Visit For more Solved Assignments: IGNOU4U.BLOGSPOT.COM IGNOU MEC-003 Solved Assignment 2012 Presented by http://www.myignou.in/ MEC-003: Quantitative Techniques Assignment Course Code: MEC-003 Assignment Code: MEC-003/TMA/2011-12 Note: Answer all the questions. While questions in Section A carry 20 marks each, those of Section B carry 12 marks each. Section A Long Answer Question 1. How do you use differential equations in economics? What type of situations can be helpfully depicted using differential equations? Discuss the role of initial condition in solving a differential equation. If your objective is to examine the stability of equilibrium, with the help of an example, show how a second-order differential equation helps in addressing your concern. Ans. Differential Equation are equations involving the derivatives (or differentials) of unknown functions. Solving a differential equations in economics means finding a function that satisfies that equation. If y = f(x) is a function for which derivatives of adequate order exist, the dy/dx = f’(x). or dy = f’(x)dx y = ∫ f’(cx)dx, Through differential equations, we solve the problems, which are related to change over time, i.e. dynamic variables. For example, suppose that a hypothetical economy’s income (y) is related to time (x). It is given in functional form: y(x) = 2x 1/2. If the income changes over time, we find the rate of change as dy/dx = x- 1/2. To find the time path if the income change, we write y = y(x). The derivative of this function will be same as that of y = y(x) + c, where c is any arbitrary constant. To determine a unique time path of the income change, it is necessary to work out a definite value of c. Additional information required for that purpose is to have the initial condition. The initial condition of the economy, say, y(0), i.e value of y at x=0, then the value of the constant c can be determined, Thus, from y(x) = 2 x1/2 + c, where x = 0, We get y(0) = 2(0)1/2 + c = c The differential equation that involves only the first derivative, has a unique solution if it has one initial condition. In addition, the differential equation that involves only the first and second derivatives, has a unique solution if it has two initial conditions. In a differential equation, if the initial value has a solution that is a constant function and hence independent of t, then the value of the constant is called an equilibrium state or stationary state of the differential equation. Marks: 100 this equation has a single equilibrium. (1-p) x n-x.a/2.Visit For more Solved Assignments: IGNOU4U. then the general solution of the equation is y(t) = Aer1t + Ber2t.d. (1-p)n-x for x = 0.…. If r1 and r2. 0. The equilibrium is stable if and only if r1 < 0 and r2 < 0. Three possible forms of the general solution of the equation to evaluate the stability of such equilibrium are as follows: Case I: Characteristics equation has two real roots. the real part of each root. That is. The function f(x) is called the probability density function (p. Stability of solutions of second order homogeneous equation Consider the above homogeneous equation y*(t) + ay’(t) + by(t) = 0 If b ≠ 0. ================================================================== 2. in which y(t) does not appear.b) ∫ f(x) = 1 a A random variable X is said to have a binomial distribution and is referred as a binomial random variable. y’(t) Take an equation of form y*(t) = F (t. Ans. Case I: Characteristics equation has complex roots. y’(t)). With a single root (say r). When the characteristic equation has complex roots.n The probability of getting ‘x’ success and ‘n-x’ failures in n trials is given by p . the only constant function that is a solution is equal to 0 for all t.COM A second order ordinary differential equation consists of time as the independent variable with the dependent variable y with its first and second derivatives. where α = . px . the characteristic equation is in stable equilibrium if and only if this root is negative. the form of the solution of the equation is Aet cos(βt + ω). viz.. .n. if and only if it’s probability distribution is given by B (x. y*(t)) = 0 for all t such that we can write it in the form y*(t) = F(t.BLOGSPOT.p) = nCx . y’(t)). y(t).2. Give examples of problems where you can make use of Poisson distribution. Does it have a probability density function? Why or why not? Discuss your answer in the context of the mean and variance of Poisson distribution. are the two roots of the characteristics equation. Equations of the form y”(t) = F(t. The equilibrium will be stable if and only if the real part of each root is negative.1. tk ert converges to 0 as t →∞. y’(t).f) provided it satisfies the following two conditions F(x) ≥ 0 b If the range of the continuous random variable is (a. Case I: Characteristics equation has a single real roots. Note that if r<0 then for any value of k. Consider for example an equation G(t. y(t). The binomial distribution written as n x n-x x B(x. The properties of the Poisson distribution are (a) Poisson distribution is a discrete probability distribution.p) = Cx (λ/n) (1-( λ/n)) = n (n-1) (n-2) (n-3)………(n-x+1)/x! x (λ/n) (1-( n-x λ/n)) x n/λ -λ -x = 1. provided (a) (b) The number of trials is very large in fact tending to infinity.