{"id":1717,"date":"2021-11-30T13:57:36","date_gmt":"2021-11-30T05:57:36","guid":{"rendered":"https:\/\/fanyuzhao.com\/?p=1717"},"modified":"2022-04-09T15:02:37","modified_gmt":"2022-04-09T07:02:37","slug":"math-tools","status":"publish","type":"post","link":"https:\/\/fanyuzhao.com\/?p=1717","title":{"rendered":"Math Tools"},"content":{"rendered":"\n<h4 class=\"wp-block-heading\">1. Homogenous of Degree \\(k\\)<\/h4>\n\n\n\n<p>Definition (Homogeneity of degree \\(k\\)). A utility function \\(u:\\mathbb{r}^n\\rightarrow \\mathbb{R}\\) is homogeneous of degree \\(k\\) if and only if for all \\(x \\in \\mathbb{R}^n\\) and all \\(\\lambda&gt;0\\), \\(u(\\lambda x)=\\lambda^ku(x)\\).<\/p>\n\n\n\n<p>$$f(\\lambda x_1,&#8230;,\\lambda x_n)=\\lambda^kf(x_1,&#8230;,x_n)$$<\/p>\n\n\n\n<p><strong>Property<\/strong><\/p>\n\n\n\n<ol><li>Constant Return to Scale: CRTS production function is homogenous of degree 1. IRTS is homogenous of degree \\(k&gt;1\\). DRTS is homogenous of degree \\(k&lt;1\\).<\/li><li>The Marishallian demand is homogeneous of degree zero. \\(x(\\lambda p,\\lambda w)=x(p,w)\\). (Maximise \\(u(x)\\) s.t. \\(px&lt;w\\). &#8220;No Money Illusion&#8221;.<\/li><li>Excess demand is also homogeneous degree of zero. Easy to prove by the Marshallian Demand.<\/li><\/ol>\n\n\n\n<p>$$CRTS:\\quad F(aK,aL)=aF(K,L) \\quad a&gt;0$$<\/p>\n\n\n\n<p>$$IRTS:\\quad F(aK,aL)&gt;aF(K,L) \\quad a&gt;1$$<\/p>\n\n\n\n<p>$$DRTS:\\quad F(aK,aL)&lt;aF(K,L) \\quad a&gt;1$$<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">2. Euler&#8217;s Theorem<\/h4>\n\n\n\n<p>Theorem (Euler&#8217;s Theorem) Let \\(f(x_1,&#8230;,x_n)\\) be a function that is <em>homogeneous of degree k. Then,<\/em><\/p>\n\n\n\n<p>$$ x_1\\frac{\\partial f(x)}{\\partial x_1}+&#8230;+ x_n\\frac{\\partial f(x)}{\\partial x_n} =kf(x) $$<\/p>\n\n\n\n<p>or, in gradient notation,<\/p>\n\n\n\n<p>$$ x\\cdot \\nabla f(x)=kf(x) $$<\/p>\n\n\n\n<p><strong>Proof:<\/strong> Differentiate \\(f(tx_1,&#8230;,tx_n)=t^k f(x_1,&#8230;,x_n)\\) w.r.t \\(t\\) and then set \\(t=1\\).<\/p>\n\n\n\n<p>P.S. We use Euler&#8217;s Theorem in the proof of the Solow Model.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">3. Envelop Theorem<\/h4>\n\n\n\n<p><strong>Motivation:<\/strong><\/p>\n\n\n\n<p>Given \\(y=ax^2+bx+c, a&gt;0, b,c \\in \\mathbb{R}\\), we need to know how does a change in the parameter \\(a\\) affect the maximum value of \\(y\\), \\(y^*\\)?<\/p>\n\n\n\n<p>We first define \\(y^*=\\max_{x} y= \\max_{x} ax^2+bx+c  \\). The solution is \\(x^*=-\\frac{b}{2a}\\), and plug it back into \\(y\\), we get \\(y^*=f(x^*)=\\frac{4ac-b^2}{4a}\\). Now, we take derivative w.r.t. \\(a\\). \\(\\frac{\\partial y^*}{\\partial a}=\\frac{b^2}{4a^2}\\). We would find that,<\/p>\n\n\n\n<p>$$\\frac{\\partial y^*}{\\partial a}= {\\frac{\\partial y}{\\partial a}}|_{x=x^*} $$<\/p>\n\n\n\n<p><strong>A Simple Envelop Theorem<\/strong><\/p>\n\n\n\n<p>$$v(q)=\\max_{x} f(x,q)$$<\/p>\n\n\n\n<p>$$=f(x^*(q),q)$$<\/p>\n\n\n\n<p>$$ \\frac{d}{dq}v(q)=\\underbrace{\\frac{\\partial}{\\partial x}f(x^*(q),q)}_{=0\\ by\\ f.o.c.