# Continuous choice

## Axioms

### Choice set

Economic agents face options from some set. This could be consumption choices, numbers of hours to work, or how much capital to invest in at a factory.

#### Utility functions

Calculating choices: Unrestricted choices

The choice of an agent is the selection which corresponds to the highest value of the utility function. Consider:

$$f=2(x-1)^2-10$$

We can easily calculate that even if the agent can choose any real number $$x$$, they will chose $$1$$.

This approach can be used if there are not meaningful constraints, or those constraints are implicit in the utility function. For example a firm can be modelled as profit maximising, where profit is a function of revenue and costs, with a single maximising value.

Other agents, such as a consumers, may instead have utility over consumption and leisure, and a separate constraint over this. This could be solved using simultaneous equations, but such an approach is not always desirable.

### Complete preferences

#### Axiom 2: An economic agent has preferences across all pairs of elements of the choice set

Where an agent prefers one option to another, we say this choice is preferred to the other. We can show this formulaically. If $$a$$ is preferred to $$b$$ then:

$$a\succ b$$

Alternatively $$b$$ may be preferred to $$a$$.

$$a\prec b$$

Finally an agent may be indifferent to the choices.

$$a\sim b$$

We use can additional symbols to denote an agent prefers or is indifferent between options:

$$a\succeq b$$

$$a\preceq b$$

Note that if:

$$a\succeq b$$

And:

$$a\preceq b$$

Then:

$$a\sim b$$

For a set of choice, we want the agent to have preferences defined across possible subsets – the power set.

Add symbols for choice across a set. include symbols for choice of a set, x =c(a,b,utility)

a>b etc only for pairwise, but preferences are across more than that

Discussion of repeated. agents don’t choose the same thing each time, interpretation?

## Choices onto $$\mathbb{R}^1$$ space

### Transitivity

#### Axiom 3: Transitivity

If $$a\preceq b$$, and $$b\preceq c$$, then $$a\preceq c$$.

For example an agent is offered apples or bananas, and prefers apples, and then bananas or carrots, and chooses bananas. If the agent is then offered apples or carrots, they axiom of transitivity says they chould choose apples.

This is not always observed. For example if a firm offers two products, a cheap $$a$$ option and an expensive $$b$$ option, then a preference may be:

$$a\prec b$$

The firm may add another product $$c$$ slightly less expensive than $$b$$, with fewer features, such if there were only $$2$$ of the goods on sale:

$$a\prec b$$

$$a\prec c$$

$$b\prec c$$

In particular, the last two imply the first, through the axiom of transitivity.

However with all three, the marketing effectively makes the consumer choose $$b$$, by making it look like a better deal, so we observe:

$$a\prec s$$

$$e\prec c$$

$$c\prec s$$

Even though the elements have not changed.

Axioms of revealed preference

Add weak axiom of revealed preference

### Independence of irrelevant alternatives

This is not always observed. For example if a firm offers two products, a cheap $$a$$ option and an expensive $$b$$ option, then a preference may be:

$$a\prec b$$

The firm may add another product $$c$$ slightly less expensive than $$b$$, with fewer features, such if there were only $$2$$ of the goods on sale:

$$a\prec b$$

$$a\prec c$$

$$b\prec c$$

In particular, the last two imply the first, through the axiom of transitivity.

However with all three, the marketing effectively makes the consumer choose $$b$$, by making it look like a better deal, so we observe:

$$a\prec s$$

$$e\prec c$$

$$c\prec s$$

Even though the elements have not changed.

Axioms of revealed preference

Add weak axiom of revealed preference

Strong axiom. can derive both (?)

### Utility functions

Given that an agent chooses a selection from a set, how can we model this?

As the outcome of the function is one of the options available to the agent, we can model it by putting a value number on each choice, and where the choice is the selection with the largest value. We can do this by applying a function to each element in the choice set.

This holds only if the agent is rational, that is, if they prefer A to B, they do not switch to B if C is offered. This is due to the transitive properties of real numbers. That is, if $$a\ge b$$ and $$b\ge c$$ then $$a\ge c$$.

If there are multiple elements with the largest value, then the agent would be indifferent between these choices.

A simple example would be choosing the number of apples to consume.

