Mathematical Economics Questions and Answers

Answer: B

The correct statement is Nash equilibrium is reached when each of them purely randomises their behavior.

Check best responses. Against Left, \(X\) prefers Bottom (2 over 1); against Right, Top (4 over 3). Against Top, \(Y\) prefers Left (2 over 1); against Bottom, Right (4 over 3). In every cell one player wants to switch, so there is no pure-strategy equilibrium. Let \(X\) play Top with probability \(p\) and \(Y\) play Left with probability \(q\):

\[ \begin{aligned} U_{Y}(L) &= 2p + 3(1 - p) = 3 - p \\ U_{Y}(R) &= p + 4(1 - p) = 4 - 3p \\ 3 - p &= 4 - 3p \Rightarrow p = \tfrac{1}{2} \end{aligned} \]

\[ \begin{aligned} U_{X}(T) &= q + 4(1 - q) = 4 - 3q \\ U_{X}(B) &= 2q + 3(1 - q) = 3 - q \\ 4 - 3q &= 3 - q \Rightarrow q = \tfrac{1}{2} \end{aligned} \]

The only equilibrium is for each player to mix 50:50, a purely random choice. There are no pure-strategy equilibria, hidden or otherwise. Battle of the Sexes has two pure-strategy equilibria, so it is unlike this game. The game shares Matching Pennies' cycle of best responses, but Matching Pennies is zero-sum, while here the payoff totals differ across cells (3, 5, 5 and 7); the key accepts only the statement about randomisation.

Answer: B

The equilibrium output is 13.

The firm maximises profit where marginal revenue equals marginal cost:

\[ \begin{aligned} MR &= \frac{dR}{dQ} = 30 - 2Q \\ MC &= \frac{dC}{dQ} = 4 \\ 30 - 2Q &= 4 \Rightarrow Q = 13 \end{aligned} \]

The second-order condition holds, since MR falls with Q while MC is constant. At Q = 13 profit is 30(13) - 169 - 20 - 52 = 149. Q = 15, where marginal revenue is zero, would maximise revenue rather than profit; the options 9, 11 and 17 do not satisfy MR = MC.

Answer: D

The non-negativity constraints are X \(\geq\) 0 and Y \(\geq\) 0.

In a linear programming problem the decision variables represent physical quantities, such as units of two products to be made, which cannot be negative. Every LP formulation therefore adds the constraints that each variable is greater than or equal to zero, which confines the feasible region to the first quadrant.

The other options require one or both variables to be zero or negative, which would describe meaningless negative production levels.

Answer: C

The area under a curve is computed using Integration.

The definite integral of a function between two limits gives the area between the curve and the horizontal axis over that interval:

\[ A = \int_{a}^{b} f(x)\, dx \]

In economics this is how consumer and producer surplus, total cost from marginal cost, and probabilities under a density function are found. Differentiation gives slopes and rates of change, not areas. Limits underlie integration but are not the method itself, and probability is one application of areas under a curve, not the means of computing them.

Answer: A

The determinant is 4.

For a 2 by 2 matrix, \(\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\):

\[ \det A = 2 \times 2 - 1 \times 0 = 4 \]

A is upper triangular, so its determinant is also just the product of the diagonal entries, \(2 \times 2 = 4\). The value 2 multiplies only one diagonal entry, and 0 or 1 come from subtracting wrongly.

Answer: B

The characteristic roots are (1, 3).

Solve \(\det(A - \lambda I) = 0\):

\[ \begin{aligned} (2 - \lambda)^{2} - 1 &= 0 \\ 2 - \lambda &= \pm 1 \\ \lambda &= 1 \text{ or } 3 \end{aligned} \]

Check: the roots add to the trace, \(1 + 3 = 4 = 2 + 2\), and multiply to the determinant, \(1 \times 3 = 3 = 4 - 1\). The negative roots fail both checks, and (1, 1) has the wrong sum.

Answer: B

\(\alpha\) measures the elasticity of output with respect to capital.

For \(y = K^{\alpha}L^{\beta}\),

\[ \begin{aligned} \frac{\partial y}{\partial K} &= \alpha K^{\alpha - 1}L^{\beta} = \alpha\frac{y}{K} \\ \frac{\partial y}{\partial K}\cdot\frac{K}{y} &= \alpha \end{aligned} \]

So \(\alpha\) is the percentage change in output from a one per cent change in capital, and similarly \(\beta\) is the output elasticity of labour. The marginal product of capital is \(\alpha y/K\), which varies with the input mix, so it is not \(\alpha\) itself.

Answer: C

The profit maximizing output is 2.

Profit is revenue minus cost, with revenue \(Pq\):

\[ \begin{aligned} \pi &= (20 - 3q)q - (3 + 2q^{2}) \\ &= 20q - 5q^{2} - 3 \\ \frac{d\pi}{dq} &= 20 - 10q = 0 \Rightarrow q = 2 \end{aligned} \]

The second derivative is \(-10 \lt 0\), so this is a maximum. Equivalently, MR = 20 - 6q equals MC = 4q at q = 2. Profit there is 40 - 20 - 3 = 17; at q = 1 it is 12, at q = 3 it is 12 and at q = 4 it is -3, so 1, 3 and 4 all give less.

Answer: A

The function is minimized at 1, the value of \(x\) (the paper writes 'value of \(y\)', but the options are values of the variable \(x\)).

\[ \begin{aligned} \frac{dy}{dx} &= 6x - 6 = 0 \Rightarrow x = 1 \\ \frac{d^{2}y}{dx^{2}} &= 6 \gt 0 \end{aligned} \]

The positive second derivative confirms a minimum. The minimum value of the function itself is \(y = 3 - 6 + 10 = 7\), which is not among the options, so the question can only be asking for \(x\).

At \(x = 0\), 2 and 3 the function takes the values 10, 10 and 19, all larger than 7.

Answer: A

\(A^{C} \cap B = \) {4, 5, 6}.

The complement of A within X is everything in X except 1, 2 and 3:

\[ A^{C} = \{0, 4, 5, 6, 7, 8, 9, 10\} \]

Intersecting with B = {4, 5, 6} keeps the elements in both sets, which is all of B, since A and B have no element in common. In general, when A and B are disjoint, \(A^{C} \cap B = B\).

{1, 2, 3, 4, 5} mixes in elements of A, which the complement excludes, and {2, 4, 5} and {5, 6, 8} include elements that are not in B or drop ones that are.