# Why care so much about expected utility?

I have a naive question about decision theory. We calculate the probabilities of various outcomes assuming particular decisions and assign utilities or costs to each outcome. We find the optimal decision by finding the one with the greatest expected utility.

But why should we reason in this way? Each decision in fact has a distribution of utility associated with it. Why do we compare the distributions of utilities for different choices only by a single summary statistic? And why do we pick the mean rather than mode or median, etc?

I can imagine cases in which two choices yield identical expected utilities but their distributions for the utility vary greatly. Surely decisions should be made based on the whole distribution and not the expectation alone?

Are we saying that for any scheme for making decisions using the whole distribution, there must exist a utility function for which maximum expected utility would give identical results? If so, shouldn't we anyway construct utilities faithfully and select a decision rule as we wish? We can later convert our faithful utilities to ones that give identical results with maximum expectation.

• See for example ch. 2.2 in C. Robert (2007) "The Bayesian Choice" for a review of the axioms underlying the existence of the utility function. – Jarle Tufto Nov 13 '17 at 10:48
• +1 for @JarleTufto. Also note that when you perform Bayesian inference, you need not choose a singular value for the utility-indexing parameter $\theta$. Rather, you task yourself with constructing a distribution $\pi$ over $\theta$ which minimizes the expected [log likelihood] loss (maximizes the expected utility) as regularized by your prior belief. – Jeremias K Jan 14 '19 at 21:44

The Von Neumann-Morgenstern utility theorem implies that under some reasonable assumptions (such as the fact that you are able to order a set of scenarios from best to worst, where each scenario stochastically resolves to some outcome), then there exists a function mapping each possible outcome to a real value (the "utility"), such that you will always prefer the scenario with the higher expected utility. Therefore it makes sense to always select the choice which maximizes expected utility.

I can imagine cases in which two choices yield identical expected utilities but their distributions for the utility vary greatly.

VNM utility takes this into account so that even if you are risk averse, the highest expected utility scenario will be the most preferable.

Are we saying that for any scheme for making decisions using the whole distribution, there must exist a utility function for which maximum expected utility would give identical results? If so, shouldn't we anyway construct utilities faithfully and select a decision rule as we wish? We can later convert our faithful utilities to ones that give identical results with maximum expectation.

I would rather say that the strategy of approximating the utility of certain outcomes via guesswork or some human heuristics leads to imperfect decision making, since the resulting utility function differs from the ideal VNM utility. Constructing utilities "faithfully" will resolve the problem and make it so that maximizing utility yields the right answer.

My answer may surprise you. I'll answer it within the expected utility theory, and beyond it.

# Beyond Expected Utility

Expected utility theory itslef is not the only way of decision making. Whether you use utility theory or not depends on the applications. For instance, in wealth management some advisors use prospect theory instead of expected utility. Kahneman got Nobel prize in Economics for his work on this theory. It brought up behavioral aspects of decision making in economics beyond the expected utility theory.

Practically, in a traditional portfolio choice approach, wealth advisors attempt to construct the client's utility function, then use it to select the best portfolio on the efficient frontier. In prospect theory approach the advisors attempt to construct the value function instead of the utility function, and use the former to pick the best portfolio.

# Within expected utility theory

I can imagine cases in which two choices yield identical expected utilities but their distributions for the utility vary greatly. Surely decisions should be made based on the whole distribution and not the expectation alone?

Now, even in traditional utility theory this is taken care of. For instance they have a notion of risk aversion and stochastic dominance. A risk averse person will not pick the decision solely based on expected utility. That would be a risk neutral person. Risk averse people will prefer decisions with lower entropy when presented with decisions having the same expected utility, for instance. This is called stochastic dominance.

The analogy would be looking at two nromal distributions with the same mean but different dispersions. Yes, these are different distributions, and the dispersion matters in many applications. However, this doesn't diminish the importance of knowing the mean. To fully define the normal distribution you need to know both mean and the dispersion, and the mean itself informs us a lot about the distribution. Similarly, expected utility is not the only thing you'll ever need to know about the agent's utility function, but it is a lot of information nevertheless.

To great extent this is really a question about expected value, that was already discussed in different place. You are right, that we are and should be interested in the whole distributions, but it is hard to compare whole distributions and comparing single point summaries is much easier. Yes, you could compare other single point summaries, and in many cases you would compare them, but expected value has several nice properties that makes it a very good single point summary for a random variable. Expected value weights the possible outcomes by their probabilities and tells you what could you "expect" in the long run. If you play against the casino, the expected value of the possible wins and losses is negative for you, so it tells you that in the long run you shouldn't expect it makes you rich.

