IGNOU MEC-003 Free Solved Assignment 2012

MEC-003: Quantitative Techniques 

Assignment

Course Code: MEC-003

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

ð       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 x^1/2 + 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.

A second order ordinary differential equation consists of time as the independent variable with the dependent variable y with its first and second derivatives. Consider for example an equation G(t, y(t), y’(t), y*(t)) = 0 for all t such that we can write it in the form

y*(t) = F(t, y(t), y’(t)).

Equations of the form y”(t) = F(t, y’(t)

Take an equation of form

y*(t) = F (t, y’(t)),

in which y(t) does not appear.

Stability of solutions of second order homogeneous equation

Consider the above homogeneous equation

            y*(t) + ay’(t) + by(t) = 0

If b ≠ 0, this equation has a single equilibrium, viz, 0. That is, the only constant function that is a solution is equal to 0 for all t. 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. If r₁ and r₂, are the two roots of the characteristics equation, then the general solution of the equation is y(t) = Ae^r1t + Be^r2t. The equilibrium is stable if and only if r₁ < 0 and r₂ < 0.

Case I: Characteristics equation has a single real roots.    With a single root (say r), the characteristic equation is in stable equilibrium if and only if this root is negative. Note that if r<0 then for any value of k, t^k e^rt converges to 0 as t →∞.

Case I: Characteristics equation has complex roots.   When the characteristic equation has complex roots, the form of the solution of the equation is Ae^t cos(βt + ω), where α = - a/2, the real part of each root. The equilibrium will be stable if and only if the real part of each root is negative.

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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.

Ans.     The function f(x) is called the probability density function (p.d.f) provided it satisfies the following two conditions

• F(x) ≥ 0                                                                       b
• If the range of the continuous random variable is (a,b) ∫ f(x) = 1

                                                                                                            a

A random variable X is said to have a binomial distribution and is referred as a binomial random variable, if and only if it’s probability distribution is given by

            B (x;n,p) = ⁿC_x . p^x . (1-p)^n-x for x = 0,1,2,…..n

The probability of getting ‘x’ success and ‘n-x’ failures in n trials is given by p^x . (1-p)^n-x.

The number of trials in the case of binomial distribution is very large and cumbersome. Then the probability distribution used to approximate binomial possibilities. This limiting form of the binomial distribution when n  → ∞ and p → 0, while n.p remains constant. Let λ = n.p, therefore p = λ/n. The binomial distribution written as

B(x,n,p) = ⁿC_x (λ/n)^x (1-( λ/n))^n-x = n (n-1) (n-2) (n-3)………(n-x+1)/x!  x  (λ/n) ^x (1-( λ/n))^n-x

              = 1.(1-1/n)(1-2/n)(1-3/n)……(1-(x-1)/n/x! x (λ) ^x x [(1 - ( λ/n))^n/λ] ^-λ x (1-(λ/n))^-x .

When n →∞ while x and p are constants

(1-1/n) (1-2/n) (1-3/n) ….(1-(x-1)/n) → 1

(1 - ( λ/n)) ^-x  → 1

(1 - ( λ/n))-^n/λ → e

Therefore, the limiting form of the binomial distribution becomes,

P (x,λ) = λ^x e ^-λ / x! for x = 0,1,2,3,……….

Thus, in the limit when n→ ∞ and p → 0, n.p = λ remains constant; the number of success is a random variable and will follow a poisson distribution with only parameter λ.

Thus Poisson distribution is a discrete probability distribution which is the limiting form of the binomial distribution, provided

            (a)        The number of trials is very large in fact tending to infinity.

            (b)        The probability of success in each trial is very small; tending to zero.

The properties of the Poisson distribution are

(a)        Poisson distribution is a discrete probability distribution, where the random variable assumes countably infinite number of values such as 0,1,2,3,…… to ∞. The distribution is completely specified if the parameter λ is known.

(b)        Mean and variance of poisson distribution are the same, both being λ.

(c)        Poisson distribution like the binomial distribution may have either one or two modes When λ is not an integer, mode is the largest value contained in λ and when λ is an integer, there are two modes, λ and (λ-1).

(d)       The poisson distribution used as an approximation to binomial distribution when n is large but np is fixed.

Example: Let X be a random variable following poisson distribution. If P(X=1) = P(X=2), find P(X=0 or 1) and E(X).

For poisson distribution, the probability mass function (p.m.f) is given by  P (x,λ) = λ^x e ^-λ / x!

Therefore,        (PX=1) = λ¹ e ^-λ / 1! =  λ e ^-λ

                        (PX=2) = λ² e ^-λ / 2! =  λ e ^-λ/2

As (PX=1) = P(X=2), from the equation λ e ^–λ  =  λ² e ^-λ / 2! , we get λ  = 2.

Therefore,        E(X) = λ = 2 and

            It does not p.d.f                     

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Section B

Medium Answer questions                                                                                      

3.         Explain the relevant considerations of making choice between one-tailed and two-tailed tests. How would you determine the level of significance in the above tests?

