2 8 Extremals and Convex cones
Advanced Math
Today we discuss some method for economics of uncertainty. The tools that we are going to use are ‘extremals’ and ‘convex cones’. The idea is the following: suppose that something is true at all of the extremes of some convex cone. In addition, suppose that there is a linear relationship between all points in the convex cone. Then, it must be true that it is true for all points in the convex cone.
Of course, we spent the end of the last lecture showing how expected utility is really linear despite utility being concave. Thus, we have the linear relationship that is used in the last paragraph. For probability, the convex cone is the set of possible combinations of outcomes (∑Pi where Pi is the probability that outcome i occurs.) For iε(1,2,), this graphically looks like:
The extremals are the points on the triangle: the points where one of the outcomes occurs for certain. The convex cone is, of course, the shaded area of possible probability combinations. All possibilities in the convex cone are convex combinations of the extremals.
We are going to use these methods to check for characteristics utility in an uncertain universe. We are, specifically, going to use it to check for characteristics that uncertain choices must have in order to be strictly preferred, and then will use the same tools to see how people with different types of utility functions react (differently) to gambles. For gamble comparison, we use the method to find first order and second order stochastic dominance. For utility, rather than using points on a plane, we use functions of a certain class.
Example one: we derive the meaning of first order stochastic dominance. Suppose that we have two gambles that we can take: Xº1 and Xº2. Furthermore, assume only that valuation of the gambles is done by some function that is increasing in the outcome (i.e. a utility function that is increasing in income.)
We want to know what the necessary and sufficient conditions are to guarantee:
Eu(Xº2)≤Eu(Xº2)
We want the extremals for the utility function (the function that we use to evaluate the gamble.) The extremal of an increasing function is called a Heavy side function, H which is defined by:
H(ξ-x)=
In other words, the heavy side function is a very simple step function. It equals one if its characterizing variable xi is less than x, the realization of the random variable in this instance. Graphically, it looks like
Of course, the derivative of the heavy side function is what is called a ‘Dyvac delta’ and looks like the following (with the spike really thin.)
Although we won’t prove it, it is simple to see that it is possible to find the utility of any outcome X from heavy side functions using the following formula:
U(x)=∫-∞∞H(x-ξ)U’(ξ)dξ
It will not be much of a problem if U’ does not exist as it could be replaced by dU(ξ). However, for the example, we assume continuity in the u function. The important aspect is that we can create any increasing utility function with a convex combination (the integral over) heavy side functions. Thus, the heavy side functions are, in fact, the extremals of increasing utility functions. (To do the actual proof, you would need to do use Leibniz rule to solve the integral and show that it is, in fact, equal to the utility function.)
Now that we have the extremals, we can now see what characteristics of the random variable lead to the characteristic we desire:
EU(Xº2)≥EU(Xº1)
If this characteristics is true for all of the extremals, then it is also true for the function that is the convex combination of the extremals.
Thus, the next step is to see what needs to be true about the random variables in order for the above comparison to be true for all of the extremals. If:
E(H(X-ξ) with ξ from Xº2)≤ E(H(X-ξ) with ξ from Xº1) for all ξ
{Note that E(H(X-ξ) with ξ from Xº2) can be written E(H(X-ξ)Xº2)}
Is true for all of the extremals, i.e. the heavyside functions, then it is true for all increasing utility functions. To see this, just note that we have utility simply from:
∫-∞∞E[H(x-ξ)Xº2]U’(ξ)dξ≤∫-∞∞E[H(x-ξ)Xº1]U’(ξ)dξ
This is a necessary characteristic as we have made up the utility function from the H functions. This is a necessary condition as we showed that all increasing functions are convex combinations of heavyside functions. If you can’t make the condition true for all heavyside functions, then there will be at least some increasing utility function for which the above inequality does not hold.
Denote Fi as the distribution function for variable Xºi. Thus, we can rewrite:
E(H(X-ξ)Xº2)≤ E(H(X-ξ)Xº1)
As
E(H(Xº2-ξ))≤ E(H(Xº1-ξ))
Where:
E(H(Xºi-ξ))=∫-∞∞H(X-ξ)dFi(X)
=Fi(X)]ξ∞=1-Fi(ξ)
The algebra from here is straightforward to see that:
EU(Xº2)≤EU(Xº1) iff EH(Xº2-ξ)≤EH(Xº1-ξ) for all ξ
Which is equivalent to:
1-F2(ξ)≤1-F1(ξ) for all ξ
In other words, the above is true iff F2 first order stochastically dominates F1:
F2(ξ)≥F1(ξ)
Example two: Second order stochastic dominance
We are going to derive the characteristics of two gambles that are necessary and sufficient conditions for one gamble to be preferred to another for all increasing and concave utility functions. In other words, we are now setting not just the first derivative, but also the second derivative of the utility function.
Just to get a sense of the pattern this will follow, we note that the extremal for this type of function has two items to worry about: the first and second derivative. Well, the second derivative we saw we could deal with by a heavy side function (actually, since we want the derivative to be negative, it is the inverse of a heavy side function, but same idea.) We drew out the heavy side function earlier. Increasing and concave… comes from some function which is the integral of heavy side functions—these are:
- the straight line: f(x)=x
- the min function: f(x)=min(x-ξ,0) where ξ describes the min function.
Graphically, min functions look like:
Once again, we do not prove that this is the set of extremals, but you can intuitively see that we can create U(x) from:
U(x)=U’(+∞)+∫-∞∞-u”(ξ)[min(x-ξ)]dξ
(The first derivative is a negative heavy side function, an the second derivative is dyvac delta.) Although we don’t do it, you could prove that this is the case by integrating by parts—or you could just believe it because it seems reasonable.)