Let me give you very game-theoretically non-rigorous example. Imagine that you are considering playing Russian roulette, you are going to take one shot towards yourself using a six-shot revolver with only one bullet in the chamber. If nothing happens, you win \$1000, otherwise you die. The mode outcome is that you win \$1000, same with the median. The expected value from this game is 5/6 $\times$ \$1000$+$1/6$\times$death, would you consider playing? Of course in game theoretic approach you would consider what is the actual utility of the money won and what is the price of dying, but I guess that without going any deeper you should see the point of using expected value as a single point summary in here. Expected value (and mean, its estimator) are sensitive to outliers and this is one of the reasons why use it so much. Would you even consider the competition if the price was \$1? What about \$1 000 000 000? Notice that if you were using the mode or median as your criteria for the "possible" outcome, you shouldn't care since in each case they tell you that you win "on average". Would you change your mind if you were shooting with blank bullets? Notice that neither the mode, nor the median do not change if you are using blanks, since they don't care about the extreme outcomes, yet the expected value changes dramatically*. Expected value (and mean) consider all the possible outcomes and weight them by probabilities, that is the reason for using it in decision scenario. More realistic example would be the lottery with 1000 coupons and only a single winning coupon. Say that the price is \$1000, so the expected value is 999/1000 $\times$ \$0$+$1/1000$\times$\$1000 = \$1, so coupon is not worth buying if its price is not less then \$1. This means that if you played the game many, many times, you would win few times and loose a lot of times and the overall balance of the invested and won money would be approximately \$1. If the prize would change to \$10 000, without changing the coupon price, the story would be different since the expected value would change to \$10. Notice that, again, the mode or median are in both cases \$0, so they are insensitive to the payoffs. This is not saying that they are useless, but it shows that the expected value is what we usually need in here.

* - To be honest, this example is misleading, since you can kill yourself with blanks, but for stake of argument lets say that you have some kind of hypothetical "safe" blanks.

• Hmm thanks but I'm looking for more general explanations. I'm quite comfortable with the Russian Roulette example – innisfree Nov 13 '17 at 9:47
• @innisfree what do you mean by more general explanations? – Tim Nov 13 '17 at 9:48
• General results and remarks about maximum expected utility versus alternative schemes for making choices. – innisfree Nov 13 '17 at 9:50
• @innisfree This is what I tried to show, please see my edit, maybe it helps. – Tim Nov 13 '17 at 10:07
• Interesting, but hardly compelling. It's true that the mean is sensitive to $p(x)$ for all $x$, but so is any other functional of the distribution. And if you consider them all, you're essentially considering all moments of the distribution, which is basically the distribution itself. – innisfree Nov 15 '17 at 1:07

In their answer, @shimao focussed on the von Neumann-Morgenstern utility theorem. The theorem indeed lies at the heart of why we consider the expected utility, rather than any other summary statistic of the utility, or indeed the whole distribution of utility.

The theorem shows, from a few axioms, that when faced with uncertainty a decision maker should choose the course of action that maximizes the expected utility. I think the relevant axiom to my question is the axiom of continuity.

We rank three possible choices in order, say, $$L \preceq M \preceq N$$, where $$A \preceq B$$ indicates that an outcome $$A$$ is worse than or no better than outcome $$B$$. The axiom of continuity states that there must exist a probability, $$p$$, such that taking option $$L$$ with probability $$p$$ and option $$N$$ with probability $$1 - p$$ must be just as good as just taking option $$M$$, i.e., there exists a $$p$$ such that $$p L + (1 - p) N \sim M$$ Without repeating the full proof, it is clear that this suggests why the variance (or any other further moments) of the utility don't matter. It doesn't matter how extreme outcomes $$L$$ and $$N$$ are, our axiom is that there must exist a probability such that the choice to take $$L$$ with probability $$p$$ and $$N$$ otherwise is just as good a choice as sticking with $$M$$. This is despite the fact that the former could have a huge variance in utility.

There are a few slight language mistakes that are creating a bit of confusion regarding your question.

But why should we reason in this way? Each decision, in fact, has a distribution of utility associated with it. Why do we compare the distributions of utilities for different choices only by a single summary statistic? And why do we pick the mean rather than mode or median, etc.?

Utilities do not have a distribution. Outcomes have a distribution and via the outcome, actions, in some cases, have a distribution. Utility is deterministic. If it were random, then your feelings regarding an outcome would constantly startle you. For example, you could have the experience of "wow, having my legs crushed in a motor vehicle accident was a surprisingly good experience!" What is uncertain is the outcome from an action.

If we exclude degenerate cases, where the integrals diverge, and a solution does not exist, I think I can also show you a case where the median maximizes expected utility.

Note that $$\mathcal{U}(\delta(x),\mu)=-\mathcal{L}(\delta(x),\mu)$$. We find it important to create a rule, which is what we are evaluating with our utility, that will find $$\mu$$ with some consistency.