…… to ∞.(1-(x-1)/n) → 1 -x (1 .→ e Therefore. therefore p = λ/n.1. (b) Mean and variance of poisson distribution are the same. the probability mass function (p. Example: Let X be a random variable following poisson distribution. the number of success is a random variable and will follow a poisson distribution with only parameter λ. Thus.λ) = λ e / x! 1 -λ -λ Therefore. while n. x -λ For poisson distribution.d.3. the limiting form of the binomial distribution becomes. find P(X=0 or 1) and E(X).p remains constant.p = λ remains constant.( λ/n)). Therefore. tending to zero.m.1. Thus Poisson distribution is a discrete probability distribution which is the limiting form of the binomial distribution. The probability of success in each trial is very small. we get λ = 2. When n →∞ while x and p are constants (1-1/n) (1-2/n) (1-3/n) …. from the equation λ e = λ e / 2! . P (x.n. E(X) = λ = 2 and It does not p. where the random variable assumes countably infinite number of values such as 0. λ and (λ-1).3. (c) Poisson distribution like the binomial distribution may have either one or two modes When λ is not an integer. (d) The poisson distribution used as an approximation to binomial distribution when n is large but np is fixed.λ) = λ e x -λ / x! for x = 0.f ================================================================== Section B .p. The distribution is completely specified if the parameter λ is known.f) is given by P (x. If P(X=1) = P(X=2).( λ/n)) ] x (1-(λ/n)) .(1-1/n)(1-2/n)(1-3/n)……(1-(x-1)/n/x! x (λ) x [(1 .( λ/n)) → 1 n/λ (1 .2.COM The number of trials in the case of binomial distribution is very large and cumbersome. both being λ. Let λ = n. n. there are two modes.Visit For more Solved Assignments: IGNOU4U.BLOGSPOT.2. Then the probability distribution used to approximate binomial possibilities. mode is the largest value contained in λ and when λ is an integer. This limiting form of the binomial distribution when n → ∞ and p → 0.………. in the limit when n→ ∞ and p → 0. (PX=1) = λ e / 1! = λ e 2 -λ -λ (PX=2) = λ e / 2! = λ e /2 –λ 2 -λ As (PX=1) = P(X=2). and (ii) the alternative hypothesis. A linear programming problem is given as Max. For example. the critical region lies entirely in the right tail of the sampling distribution of x.e. while for the lefttest (H1 : μ < μ0). Thus. In the right-tailed test (H1 : μ > μ0). against the alternative hypothesis H1 : μ ≠ μ0 (μ > μ0 and μ < μ0) is known as ‘twotailed test’ and in such a case the critical region is given by the portion of the area lying in both the tails of the probability curve of the test statistic. whether one-tailed or two-tailed test is to be applied depends entirely on the nature of the alternative hypothesis. the critical region is entirely in the left tail of the distribution. then apply one-tailed test. z 30x1 50x 2 subject to x1 x 2 9 x1 2x 2 12 Where x1 0. It depends upon: (i) the level of significance used.COM Medium Answer questions 3. If the alternative hypothesis is two-tailed then apply the two-tailed test and if alternative hypothesis is one-tailed. Thus. In a particular problem. zα is the value such that area to the right of zα is α/2 and to the left of -zα is α/2. is a ‘single-tailed test”. A test of any statistical hypothesis where the alternative hypothesis is one-tailed (right-tailed or left-tailed) is called a ‘one-tailed test’. which separates the critical (or rejection) region and the acceptance region is called the ‘critical value’ or ‘significant value’. Find its optimal solution. A test of any statistical hypothesis where the alternative hypothesis is two-tailed such as: H0 : μ = μ0. The value of the test statistic. ================================================================== 4. Since normal probability curve is symmetrical curve.BLOGSPOT. the significant or critical value of Z for a single-tailed test (left or right) at level of significance ‘α’ is the same as the critical value of Z for two-tailed test at level of significance ‘2α’. Explain the relevant considerations of making choice between one-tailed and two-tailed tests. x 2 0. . so P[|Z| > zα] = α can be written as. How would you determine the level of significance in the above tests? Ans. P[Z > zα] + P[Z < -zα] = α P[Z > zα] + P[Z < zα] = α 2P[Z > zα] = α P[Z > zα] = α/2 i. the area of each tail is α/2.Visit For more Solved Assignments: IGNOU4U. whether it is two-tailed or single-tailed.e zα is the value so that the total area of the critical region on both tails is α. The critical value of test statistic at level of significance α for a two -tailed test is given by zα where zα is determined by the equation P[|Z| > zα] = α i. a test for testing the mean of a population H0 : μ = μ0 against the alternative hypothesis H1 : μ > μ0 (right-tailed) or H1 : μ < μ0 (left-tailed). 