If the agent always prefers more apples, we can have a function which is always increasing when the number of apples increases.

This could be modelled by:

$$f=x$$

$$f=(x-1)^2+1$$

$$f=2(x-1)^2-10$$

$$f=cos(x)$$

$$f=c$$

These correspond to different preferences. In the first the agent prefers more and more apples. In the second and third the agent prefers one apple. These two formulas are monotonic transformations of each other and so are identical for describing preferences. The fourth describes an infinite number of optimal numbers of apples, but is unlikely to correspond to any real preferences, and the fifth shows that agent doesn’t care about apples.

The last gives a real number as output, but doesn’t necessarily take in a real number. Utility functions generally take real numbers, and always give out real numbers.

#### Solving

Option which maximises utility

## Choices from $$\mathbb{R}^1$$ space

### Continuity

#### Transitivity

Axioms of continuity and transitivity

Continuous

Independence of irrelevant alternatives

#### Axiom 3: Continuity

In order to find optimal points for a utility function we want these functions to be differentiable. This requires complete sets of choices. That is, if $$a$$ if preferred to $$b$$, points very close to $$a$$ will also be preferred to $$b$$.

Agents often make choices discretely. How much of good $$x$$ to consume, whether to go on a holiday. We treat these as continuous. This is generally not problematic as agents choices become less discrete over longer time spans, and most economic areas of interest do not rest on discrete consumption.

#### Marginal utility

We can differentiate our utility function with respect to a good. For example, for:

$$f=2(x-1)^2-10$$

$$\dfrac{\delta f}{\delta x}=4x-1$$

This last term is the marginal utility of $$x$$, often shown as $$MU_x$$.

## Choices from $$\mathbb{R}^n$$ space

### Indifference curves

#### Marginal rate of substitution

We can consider how much of one good a consumer is willing to give up to get one of another.

Consider the utility function:

$$U=f(x,y)$$

We can examine the change in utility following changes in inputs by taking the total differential.

$$dU=\dfrac{\delta f}{\delta x}dx+\dfrac{\delta f}{\delta y}dy$$

We want to examine changes where $$dU$$ is $$0$$, so:

$$\dfrac{\delta f}{\delta x}dx+\dfrac{\delta f}{\delta y}dy=0$$

$$\dfrac{\delta f}{\delta x}dx+\dfrac{\delta f}{\delta y}dy=0$$

$$MU_xdx+MU_ydy=0$$

$$\dfrac{dy}{dx}=-\dfrac{MU_x}{MU_y}$$

This is the form of the indifference curve. We refer to $$\dfrac{MU_x}{MU_y}$$ as the marginal rate of substitution.

#### Multiple choices

If the set of choices is more complex, say there are now apples and bananas, we have to be more careful with a representative function.

The agent still prefers more apples and more bananas, but the following imply different choices:

$$f=x^2.y^2$$

$$f=ln(x)+ln(y)$$

In the first example, the agent would always swap an apple for a banana, if they had more apples, whereas the opposite is true in the second case. Functional form is important with multiple goods.

### Specific function results

#### Cobb Douglas

$$U=f(x,y)$$

$$U=x^\alpha y^\beta$$

$$MU_x=\alpha x^{\alpha -1}y^\beta$$

$$MU_x=\alpha \dfrac{y^\beta }{x^{1- \alpha }}$$

$$MU_y=\beta \dfrac{x^\alpha}{y^{1- \beta }}$$

So:

$$\dfrac{MU_x}{MU_y}=\dfrac{\alpha \dfrac{y^\beta }{x^{1- \alpha }} }{\beta \dfrac{x^\alpha}{y^{1- \beta }}}$$

$$\dfrac{MU_x}{MU_y}=\dfrac{\alpha }{\beta }\dfrac{y}{x}$$

#### Constant elasticity of substitution

$$U=f(x,y)$$

$$U=(ax^b+(1-a)y^b)^{\dfrac{1}{b}}$$

$$MU_x=\dfrac{1}{b}abx^{b-1}(ax^b+(1-a)y^b)^{\dfrac{1}{b}-1}$$

$$MU_y=\dfrac{1}{b}(1-a)by^{b-1}(ax^b+(1-a)y^b)^{\dfrac{1}{b}-1}$$

So:

$$\dfrac{MU_x}{MU_y}=\dfrac{\dfrac{1}{b}abx^{b-1}(ax^b+(1-a)y^b)^{\dfrac{1}{b}-1}}{\dfrac{1}{b}(1-a)by^{b-1}(ax^b+(1-a)y^b)^{\dfrac{1}{b}-1}}$$

$$\dfrac{MU_x}{MU_y}=\dfrac{ax^{b-1}}{ (1-a)y^{b-1}}$$

#### Constant elasticity of substitution

$$U=f(x,y)$$

$$U=ax+by$$

$$MU_x=a$$

$$MU_y=b$$

So

$$\dfrac{MU_x}{MU_y}=\dfrac{a}{b}$$

#### Constant elasticity of substitution

$$U=f(x,y)$$

$$U=\alpha .ln(x)+\beta .ln(y)$$

$$MU_x=\dfrac{\alpha }{x}$$

$$MU_y=\dfrac{\beta }{y}$$

So:

$$\dfrac{MU_x}{MU_y}=\dfrac{\alpha }{\beta }\dfrac{y}{x}$$

This is the same as Cobb Douglas. This reflects that the logarithmic function preseres positive inflections.

## Budget constraints

### Budget constraints

Economic agents can choose elements from a set. We can define this set both by what is included, and what is excluded.

An agent could choose inputs from an interval of real numbers. In this case the finding the value which maximises the utility function across all real numbers may not be a valid solution. For example an agent may optimally wish to consume $$20$$ apples, but only have $$10$$.

Similarly an agent may be limited in combinations of choices. For example an agent could choose how much to work and how much to enjoy leisure, but be constrained by the amount of time in the day.

### Budget optimisation

Calculating choices: Restricted choices

But what about where there is not clear maximum, like:

$$f=ln(x)+ln(y)$$

Here the agent would always prefer more of $$x$$ and $$y$$. In practice agents are often limited in their choices by budget constraints. That is, they cannot choose all combinations of inputs.

Here we can use a Lagrangian. This maximises the value of a function subject to constraints on inputs. This may not always be appropriate. The budget constraint for an agent is often an inequality, for example consumption is less than or equal to income, but the Lagrangian takes this to be binding.

Fortunately, this can be resolved. The value of $$\lambda$$ in the Lagrangian corresponds to the marginal effect of weakening the constraint. This is positive where the constraint is binding on the agent, but not positive if it is not. Therefore if we find the constraint is not binding, we can remove it from the optimisation.

Under some conditions, constraints will always be binding. These are useful for specific cases of agents later.

In order for the constraint to be binding we make an additional assumption:

#### Condition 1: Non-satiation

The marginal utility of a good is always positive.

Note that we can “do economics” without this, but we want rely on Lagrangians.

#### Condition 2: Decreasing marginal utility

This ensures that we do not get corner solutions, for example consuming all apples.

These two assumptions allow the use of the Lagrangian.

We know that for the Lagrangian the following is true:

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta g}{\delta x}}=\dfrac{\dfrac{\delta f}{\delta y}}{\dfrac{\delta g}{\delta y}}$$

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta f}{\delta y}}=\dfrac{\dfrac{\delta g}{\delta x}}{\dfrac{\delta g}{\delta y}}$$

Where $$f$$ is the utility function and $$g$$ is the budget constraint.

#### Budget testing

Calculating choices: Restricted choices

But what about where there is not clear maximum, like:

$$f=ln(x)+ln(y)$$

Here the agent would always prefer more of $$x$$ and $$y$$. In practice agents are often limited in their choices by budget constraints. That is, they cannot choose all combinations of inputs.

Here we can use a Lagrangian. This maximises the value of a function subject to constraints on inputs. This may not always be appropriate. The budget constraint for an agent is often an inequality, for example consumption is less than or equal to income, but the Lagrangian takes this to be binding.

Fortunately, this can be resolved. The value of $$\lambda$$ in the Lagrangian corresponds to the marginal effect of weakening the constraint. This is positive where the constraint is binding on the agent, but not positive if it is not. Therefore if we find the constraint is not binding, we can remove it from the optimisation.