We want to solve: $$\min_{\delta}\mathcal{L}(\delta,\mu)=|\delta(x)-\mu|$$ subject to $$f(x|\mu)=\frac{1}{\pi}\frac{1}{1+(x-\mu)^2}.$$

If we assume that $$\Pr(\mu)\propto{1},$$ then the risk is $$\int_{-\infty}^\infty|\delta(x)-\mu|\prod_{i=1}^n\frac{1}{\pi}\frac{1}{1+(x_i-\mu)^2}\mathrm{d}x$$

and the integrated risk minimizes when $$\int_{-\infty}^\infty\int_{-\infty}^\infty|\delta(x)-\mu|\prod_{i=1}^n\frac{1}{\pi}\frac{1}{1+(x_i-\mu)^2}\mathrm{d}x\mathrm{d}\mu$$ is at a minimum. It minimizes when $$\delta(x)$$ is the median.

You maximize expected utility when you find the median of the data. You cannot find a mean for $$f(x|\mu)=\frac{1}{\pi}\frac{1}{1+(x-\mu)^2},$$ as it does not exist. Because it has no mean, it also has no variance. Since it has no variance, you cannot minimize quadratic loss. Consequently, quadratic utility, if it were the true case, would be minimized by any value in the real numbers.

If you ignore the degenerate cases, like the above case, expected utility has an unexpected advantage over other methods. Considering all possible decision rules and actions that could be taken, when you use expected utility, then you end up with a total ordering. You are correct, there could be ties, but because the impact of all parameters has been accounted for, you would be indifferent between the choices with tied utility.

The alternative, which is used in Frequentist decision theory, is to order the risk function through stochastic dominance. A Frequentist decision is said to be admissible if it cannot be stochastically dominated. This doesn't permit a total ordering. Nonetheless, if $$\delta(x)$$ first-order stochastically dominates $$\delta'(x)$$, then it is also true that the expected utility of choosing $$\delta>\delta'$$. So the alternative gets you the same result.

There are a few other solutions that can be used, but they either map to maximizing expected utility, or they beg the question of why you would use them in the cases where they do not. To give another statistical example, imagine you read a research study that had a sample size of one million observations using maximum likelihood methods or Bayesian methods. You replicate the study with a sample size of 100 and estimate the mean and the variance using an unbiased estimator. Neither Bayesian nor maximum likelihood estimators are unbiased in the general case.

You insist that you will not combine your estimates because the other estimate is biased, whereas yours in unbiased. Bayesian methods offer a disciplined method to combine the samples into a single point estimator maximizing your utility. You insist on losing the information in the one million person sample in favor of unbiasedness.

Now, if your utility had a very very strong bias toward unbiased estimators, then you would be maximizing your utility by not maximizing the utility of your estimator. But in the absence of that, the biased estimator will be far more accurate than that from your small sample alone. If accuracy maximizes your utility, then you end up choosing an estimator that is utility maximizing.

Don't confuse the expectation of the utility with the expected value of the action. Those are different things.

Also, consider maximizing expected utility versus median utility. You take the utility of every outcome times its probability and sum it. $$\mathbb{E}[\mathcal{U}(\tilde{x})]=\int_{\tilde{x}\in\chi}\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}$$

Now let's think about the median utility.$$\mathbb{M}[\mathcal{U}(\tilde{x})]=c$$ if $$\int_a^c\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}=\int_c^b\mathcal{U}(\tilde{x})\Pr(\tilde{x})\mathrm{d}\tilde{x}.$$

What would that mean? You would be just as happy if you landed to the left as you landed to the right of $$c$$? Why would you care about that?

If you chose an action that maximized the expected utility, then there is no action that you could take that you believe would make you happier. The median utility doesn't permit a maximization as the action is chosen by force of being in the center. You would always take the action that gives you a fifty percent chance of being happier than usual or sadder than usual. What a strange thing to do!

EDIT From Kolmogorov's axioms, the sum of a distribution must equal one. Consider a case with two sets of actions, $$a$$ and $$a'$$, where $$a'$$ is the set of actions which are not $$a$$.

Focusing in on $$a$$, let us assume that the utility function is $$-x^2$$. Let us assume that $$x$$, when the action is $$a$$, is drawn from $$f(x)=\exp(-x),x>0$$.

Noting that $$\int_0^\infty\exp(-x)\mathrm{d}x=1,$$ we can readily confirm that it is a probability density function. Including utility results in $$\int_0^\infty{x^2}\exp(-x)\mathrm{d}x=-2,$$ which confirms it is not a distribution. $$\mathbb{E}(\mathcal{U}(a))=-2.$$

While it would be possible to constuct a distribution by utilities, it won't necessarily be a function since if $$g(x)=\mathcal{U}(x)\Pr(x)$$, then $$g^{-1}(x)$$ isn't guaranteed to be a function.

• Utilities of course have a distribution!? I’ve no idea what you’re talking about – innisfree Jan 22 '19 at 2:38
• Ah, OK. The utility of a particular sure-thing outcome $A$, $U(A)$, is indeed 'deterministic'. The outcome though is a priori unknown, so since outcome $O\sim \text{some distribution}$, we also have a distribution of the a priori unknown $U(O)$. – innisfree Jan 22 '19 at 9:17