Ans.     A test of any statistical hypothesis where the alternative hypothesis is one-tailed (right-tailed or left-tailed) is called a ‘one-tailed test’. For example, 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), is a ‘single-tailed test”. In the right-tailed test (H₁ : μ > μ₀), the critical region lies entirely in the right tail of the sampling distribution of x, while for the left-test (H₁ : μ < μ₀), the critical region is entirely in the left tail of the distribution.

A test of any statistical hypothesis where the alternative hypothesis is two-tailed such as: H₀ : μ = μ_0, against the alternative hypothesis H₁ : μ ≠ μ₀ (μ > μ₀ and μ < μ₀) is known as ‘two-tailed 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.

In a particular problem, whether one-tailed or two-tailed test is to be applied depends entirely on the nature of the alternative hypothesis. If the alternative hypothesis is two-tailed then apply the two-tailed test and if alternative hypothesis is one-tailed, then apply one-tailed test.

The value of the test statistic, which separates the critical (or rejection) region and the acceptance region is called the ‘critical value’ or ‘significant value’. It depends upon: (i) the level of significance used, and (ii) the alternative hypothesis, whether it is two-tailed or single-tailed.

The critical value of test statistic at level of significance α for a two-tailed test is given by z_α where z_α is determined by the equation P[|Z| > z_α] = α i.e z_α is the value so that the total area of the critical region on both tails is α. Since normal probability curve is symmetrical curve, so P[|Z| > z_α] = α can be written as,

P[Z > z_α] + P[Z < -z_α] = α     

ð        P[Z > z_α] + P[Z < z_α]  = α

ð        2P[Z > z_α]  = α

ð        P[Z > z_α]   = α/2

i.e, the area of each tail is α/2.

Thus, z_α is the value such that area to the right of z_α is α/2 and to the left of -z_α is α/2. Thus, 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’.

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4.         A linear programming problem is given as

                                                  

      Find its optimal solution.

Ans     

Simplex Table

				30	50	0	0
	C	Basic Variable	Values of the Basic Variables	x	y	S1	S2	Ratio
	(1)	(2)	(3)	(4)	(5)	(6)	(7)	(8)
Table 1	0	S1	9	1	1	1	0	9/1 = 9
	0	S2	12	1	2	0	1	12/2 = 6
		zj	0	0	0	0	0
		Cj  - zj	-	30	50	0	0
Table 2	0	S1	3	½	0	1	- ½	3/ ½ = 6
	50	x2	6	½	1	0	½
		zj	300	25	50	0	25
		Cj  - zj	-	5	0	0	- 25
Table 3	30	x1	6	1	0	2	- 1
	50	x2	3	0	1	- 1	1
		zj	330	30	50	10	20
		Cj  - zj	-	0	0	- 10	- 20

            In the net evaluation row of the table all the elements are either zero or negative. This means the optimum program has been attained and there is no scope for further improvement. Hence, the required optimum solution is x₁ = 6 and x₂ = 3 and the corresponding value of z = 330.

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5.         How would you determine linear dependence of a matrix? Define the rank of a matrix on terms of its linear independence.

Ans.    A number ‘r’ is said to be the rank of matrix A, if

            (i)         there is at least one (r x r) sub-matrix of A whose determinant is not equal to zero; and

            (ii)        the determinant of every (r+1) – rowed square sub-matrix of A is zero.

In other words, rank of a matrix is equal to the order of the highest order non-singular square matrix contained in A. Moreover, rank of a matrix can also be defined as the maximum number of rows (or columns) in a matrix. The rank of a matrix cannot exceed the number of its rows or columns – whichever is less.

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). The rows (columns), which are not linearly dependent, are said to linearly independent.

Consider the matrix:

                                    a₁₁        a₁₂        a₁₃

                                    a₂₁        a₂₂        a₂₃

                                    a₃₁        a₃₂        a₃₃

                                               

A linear combination of the rows 1, 2, and 3 are obtained by multiplying the 1^st, 2^nd and 3^rd rows by any constants k₁, k₂ and k₃ and adding them, where at least one constant is not zero. Thus, the new role will be

[k₁ a₁₁ + k₂ a₂₁ + k₃ a₃₁                 k₁ a₁₂ + k₂ a₂₂ + k₃ a₃₂                   k₁ a₁₃ + k₂ a₂₃ + k₃ a₃₃] =  [c₁            c₂      c₃] 

The rows are linearly dependent if all the three elements c₁, c₂ and c₃ are equal to zero for some values of k₁, k₂, k₃.  Of course, k₁, k₂, k₃ should all not be equal to zero, at least one of these must be non-zero.

With regard to the concept of linear dependence (independence), rank of a matrix is defined as follows:

The rank of matrix A (denoted by ρ(A), is the maximum number of linearly independent rows (or columns) in A.

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6.         The correlation coefficient between nasal length and stature for a group of 20 Indian adult males was found to be 0.203. Test whether there is any correlation between the characteristics in the population.

Ans.                

