$$的符号，在这种情况下我们没法确定$$dy/dx$$的符号，那么契约曲线也就有可能递减. 如果$$u_{ixy}\geq 0$$,比如Cobb-Douglass，或者任何x y可分离的情况，那么我们可以确定$$dy/dx\geq 0$$，这样也就不可能单调递减. 除此之外，似乎在这个general form下没有什么办法可以确定dy/dx的符号.

如果我们进一步假定两个人的效用函数相同，那么：


 * $$\frac{dy}{dx}=\frac{u_{xx}u_y-u_{xy}u_x}{u_{yy}u_x-u_{xy}u_y}$$

这样的话这个问题就可以有一个确定的答案. 如果$$u_{xy}\geq 0$$，那么如前所述，必有$$dy/dx\geq 0$$，不可能出现扭曲;所以我们只需要讨论$$u_{xy}<0$$. 给同样的假定，即凸性和单调性，我们现在有：


 * $$u_{xx}u_{yy}-u_{xy}^2\leq 0$$

这个地方有点不严谨，因为我直接用了海赛阵而没有用加边海赛；结果应该差不多的，就偷懒了:)

这个条件等价于:


 * $$u_{yy}\geq \frac{u_{xy}^2}{u_{xx}}$$

这里面不等号变换了方向，因为每一项都是负的. 那么，如果在dy/dx那个式子里面的分子大于零，那么一定有分母也大于零；换句话说，上下同号，故我们有$$dy/dx\geq 0$$,不可能出现扭曲：


 * $$u_{xx}u_y-u_{xy}u_x\geq 0 \implies u_{yy}u_x-u_{xy}u_y\geq 0$$

最后这个不等式是这样的推导的：


 * $$u_{yy}\geq \frac{u_{xy}^2}{u_{xx}}$$
 * $$\frac{u_{yy}}{u_y}\geq \frac{u_{xy}^2}{u_{xx}u_y}$$

从分子大于零我们知道：


 * $$\frac{u_{xy}}{u_{xx}}>\frac{u_y}{u_x} $$

将这个式子带入上式，我们有：


 * $$\frac{u_{yy}}{u_y}\geq \frac{u_{xy}^2}{u_{xx}u_y}\geq \frac{u_y}{u_x}\frac{u_{xy}}{u_y}=\frac{u_{xy}}{u_x}$$

最后这个式子也就等价于分母大于零了.

总结一下：

如果两个人的效用函数不相同，那么没有扭曲的充分条件是偏好的凸性，单调性，以及$$u_{ixy}\geq 0$$. 我觉的最后这个条件应该是有个名字的，跟complementary 有没有关系？不知道了，那位大哥指导一下. . . . . 因为最后这个条件并不是最基本的条件，那么有可能出现左上到右下的扭曲

如果两个人的效用函数相同，那么只要偏好满足凸性和单调性（这点不太严谨，我证明的是效用函数满足凹性和单调性，关于凸性的问题，我觉的应该是可以成立的，不过懒得弄加边海赛了……），那么就不可能出现扭曲的情况.

Will Monopoly Produce When Demand is Inelastic?
一个垄断厂商解决这个最优化问题:

$$\max_q p(q)\cdot q -c(q)$$

First Order Condition:


 * $$\frac{dp}{dq}q+p=c'(q)$$
 * $$p(\frac{dp}{dq}\frac{q}{p}+1)=c'(q)$$
 * $$p(\frac{1}{e}+1)=c'(q)$$

如果需求缺乏弹性，那么$$e>-1$$，故我们有：


 * $$p(\frac{1}{e}+1)<0$$

但是如果生产可能性集合是凸的，那么我们总有 $$c'(q)>0$$，所以这里就矛盾了，故垄断厂商不会在需求缺乏弹性的时候生产.

WikiProject Economics census
Hello there. Sorry to bother you, but you are (titularly at least) a member of WP:WikiProject Economics, as defined by this category. If you don't know me, I'm a Wikipedia administrator, but an unqualified economist. I enjoy writing about economics, but I'm not very good at it, which is why I would like to support in any way I can the strong body of economists here on Wikipedia. I'm only bothering you because you are probably one of them. Together, I'd like us to establish the future direction of WikiProject Economics, but first, we need to know who we've got to help.

Whatever your area of expertise or level of qualification, if you're interested in helping with the WikiProject (even if only as part of a larger commitment to this wonderful online encyclopedia of ours), would you mind adding your signature to this page? It only takes a second. Thank you.

Message delivered on behalf of User:Jarry1250 by LivingBot.
 * Firstly, thank you for signing the census, and an apology if you are one of those editors who dislike posts such as this one for messaging you again in this way. I've now got myself organised and you can opt-out of any future communication at WP:WikiProject Economics/Newsletter. Just remove your name and you won't be bothered again.


 * Secondly, and most importantly, I would like to invite your comments on the census talk page about the project as a whole. I've given my own personal opinion on a range of topics, but my babbling is essentially worthless without your thoughts - I can't believe for one moment that everyone agrees with me in the slightest! :)


 * All your comments are welcomed. Thanks, - Jarry1250 [Humorous? Discuss.] 17:56, 21 April 2010 (UTC)

Michigan Wikipedians
Greetings Malin84! I noticed that you made mention of the University of Michigan or Ann Arbor on your userpage. If you are a current student, faculty, or other affiliate at the University of Michigan, I would like to welcome you, on behalf of the Michigan Wikipedians, to our next weekly meeting on Monday September 30 (and every Monday thereafter). The meetings are held at 8:00 PM (EDT) in the University of Michigan Shapiro Library, room 4041. New and experienced editors alike are most welcome. Do not hesitate to leave me a message if you have any questions, and feel free to stop by the MWiki talk page. The Michigan Wikipedians are excited to meet you!   A rbitrarily 0   ( talk ) 00:20, 24 September 2013 (UTC)

Asian 10,000 Challenge invite
Hi. The WikiProject Asia/The 10,000 Challenge has recently started, based on the UK/Ireland The 10,000 Challenge and WikiProject Africa/The 10,000 Challenge. The idea is not to record every minor edit, but to create a momentum to motivate editors to produce good content improvements and creations and inspire people to work on more countries than they might otherwise work on. There's also the possibility of establishing smaller country or regional challenges for places like South East Asia, Japan/China or India etc, much like The 1000 Challenge (Nordic). For this to really work we need diversity and exciting content and editors from a broad range of countries regularly contributing. At some stage we hope to run some contests to benefit Asian content, a destubathon perhaps, aimed at reducing the stub count would be a good place to start, based on the current WikiProject Africa/The Africa Destubathon which has produced near 200 articles in just three days. If you would like to see this happening for Asia, and see potential in this attracting more interest and editors for the country/countries you work on please sign up and being contributing to the challenge! This is a way we can target every country of Asia, and steadily vastly improve the encyclopedia. We need numbers to make this work so consider signing up as a participant! Thank you. -- Ser Amantio di Nicolao Che dicono a Signa?Lo dicono a Signa. 05:20, 20 October 2016 (UTC)