Under some conditions, constraints will always be binding. These are useful for specific cases of agents later.

In order for the constraint to be binding we make an additional assumption:

#### Condition 1: Non-satiation

The marginal utility of a good is always positive.

Note that we can “do economics” without this, but we want rely on Lagrangians.

#### Condition 2: Decreasing marginal utility

This ensures that we do not get corner solutions, for example consuming all apples.

These two assumptions allow the use of the Lagrangian.

We know that for the Lagrangian the following is true:

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta g}{\delta x}}=\dfrac{\dfrac{\delta f}{\delta y}}{\dfrac{\delta g}{\delta y}}$$

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta f}{\delta y}}=\dfrac{\dfrac{\delta g}{\delta x}}{\dfrac{\delta g}{\delta y}}$$

Where $$f$$ is the utility function and $$g$$ is the budget constraint.

### Multiple restrictions

Calculating choices: Restricted choices

But what about where there is not clear maximum, like:

$$f=ln(x)+ln(y)$$

Here the agent would always prefer more of $$x$$ and $$y$$. In practice agents are often limited in their choices by budget constraints. That is, they cannot choose all combinations of inputs.

Here we can use a Lagrangian. This maximises the value of a function subject to constraints on inputs. This may not always be appropriate. The budget constraint for an agent is often an inequality, for example consumption is less than or equal to income, but the Lagrangian takes this to be binding.

Fortunately, this can be resolved. The value of $$\lambda$$ in the Lagrangian corresponds to the marginal effect of weakening the constraint. This is positive where the constraint is binding on the agent, but not positive if it is not. Therefore if we find the constraint is not binding, we can remove it from the optimisation.

Under some conditions, constraints will always be binding. These are useful for specific cases of agents later.

In order for the constraint to be binding we make an additional assumption:

#### Condition 1: Non-satiation

The marginal utility of a good is always positive.

Note that we can “do economics” without this, but we want rely on Lagrangians.

#### Condition 2: Decreasing marginal utility

This ensures that we do not get corner solutions, for example consuming all apples.

These two assumptions allow the use of the Lagrangian.

We know that for the Lagrangian the following is true:

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta g}{\delta x}}=\dfrac{\dfrac{\delta f}{\delta y}}{\dfrac{\delta g}{\delta y}}$$

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta f}{\delta y}}=\dfrac{\dfrac{\delta g}{\delta x}}{\dfrac{\delta g}{\delta y}}$$

Where $$f$$ is the utility function and $$g$$ is the budget constraint.

### ??

Calculating choices: Restricted choices

But what about where there is not clear maximum, like:

$$f=ln(x)+ln(y)$$

Here the agent would always prefer more of $$x$$ and $$y$$. In practice agents are often limited in their choices by budget constraints. That is, they cannot choose all combinations of inputs.

Here we can use a Lagrangian. This maximises the value of a function subject to constraints on inputs. This may not always be appropriate. The budget constraint for an agent is often an inequality, for example consumption is less than or equal to income, but the Lagrangian takes this to be binding.

Fortunately, this can be resolved. The value of $$\lambda$$ in the Lagrangian corresponds to the marginal effect of weakening the constraint. This is positive where the constraint is binding on the agent, but not positive if it is not. Therefore if we find the constraint is not binding, we can remove it from the optimisation.

Under some conditions, constraints will always be binding. These are useful for specific cases of agents later.

In order for the constraint to be binding we make an additional assumption:

#### Condition 1: Non-satiation

The marginal utility of a good is always positive.

Note that we can “do economics” without this, but we want rely on Lagrangians.

#### Condition 2: Decreasing marginal utility

This ensures that we do not get corner solutions, for example consuming all apples.

These two assumptions allow the use of the Lagrangian.

We know that for the Lagrangian the following is true:

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta g}{\delta x}}=\dfrac{\dfrac{\delta f}{\delta y}}{\dfrac{\delta g}{\delta y}}$$

$$\dfrac{\dfrac{\delta f}{\delta x}}{\dfrac{\delta f}{\delta y}}=\dfrac{\dfrac{\delta g}{\delta x}}{\dfrac{\delta g}{\delta y}}$$

Where $$f$$ is the utility function and $$g$$ is the budget constraint.