Just as before, we want to find the expected value of each min function given each random variable. Algebraically, we do:
∫-∞∞Min(x-ξ,0)dFi(x)=∫-∞ξ(X-ξ)dFi(x)
Integrate by parts:
=F1(X)(X-ξ)]-∞ξ-∫-∞ξ1F1(x)dx
The first term is zero. One method of proving it is zero is to take the limit as x goes to negative infinity, noticing that the value of the low probability of a low value must go to zero by Chebyshev’s inequality. However, for economics problems, it is generally enough to simply state that the CDF of X has a finite lower bound.
Anyway, we have:
∫-∞∞Min(x-ξ,0)dFi(x)= -∫-∞ξF1(x)dx
which means that EU(Xº2)≤EU(Xº1) iff
-∫-∞ξF2(x)dx≤-∫-∞ξF1(x)dx
and we also need to make sure that it is also true for the final extremal: linear utility. Thus, we also require that:
E(Xº2)≤E(Xº1)
This last condition is not important for this second order stochastic dominance. However, it would prove to be important for the third order.
Let’s consider third order stochastic dominance. What does it take for one gamble to be preferred to another one when utility is increasing, concave, and has a positive third derivative? Of course, the key is once again that the lowest derivative with assigned sign is characterized by a heavy side function, the next derivative up is characterized by a min (actually, here it will be a max function), and the next one is more curved—a quadratic.
Graphically, you see that the quadratic will look like:
In other words, the extremals will be:
- g(X,ξ)=[Min(X-ξ)2,0]
- U(x)=X2 (this is for the limit case as ξ goes to ∞)
We need to find the expected value of this for X being a random variable, and thus:
∫-∞∞Min[(x-ξ,0)2]dFi(x)=∫-∞ξ[(X-ξ)2]dFi(x)
(for the a type extremals.)
Integrate by parts:
=2F1(X)(X-ξ)]-∞ξ-∫-∞ξ2(X-ξ)F1(x)dx
Of course, the first term is zero again. This leaves us:
=-∫-∞ξ2(X)F1(x)dx+∫-∞ξξF1(x)dx
Integrate the first by parts:
=-2X∫-∞ξF1(x)dx]-∞ξ+2∫-∞ξ∫-∞δF1(x)dxdδ+ξ∫-∞ξF1(x)dx
I’m sure there is an algebra mistake, but so be it. I believe that is the correct approach. However, that’s a start. Furthermore, notice that for part b, you also need to guarantee that E([Xº2]2), i.e. the variance.
Moving on to risk aversion signing: Suppose that we want to consider only the situation in which we have increasing risk aversion. We will check utility in logs as it simplifies matters, so we write this as the following inequality (risk aversion is decreasing in X):
means:
which is rewritten:
Where we know the RHS is positive (risk averse) and u”<0. Therefore, u”’>0 in order for the equality to work out. Thus, when you assume decreasing relative risk aversion, you assume positive third derivative. This positive third derivative really does seem reasonable as wealthier people do tend to invest more in riskier assets.
Let’s move on; we have three derivatives, let’s go for four. Suppose that we think one of two items:
- wealthier will respond less than poor people(lesser precautionary savings) for a risk of a given size; i.e. the possibility of a loss of $10,000.
- People increase holding of the risky asset less for a given increase in income as their wealth builds.
If we buy into either of those conditions, then we are assuming that:
The math is the same, and thus taking the actual derivative leads to:
which means that the fourth derivative needs be >0.
We see that the four derivatives that we have signed thus far are alternating in sign. It turns out that there is a particular function that is the extremal as the limit of the number of alternating sign derivatives goes to infinity. That function is:
U(x)=-e-ax
While there certainly is no reason to go a heck of a lot further then fourth derivative, it is nice to note that such a simple function has the signs of all derivatives we might want correct. The infinite alternating sign aspect is not useful for economics. However, the fact that it gets the first four signs correct means that it may be a reasonable check for some things. I.e. If you think some result may be true, you could plug in this utility function and see if it holds. Because it has the correct signs (for the important derivatives), we may be inclined to think we are on to something.
Next class’s problems:
During the next lecture, we are going to consider problems in which the following must be true:
If Ef(Xº)≤f(0) then Eg(Xº)≤g(0)
(Note that 0 is the outside option which we simply normalize to 0—we do not mean the null set.)
In other words, we are going to be interested in ‘ordering’ people or their preferences, their things, etc.
Let’s consider some of the interesting possibilities:
- If F(x)=x, then we can consider how strongly people ‘like’ given things
- If f(x)=XU’(w+x), then, as we found in the last lecture, we are thinking about how much risky asset people of a type like. (I.e. it is coming from maxx EU(w+αx))
- If we want to consider how much precautionary savings one type of person is going to do, then f(x) is
f(x)=
If f(x) is greater than 0, then we know:
Multiply through by u” and do a bit of simple algebra (note that the sign of the inequality changes when we multiply a negative number from one side to the other), and we this transforms into this inequality:
The Xºhere is ‘background risk’. I.e. suppose that xº and Yº are two statistically independent risks. We want to know how the existence of the one risk effects reaction to the other risk.
Consume one of the risks (xº) into a new utility function:
Û(w)=EU(w+Xº)
We therefore have:
ExEyU(W+Xº+Yº)=EyÛ(w+Yº)
EU(w+Xº+Yº)=E Û(w+Yº)
In other words, we are now able to sign the risk aversion for one risk—given the effect on another risk. Specifically, we have:
where the inequality comes from risk aversion.
We provide a unifying method for dealing with all of these risk-related items. See p. 77 in the course pack for a chart of which f functions and which g functions are important for various risk-related characteristics.
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