Tutor profile: Jerome S.

Find the solution of the following system of simultaneous linear equations, with 3 variables y 3 equations: $(x_1+3x_2+3x_3=2$)$({2x}_1+x_2+{6x}_3=4$)$(x_1-x_2-3x_3=3$)

The system of linear equations is equivalent to a matrix equation of the form: Ax = y where A is the matrix of coefficients: $(A=\left[\begin{matrix}1&3&3\\2&1&6\\1&-1&-3\\\end{matrix}\right]$) x is the unknown vector we are seeking: $(x=\left[\begin{matrix}x_1\\x_2\\x_3\\\end{matrix}\right]$) y is the vector of coefficients on the right hand side of the system of equations: $(y=\left[\begin{matrix}2\\4\\3\\\end{matrix}\right]$)If we multiply both sides of the matrix equation by the inverse matrix of A: $$A^{-1}$$, we get: $(A^{-1}Ax=A^{-1}y$)$(\therefore x=A^{-1}y$) This is the solution. We simply need to compute the inverse matrix of A and multiply it by y. We can do this easily in Excel or R. In Excel: MMULT(MINVERSE(A1:C3); E1:E3) where A1:C3 and E1:E3 are the Excel ranges where we write A and y respectively. Of course, these can be any other ranges, so long as they comply with the dimensions of A and y: 3x3 and 3x1 respectively. In R: A = rbind(c(1,3,3), c(2,1,6), c(1,-1,-3)) y = c(2,4,3) solve(A) %*% y Answer: $(x=\left[\begin{matrix}\frac{5}{2}\\0\\\frac{1}{6}\\\end{matrix}\right]$)

The Solow growth model represents growth of an economy in the long run. The differential equation that describes the change in capital per capita as a function of time is:$(k'\left(t\right)=sk^\beta(t)-\left(\delta+g_N\right)k(t)$)where $$\beta<1$$, s is the saving rate, $$\delta$$ is the capital depreciation rate and $$g_N$$ is the rate of population growth. Let $$\beta=0.3,\ \delta=12%,\ g_N=2%,\ s=7%$$. Find the value of k at the locally stable equilibrium.

The differential equation is explicit and the function f of k is: $(f\left(k\right)=sk^\beta-\left(\delta+g_N\right)k$)The roots of f(k) are the equilibria. To find them, we put f(k) = 0: $(f\left(k\right)=sk^\beta-\left(\delta+g_N\right)k=0$)$(\therefore k[sk^{\beta-1}-\left(\delta+g_N\right)]=0$)Clearly, one root is:$({k}_{1}={0}\ldots(\mathbf{1})$)The other root is given by:$(sk_2^{\beta-1}-\left(\delta+g_N\right)=0$)$(\therefore k_2^{\beta-1}=\frac{\left(\delta+g_N\right)}{s}$)$(\therefore{k}_{2}=\left[\frac{\left({\delta}+{g}_{N}\right){c}}{{s}}\right]^\frac{\mathbf{1}}{{\beta}-\mathbf{1}}\ldots(\mathbf{2})$)The derivative of f as a function of k is:$(f\prime(k)=\beta sk^{\beta-1}-\left(\delta+g_N\right)$)We need to check the the sign of $$f\prime(k)$$ at the equilibria $$k_1$$ and $$k_2$$ to know if they are locally stable or unstable. $$f\prime(k_1)=f^\prime\left(0\right)$$ is not defined because $$\beta-1=-0,6<0$$. However, at the limit, $$f^\prime\left(k\right)\rightarrow+\infty$$ when $$k\rightarrow0^+$$. Then $$f^\prime\left(k\right)>0$$ for all k sufficiently close to zero, and hence we can conclude that $$k_1$$ is an unstable equilibrium. To check the sign of $$f^\prime\left(k_2\right)$$: $(f^\prime\left(k_2\right)=\beta s\left[\left[\frac{\left(\delta+g_N\right)}{s}\right]^\frac{1}{\beta-1}\right]^{\beta-1}-\left(\delta+g_N\right)$)$({\therefore f}^\prime\left(k_2\right)=\beta\left(\delta+g_N\right)-\left(\delta+g_N\right)$)$({\therefore f}^\prime\left(k_2\right)=(\beta-1)\left(\delta+g_N\right)$) $$\beta<1$$ and $$\delta{,g}_N>0$$ imply that $$f^\prime\left(k_2\right)<0$$. Therefore, $$k_2$$ is a stable equilibrium, and hence it is the one that we are seeking to solve the problem. Substitute the given values in equation (2): $(k_2=\left[\frac{\left(0,12+0,02\right)}{0,07}\right]^\frac{1}{0,3-1}$)$(\therefore k_2=0,371499$) Answer: 0.3715 (to 4 significant figures)

A start-up company is launching a new internet browser, with attractive features that would make many users prefer it over existing browsers. Assume that 15% of the population would switch to this new browser. However, this is a population parameter and hence unknown. In order to estimate this percentage, you interview 10 people chosen at random and ask them if they would switch. What is the standard deviation of this estimator?

Let p = 15% be the (unknown) percentage of the population that would switch to the new browser, and n = 10 be the sample size. Each person in the survey sample can answer “yes” or “no”. We can model this as a sequence of indentical and independent Bernoulli random variables: $$B_1,B_2...B_n$$ in which 1 represents “yes” and 0 represents “no”. The sample percentage is given by: $(\bar{X}=\frac{1}{n}\sum_{i=1}^{n}B_i$)Now the problem is clearly defined: we wish to find the population standard deviation of the random variable $$\bar{X}$$, or equivalently, its variance:$(\therefore var\left(\bar{X}\right)=var\left(\frac{1}{n}\sum_{i=1}^{n}B_i\right)$)Recall the variance rules:$(var\left(cX\right)=c^2var(X)$)where c is a constant, and:$(var\left(X+Y\right)=var\left(X\right)+var\left(Y\right)+2cov(X,Y)$)$(\therefore var\left(\bar{X}\right)=\frac{1}{n^2}var\left(\sum_{i=1}^{n}B_i\right)$)$(\therefore var\left(\bar{X}\right)=\frac{1}{n^2}\left[\sum_{i=1}^{n}{var(B_i)}+2\sum_{i\neq j}^{n}{cov(B_i,B_j)}\right]$) We know that the variance of a Bernoulli variable is p(1-p). Moreover, the Bernoulli variables in our problem are independent, and hence: $$cov\left(B_i,B_j\right)=0\ \forall i\neq j$$. $(\therefore var\left(\bar{X}\right)=\frac{1}{n^2}\sum_{i=1}^{n}{p(1-p)}$)$(\therefore var\left(\bar{X}\right)=\frac{1}{n^2}np(1-p) $) Substitute p = 0.15 and n = 10: $(var(\bar{X})=\frac{1}{{10}^2}\bullet10\bullet0.15\bullet(1-0.15)$)$(\therefore var(\bar{X})=0.01275$)$$\therefore$$ Standard Deviation of $$\bar{X}=\sqrt0.01275\ =0.1129$$ (to 4 significant figures) Answer: 0.1129