Topic 6: Tools for comparative statics Flashcards
When can implicit differentiation be used with multivariate functions (1)
-Implicit differentiation can be used if c = f(x, y), c ∈ ℝ, and we want to see how y changes when x changes to keep c constant
How can we use f(x, y) = xy and z = 5 to create an implicit function and find y’ (3,3)
-Take f(x, y) = xy, z = 5
-xy = 5
-xg(x) = 5, where g is the implicit function
Differentiate with respect to x
-g(x) + x(dg(x)/dx) = 0
-dg(x)/dx = -g(x)/x
-y’ = -y/x
How do we use implicit functions to find the slope of a level curve (4)
-With level curves c = f(x, y), we need to turn it into an implicit function to find the slope
-for f(x, y) = c, there can be g = g(x)
-f(x, g(x)) = c
-We then take the derivative with respect to x
What is the formula for the slope of a level curve (1)
-y’ = dy/dx = -f’1(x,y)/f’2(x,y)
What does the slope of a level curve tell us (1)
-How much Y has to change in response to a change in x so that the function maintains its value C
What is the total derivative of a bivariate function with respect to t (2)
-Suppose we have bivariate function z = f(x, y) and suppose x = g(t) and y = h(t), so that z = f(g(t), h(t))
-dz/dt = ∂f(x, y)/∂x dx/dt + ∂f(x, y)/∂y dy/dt
What is the value function and why does it matter (2, 3)
-The value function is the value of a function when evaluated at its optimum
-F(r) = f(x(r), r)
-This matters as we want to know how the functions representing the extreme points change when parameters (fixed numbers) change
-Consider the problem max x f(x, r), where r is a parameter
-If we change r, x* might change, so we have x*(r)
How do we differentiate the value function with respect to r (3)
-df(r)/dr = f’1(x(r), r) dx(r)/dr + f’2(x(r), r)
-This shows how r influences changes in the value function when r changes indirectly through x(r), or directly through r
-However, since x(r) is an extreme point, then f’1(x*(r), r) = 0
What is the envelope theorem (2)
-df(r)/dr = f’2(x(r), r)
-Only the direct effect of the change in r
What do we know about k curves vs the value function (2,2,2)
-For a given point x0, there is curve Kx0 representing y = f(xo, r)
-We have many of these K curves, and also our value function g = f*(r)
-By definition of our value function, f(x, r) ≤ maxxf(x, r) = f*(r)
-Therefore, none of the K curves can lie above the value function
-For each r there is at least one x(r)
-Thus, we know Kx*(r) will touch the curve of y = f(r) in the point (r, f(x(r))) = (r, f(x*(r)), r)) and will have the same tangent in that point
How do we draw the relationship between the value function and the k curves (3)
-Have r on the x axis, and y on the y axis
-Draw the value function as an upwards sloping curve, going from concave to convex
-All the K curves are convex, small, and have one tangential point with the value function
What is the general version of the envelope theorem (4)
-The general version considers the case where there are n variables denoted in vector notation as n, and k parameters denoted in vector notation as r
-Assume f(r) = maxxf(x, r), and x(r) is the max value
-df(r)/drj = [∂f(x, r)/∂rj]_x_ = _x_*(_r_) for any j = 1, …, k
-The subscript of the square brackets means evaluated at x(r)
What is the differential of f(x,y) (2)
-dz = f’x(x,y) dx + f’y dy
-This considers small changes in the variables x and y denoted by dx and dy
What is the differential of z = x2y3 (1)
-dz = 2xy3 dx + 3x2y2dy
What are some rules with 2 functions and differentiation (4)
Let f(x, y) and g(x, y) be functions, a and b be parameters, and d be the differential
-d(Af + by) = adf + bdy
-d(fg) = gdf + fdg
-d(f/g) = (gdf - fdg)/g2 (similar to product rule)
-If z = g(f(x,y)), dz = g’(f,x, y))df (chain rule)
What is the formula for a linear approximation (1)
f(x) at x = x0
-f(x) ≈ f(x0) + f’(x0)(x - x0)
How can we form a linear approximation of f(x) = 5x3 - 7x at x0 = 4, then evaluate the function at x = 5 (2,1)
-f(x) ≈ f(4) + f’(4)(x-4)
-f(x) ≈ 292 + 233(x-4)
-Evaluating the function at x = 5 gives us a true value of 590, and an approximate of 525, and thus a -65 error
How do you evaluate the function at x = x0 (2,1)
-Sub x0 into both the approximation and actual function
-The error is approximate - true value
-It is typical the further from the approximation point we evaluate, the less precise the approximation becomes
What is the linear approximation to a multivariate function (1)
The approximation occurs to z = f(x) at (x1, …, xn) around x0 = (x10, …, xn0)
-f(x) ≈ f(x0) + f’1(x0)(x1 - x10) + … + f’n(x0)(xn - xn0)
What is the formula for a quadratic approximation to f(x) around x = x0 (1)
-f(x) ≈ f(x0) + f’(x0)(x - x0) + (1/2)f’‘(x0)(x - x0)2
What is a taylor approximation (2)
A Taylor approximation of order n to f(x) around x = x0 is
-f(x) ≈ f(x0) + f’(x0)(x - x0) + … + (1/n!)(fn(x0)(x - x0)n
-The approximation tends to be more accurate the higher the n value, and is exact if n ≥ the highest polynomial
What does it mean for a function to be homogeneous (2)
A function of n variables (x1, …, xn) is homogeneous of degree k ∈ ℝ if for any t > 0
-f(tx1, …, txn) = tkf(x1, …, xn)
-if k = 1, we get the same proportion out what we put in
How can we continue for homgeneousity of a cobb-douglas function, and the returns to scale parameter (2, 3)
Consider cobb-douglas f(K, L) = AKaLb
-F(tK, tL) = ta+bf(K, L)
-The function is homogeneous of degree a + b
-a+b > 1 = increasing returns to scale
-a + b = 1 = constant returns to scale
-a + b < 1 = decreasing returns to scale
What are properties of homogeneous function (2)
Let f(x, y) be homogeneous of degree k
-f’x and f’y are both homogeneous of degree k-1
-f(x, y) = xkf(1, y/x) = ykf(x/y, 1) for all x > 0, y > 0