10 (7) 0 1 0 0 -½ ½ 25 . x 2 . s1 .25 -1 1 20 .BLOGSPOT.COM Ans Max.20 (8) 9/1 = 9 12/2 = 6 Table 2 0 50 3/ ½ = 6 Table 3 30 50 .zj x1 x2 zj Cj . z 30x1 50x 2 0s1 0s 2 subject to x1 x 2 s1 0s 2 9 x1 2x 2 0s1 s 2 12 Where x1 .Visit For more Solved Assignments: IGNOU4U.zj Table 1 (1) 0 0 Values of the Basic Variables (3) 9 12 0 3 6 300 6 3 330 - Ratio (4) 1 1 0 30 ½ ½ 25 5 1 0 30 0 (5) 1 2 0 50 0 1 50 0 0 1 50 0 (6) 1 0 0 0 1 0 0 0 2 -1 10 . Simplex Table 30 x 50 y 0 S1 0 S2 C Basic Variable (2) S1 S2 zj Cj .zj S1 x2 zj Cj . s 2 0. the required optimum solution is x1 = 6 and x2 = 3 and the corresponding value of z = 330. where at least one constant is not zero. The rank of a matrix cannot exceed the number of its rows or columns – whichever is less. (i) zero.COM In the net evaluation row of the table all the elements are either zero or negative. k3 should all not be equal to zero. Thus. How would you determine linear dependence of a matrix? Define the rank of a matrix on terms of its linear independence. k1. rank of a matrix can also be defined as the maximum number of rows (or columns) in a matrix. This means the optimum program has been attained and there is no scope for further improvement. rank of a matrix is defined as follows: . if there is at least one (r x r) sub-matrix of A whose determinant is not equal to the determinant of every (r+1) – rowed square sub-matrix of A is zero. c2 and c3 are equal to zero for some values of k1. The vector of rows (columns) of a matrix are said to be linearly dependent if and only if some linear combination of them is a null vector or null row (column). A number ‘r’ is said to be the rank of matrix A. k2. 2nd and 3rd rows by any constants k1. rank of a matrix is equal to the order of the highest order non-singular square matrix contained in A. ================================================================== 5. Consider the matrix: a11 a21 a31 a12 a22 a32 a13 a23 a33 A linear combination of the rows 1. Of course. Ans. k3. Moreover. The rows (columns). at least one of these must be non-zero. k2 and k3 and adding them. which are not linearly dependent. 2.Visit For more Solved Assignments: IGNOU4U. Hence. are said to linearly independent. k2.BLOGSPOT. and (ii) In other words. With regard to the concept of linear dependence (independence). and 3 are obtained by multiplying the 1st. the new role will be [k1 a11 + k2 a21 + k3 a31 c2 c3] k1 a12 + k2 a22 + k3 a32 k1 a13 + k2 a23 + k3 a33] = [c1 The rows are linearly dependent if all the three elements c1. For each eigenvector. but not all authors also require x to be non-zero. after being multiplied by the matrix. Then x is an eigen-vector of T with eigen-value λ if the following equation holds: This equation is called the eigen-value equation. Most. The correlation coefficient between nasal length and stature for a group of 20 Indian adult males was found to be 0. ================================================================== 7. the corresponding eigen-value is the factor by which the eigen-vector is scaled when multiplied by the matrix. To extract this behavior it is possible to write . and let T be a linear transformation mapping V into V. is the maximum number of linearly independent rows (or columns) in A. remain parallel to the original vector. Write short notes on the following: (a) (b) (c) (d) Eigen-vectors and Eigen-values Taylor’s expansion Mixed strategy Kuhn-Tucker condition Ans. let x be a vector in that vector space. the action of the transformation T on x.Visit For more Solved Assignments: IGNOU4U. while λx means the product of the number λ times the vector x. we may be faced with a very complicated function f(x) whose behavior in general may not interest us but we may be interested in its properties at or near a point x=a. a more general definition is given: Let V be any vector space. Test whether there is any correlation between the characteristics in the population. Ans.BLOGSPOT. ================================================================== Ans (b) The Taylor Expansion Suppose to approximate a function f(x) at some arbitrary point x=a by a polynomial of the form. Note that Tx means T of x. The set of eigen-values of T is sometimes called the spectrum of T.203.