}\\frac{\\partial}{\\partial q}x^*(q)+\\frac{\\partial}{\\partial q} f(x^*(q),q) $$<\/p>\n\n\n\n<p>$$  \\frac{d}{dq}v(q) =\\frac{\\partial}{\\partial q}f(x^*(q),q) $$<\/p>\n\n\n\n<p>Think of the ET as an application of the chain rule and then F.O.C., our goal is to find how does parameter affect the already maximised function \\(v(q)=f(x^*(q),q)\\).<\/p>\n\n\n\n<p><strong>A formal expression<\/strong><\/p>\n\n\n\n<p>Theorem (Envelope Theorem). Consider a constrained optimisation problem \\(v(\\theta)=\\max_x f(x,\\theta)\\) such that \\(g_1(x,\\theta)\\geq0,&#8230;,g_K(x,\\theta)\\geq0\\).<\/p>\n\n\n\n<p>Comparative statics on the value function are given by: (\\(v(\\theta)=f(x,\\theta)|_{x=x^*(\\theta)}=f(x^*(\\theta),\\theta)\\))<\/p>\n\n\n\n<p>$$ \\frac{\\partial v}{\\partial \\theta_i}=\\sum_{k=1}^{K}\\lambda_k \\frac{\\partial g_k}{\\partial \\theta_i}|_{x^*}+{\\frac{\\partial f}{\\partial \\theta_i}}|_{x^*}=\\frac{\\partial \\mathcal{L}}{\\partial \\theta_i}|_{x^*} $$<\/p>\n\n\n\n<p>(for Lagrangian \\(\\mathcal{L}(x,\\theta,\\lambda)\\equiv f(x,\\theta)+\\sum_{k}\\lambda_k g_k(x,\\theta)\\)) for all \\(\\theta\\) such that the set of binding constraints does not change in an open neighborhood.<\/p>\n\n\n\n<p>Roughly, the derivative of the value function is the derivative of the Lagrangian w.r.t. parameters, \\(\\theta\\), while argmax those unknows (\\(x=x^*\\)).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">4. Hicksian and Marshallian demand + Shepherd&#8217;s Lemma<\/h4>\n\n\n\n<p>To be continued.<\/p>\n\n\n\n<p><a href=\"https:\/\/www.bilibili.com\/video\/BV1VJ411J7ZL?spm_id_from=333.999.0.0\">https:\/\/www.bilibili.com\/video\/BV1VJ411J7ZL?spm_id_from=333.999.0.0<\/a><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">5. KKT<\/h4>\n\n\n\n<h4 class=\"wp-block-heading\">6. Taylor Series<\/h4>\n\n\n\n<p>A Taylor series is a&nbsp;series expansion&nbsp;of a&nbsp;function&nbsp;about a point. A one-dimensional Taylor series is an expansion of a&nbsp;real function \\(f(x)\\) about point \\(x=a\\) is given by,<\/p>\n\n\n\n<p>$$ f(x)=f(a)+f'(a)(x-a)+\\frac{f&#8221;(a)}{2!}(x-a)^2\\\\+\\frac{f^{(3)}(a)}{3!}(x-a)^3+&#8230;+\\frac{f^{(n)}(a)}{n!}(x-a)^n+&#8230; $$<\/p>\n\n\n\n<p>Taylor expansion is a way to approximate a functional curve <strong>around <\/strong>a certain point, by taking derivatives. <em>We focus the function around this point<\/em>.<\/p>\n\n\n\n<p><strong>For example,<\/strong> we approximate \\(f(x)=x^3\\) around \\(x=2\\).<\/p>\n\n\n\n<p>$$ f(x)\\approx f(2)+\\frac{f'(2)}{1!}(x-2)+\\frac{f&#8221;(2)}{2!}(x-2)^2\\\\+frac{f^{(2)}(1)}{3!}(x-2)^3+&#8230; $$<\/p>\n\n\n\n<p>$$ f(x)\\approx 8+\\frac{12}{1}(x-2)+\\frac{12}{2}(x-2)^2+\\frac{6}{3\\times2}(x-2)^3 $$<\/p>\n\n\n\n<p>Simplifying it, we get \\(f(x)\\) around \\(x=2\\) is,<\/p>\n\n\n\n<p>$$ f(x)=x^3 $$<\/p>\n\n\n\n<p>That is a coincidence that the original function and the Taylor polynomial are exactly the same if \\(f(x)=x^3\\).