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7.         Write short notes on the following:

            (a)        Eigen-vectors and Eigen-values

            (b)       Taylor’s expansion

            (c)        Mixed strategy

            (d)       Kuhn-Tucker condition

Ans.(a)                        The eigen-vectors of a square matrix are the non-zero vectors that, after being multiplied by the matrix, remain parallel to the original vector. For each eigenvector, the corresponding eigen-value is the factor by which the eigen-vector is scaled when multiplied by the matrix.

In abstract mathematics, a more general definition is given:

Let V be any vector space, let x be a vector in that vector space, and let T be a linear transformation mapping V into V. 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. Note that Tx means T of x, the action of the transformation T on x, while λx means the product of the number λ times the vector x. Most, but not all authors also require x to be non-zero. The set of eigen-values of T is sometimes called the spectrum of T.

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Ans (b)  The Taylor Expansion

Suppose to approximate a function f(x) at some arbitrary point x=a by a polynomial of the form, 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. To extract this behavior it is possible to write

f(x) = A₀ + A₁(x-a) + A₂(x-a)² + A₃(x-a)³ + ... = ¥ A_n(x-a)ⁿ                                (1.1)

                                                                                                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. Thus the function has now been approximated by a polynomial. The catch however lies in obtaining the coefficients of the expansion. A trick catch however lies in obtaining the coefficients of the expansion. A trick was provided by Taylor and proceeds as follows:

(a) In eq (1.1) everywhere put x=a, then all terms vanish except A₀

\ A₀ = f(a)                                                                             (1.2)

(b) differentiate eq. (1.1) once with respect to x

(x) / dx = A₁ + 2A₂(x-a) + 3A₃(x-a)² + ...                                          (1.3)

Now set, in eq. (1.3), x=a everywhere

df(x) / dx |_x=a= A₁ = 1!A₁                                                                   (1.4)

(c) Differentiate eq. (1.1) twice or what amounts to differentiating eq. (1.3) once

d²f(x) / dx² = 2A₂ + 6A₃(x-a) + ...                                                     (1.5)

Now set x=a and

d²f(x) / dx² |_x=a= 2A₂ = 2!A₂                                                  (1.6)

If you continue in this way you will soon discover that

dⁿf(x) / dxⁿ |_x=a= n!A_n                                                                         (1.7)

\ A_n = 1/n! (dⁿf(x)/ dxⁿ) |_x=a

                                          

Note the meaning of this: FIRST DIFFERENTIATE

 n TIMES THEN TAKE x=a

Thus eq. (1.1) now becomes

f(x) =   ¥ 1/n! [dⁿf(x)/dxⁿ] _x=a(x-a)ⁿ                                                 (1.8)

n=0

                                     

This is called the Taylor expansion

           

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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. Formally, a mixed strategy is a probability distribution that assigns to each available action a likelihood of being selected. If only one action has a positive probability of being selected, the player is said to use a pure strategy.

A mixed strategy profile is a list of strategies, one for each player in the game. A mixed strategy profile induces a probability distribution or lottery over the possible outcomes of the game. A Nash equilibrium (mixed strategy) is a strategy profile with the property that no single player can, by deviating unilaterally to another strategy, induce a lottery that he or she finds strictly preferable.

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Ans (d)The Kuhn–Tucker conditions are necessary for a solution in nonlinear programming to be optimal, provided that some regularity conditions are satisfied. Allowing inequality constraints, the KKT approach to nonlinear programming generalizes the method of Lagrange multipliers, which allows only equality constraints. The KKT conditions were originally named after Harold W. Kuhn, and Albert W. Tucker, who first published the conditions

Suppose that the objective function and the constraint functions and are continuously differentiable at a point x ^* . If x ^* is a local minimum that satisfies some regularity conditions (see below), then there exist constants and , called KKT multipliers, such that

Stationarity                                                   

Primal feasibility

Dual feasibility

Complementary slackness

In the particular case m = 0, i.e., when there are no inequality constraints, the KKT conditions turn into the Lagrange conditions, and the KKT multipliers are called Lagrange multipliers.

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            The linear regression of Y on X and X on Y are:

            Y = a + bX                                          and                              X = a + bY                 

The two regression equations are

                                                       

ð  Y_i – Y = b_yx (X_i – X)                              and  X_i – X = b_xy ( Y_i – Y) 

Where b_yx and b_xy are coefficients of regression

Mean of X = 30/5 = 6                                     and      Mean of Y = 40/5 = 8

Estimated value of Y_i = (X_i – X) + Y

                                    = 26/5 x 1/8 (2-6) + 8 = 5.4

Similarly, estimated value of X_i = (Y_i – Y) + X

                                                    = 26/5 x 1/4 (5-8) + 6 = 2.1

            The value of b_yx = 0.65 and b_xy = 1.3

            y = a + bX  => y = 0.65x + a       for y =5, x=2 we get a = 3.65 > 1 which is impossible

Again x = a + bY => x = 1.3y + a         for y = 5, x = 2 we get a = -4.5 <1 which is possible

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