(a) The eigen-vectors of a square matrix are the non-zero vectors that.COM The rank of matrix A (denoted by ρ(A). ================================================================== 6. In abstract mathematics. (1.Visit For more Solved Assignments: IGNOU4U.6) (1. The catch however lies in obtaining the coefficients of the expansion.. = An(x-a)n (1. A Nash equilibrium (mixed strategy) is a strategy profile with (1. (1..1) now becomes f(x) = 1/n! [dnf(x)/dxn] x=a(x-a)n n=0 This is called the Taylor expansion ================================================================== Ans (c) In the theory of games a player is said to use a mixed strategy whenever he or she chooses to randomize over the set of available actions. A trick was provided by Taylor and proceeds as follows: (a) In eq (1. then all terms vanish except A0 A0 = f(a) (1.1) once with respect to x (x) / dx = A1 + 2A2(x-a) + 3A3(x-a)2 + . the player is said to use a pure strategy. A mixed strategy profile is a list of strategies. (1... Now set x=a and d2f(x) / dx2 |x=a= 2A2 = 2!A2 If you continue in this way you will soon discover that dnf(x) / dxn |x=a= n!An An = 1/n! (dnf(x)/ dxn) |x=a Note the meaning of this: FIRST DIFFERENTIATE n TIMES THEN TAKE x=a Thus eq. If only one action has a positive probability of being selected.1) n=0 If we are interested in the function near a then the quantities (x-a)n will become rapidly smaller and smaller and ultimately vanishing after some value of n.8) (1.4) (c) Differentiate eq. A trick catch however lies in obtaining the coefficients of the expansion. a mixed strategy is a probability distribution that assigns to each available action a likelihood of being selected.COM f(x) = A0 + A1(x-a) + A2(x-a)2 + A3(x-a)3 + . one for each player in the game.1) everywhere put x=a. Thus the function has now been approximated by a polynomial. (1..BLOGSPOT. (1. (1. in eq.3).5) (1. Formally..3) Now set.1) twice or what amounts to differentiating eq.2) (b) differentiate eq. x=a everywhere df(x) / dx |x=a= A1 = 1!A1 once d2f(x) / dx2 = 2A2 + 6A3(x-a) + . A mixed strategy profile induces a probability distribution or lottery over the possible outcomes of the game.7) .3) (1. e. which allows only equality constraints. the KKT approach to nonlinear programming generalizes the method of Lagrange multipliers. provided that some regularity conditions are satisfied.Visit For more Solved Assignments: IGNOU4U. Allowing inequality constraints. Kuhn. then there exist constants such that Stationarity and ..COM the property that no single player can. who first published the conditions Suppose that the objective function and the constraint functions and are continuously differentiable at a point x * . and Albert W. If x * is a local minimum that satisfies some regularity conditions (see below). Tucker. ================================================================== The linear regression of Y on X and X on Y are: Y = a + bX The two regression equations are and X = a + bY . and the KKT multipliers are called Lagrange multipliers. Primal feasibility Dual feasibility Complementary slackness In the particular case m = 0. i.BLOGSPOT. when there are no inequality constraints. the KKT conditions turn into the Lagrange conditions. induce a lottery that he or she finds strictly preferable. by deviating unilaterally to another strategy. called KKT multipliers. ================================================================== Ans (d)The Kuhn–Tucker conditions are necessary for a solution in nonlinear programming to be optimal. The KKT conditions were originally named after Harold W. BLOGSPOT. Y ) (Xi – X) [Y on X] and Xi – X = (Yi – Y) var( X ) var(Y ) [X on Y] Yi – Y = byx (Xi – X) and Xi – X = bxy ( Yi – Y) Where byx and bxy are coefficients of regression Mean of X = 30/5 = 6 and cov( X . estimated value of Xi = = 26/5 x 1/4 (5-8) + 6 = 2. Y ) (Xi – X) + Y var( X ) Mean of Y = 40/5 = 8 Estimated value of Yi = = 26/5 x 1/8 (2-6) + 8 = 5.65 > 1 which is for y = 5.1 The value of byx = 0.Visit For more Solved Assignments: IGNOU4U.65x + a impossible Again x = a + bY => x = 1.COM Yi – Y = cov( X .3 y = a + bX => y = 0.3y + a for y =5.5 <1 which is possible ================================================================== ================================================================== ================================================================= . Y ) (Yi – Y) + X var(Y ) Similarly.65 and bxy = 1. Y ) cov( X . x=2 we get a = 3.4 cov( X . x = 2 we get a = -4.