<\/p>\n\n\n\n<p><strong>Another Example,<\/strong> we take first order Taylor approximation to \\(f(k)=ln(k)\\) at \\(k^*\\),<\/p>\n\n\n\n<p>$$ ln(k)\\approx ln(k^*)+\\frac{1}{ln(k^*)}(k-k^*) $$<\/p>\n\n\n\n<p>Thus, we know,<\/p>\n\n\n\n<p>$$ ln(k)- ln(k^*)\\approx\\frac{k-k^*}{k^*}   $$<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">New Study and Idea of Taylor Expension<\/h4>\n\n\n\n<iframe loading=\"lazy\" title=\"Taylor series | Chapter 11, Essence of calculus\" width=\"660\" height=\"371\" src=\"https:\/\/www.youtube.com\/embed\/3d6DsjIBzJ4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe>\n\n\n\n<p>Taylor Expansion aims to use polynomial to approximate a certain function.<\/p>\n\n\n\n<p>For example, in order to describe the shape of function \\(cos(x)\\) at x=0, we would first construct a polynomial.<\/p>\n\n\n\n<p>(P.S. We let \\(c_0=1\\) as we need to pin the polynomial equal to 1 at x=0.)<\/p>\n\n\n\n<p>$$ P(x)=c_0+c_1x+c_2x^2 \\ and\\ at\\ x=0\\ P(0)=c_0 $$<\/p>\n\n\n\n<p>, where those coefficients are free to change, and the magnitude of those coefficients would affect how the approcimated curve looks like.<\/p>\n\n\n\n<p>To get a better approximation, we would adjust those coefficients. Thus, we consider using different orders of derivatives to simulate our target function. <\/p>\n\n\n\n<p>We need the first order derivate of \\(cos'(x)|_{x=0}=sin(x)|_{x=0}\\) to be zero, so we set the first-order derivative of our polynomial function to equal to zero as well!<\/p>\n\n\n\n<p>$$\\frac{\\partial P(x)}{\\partial x}|_{x=0}=c_1 \\times 1 |_{x=0}=c_1$$<\/p>\n\n\n\n<p>Therefore, \\(c_1\\) must be zero.<\/p>\n\n\n\n<p>Let&#8217;s go one more step. As the second derivative of \\(cos^{(2)}(x)=-1\\), we need the second derivative (, which is also the second derivate of the second-order term of our constructed polynomial function) of our polynomial function to be also -1.<\/p>\n\n\n\n<p>$$\\frac{\\partial^2 P(x)}{\\partial x^2}|_{x=0}=2\\times c_2$$<\/p>\n\n\n\n<p>We adjust that to be negative one, so \\(c_2=-\\frac{1}{2}\\).<\/p>\n\n\n\n<p>Therefore, we get,<\/p>\n\n\n\n<p>$$cos(x)|_{x=0}\\approx P(x)=c_0+c_1 x +c_2 x^2 = 1-\\frac{1}{2}x^2$$<\/p>\n\n\n\n<p>Great! If we need a more accurate approximation, then we keep on going to more derivates and calculate the coefficient of the higher-order term. However, I would do that, so I just simply add a term \\(O(x^3)\\) to represent there are other terms that are less equal than \\(x^3\\). (There are accurate descriptions that I will update in later posts).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>1. Homogenous of Degree \\(k\\) Definition (Homogeneity of degree \\(k\\)). A utility function \\(u:\\mathbb{r}^n\\rightarrow \\mathbb{R}\\) is homogeneous of degree \\(k\\) if and only if for all \\(x \\in \\mathbb{R}^n\\) and all \\(\\lambda&gt;0\\), \\(u(\\lambda x)=\\lambda^ku(x)\\). $$f(\\lambda x_1,&#8230;,\\lambda x_n)=\\lambda^kf(x_1,&#8230;,x_n)$$ Property Constant Return to Scale: CRTS production function is homogenous of degree 1. 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