<p>You might be looking for the concept of <a href="https://en.wikipedia.org/wiki/Stochastic_dominance" rel="nofollow noreferrer">stochastic dominance</a>:</p>

<p>Roughly speaking, first order stochastic dominance of distribution A over distribution B, means that for every possible outcome gamble A pays out weakly more and in at least one state it pays out strictly more. Roughly speaking, second order stochastic dominance means that distribution A has the same risk as distribution B but a higher mean (but not necessarily weakly-higher in every state of the world. Alternatively, you preserve the mean of distribution B and you "scrunch" the probability mass to create distribution A. "All risk-averse expected-utility maximizers (that is, those with increasing and concave utility functions) prefer a second-order stochastically dominant gamble to a dominated one." </p>

<p>The seminal paper is <em>Rules for Ordering Uncertain Prospects</em> (<a href="https://www.jstor.org/stable/1811090?seq=2#metadata_info_tab_contents" rel="nofollow noreferrer">Hadar and Russell (1969)</a>)</p>

CC BY-SA 4.0
2020-01-30T19:53:11.153Z
2020-01-30T19:53:11.153Z
BKay
<p>Risk is a subjective measure. Perhaps climbing a tree is very risky for me, while you think it is a child's game and entails no risk. Intuitively this is why there are many ways to define risk, for example, the variance is a common one. </p>

<p>The variance has limitations of course, for example, if the distribution is very concentrated in two different outcomes  (bimodal) the variance becomes harder to interpret since it will be a large number, even though the lottery yields one of the two outcomes with very high probability. </p>

<p>The other approach suggested by @BKay is that of stochastic dominance. Let me expand on it. </p>

<p>This approach takes seriously the fact that people might disagree in their risk preferences, but if lottery A is preferred to lottery B by any agent that prefers to have more money rather than less, we say that A first-order stochastic dominates (FSD) B (beware there are other equivalent definitions out there, but I find this one very intuitive), so without specifying a utility function you can say that A is better than B. Do you want to say that A is less risky than B? - maybe, maybe not. For many pairs of lotteries, however, you will not be able to say that A FSD B, because the requirement is strong: every money maximizer should prefer A to B. </p>

<p>If you are willing to be less strict you can ask, will at least every risk-averse agent prefer A to B? If the answer is yes, we say that A second-order stochastically dominates (SSD) B. Here every agent that dislikes risk prefers A to B regardless of the specific shape of their utility. Do you now want to say that A is less risky than B? It would not be crazy to say so, because even if A has a higher variance than B, since A SSD B, it must be that in average A gives a better payoff. So it is less risky not in terms of having less uncertainty about what the outcome will be, but it is less risky in that it is likely to give a better outcome than the other lottery. </p>

<p>In your example, <span class="math-container">$L_1 \succsim_{SSD} L_2$</span> because they have the same expected value, but L2 has a higher variance - These two properties are sufficient conditions for SSD. Any risk-averse agent will prefer <span class="math-container">$L_1$</span> to <span class="math-container">$L_2$</span>. </p>

<p>One of the limitations of stochastic dominance is that it cannot possibly meassure how much more "risky" a lottery is, it simply can say which one is more and which one is less. Another very important limitation is that there are still a lot of pairs of lotteries where some risk-averse agents will prefer one of them and the others will prefer the other one, so you cannot say which one is riskier using this definition. </p>

CC BY-SA 4.0
2020-01-31T20:42:18.083Z
2020-01-31T20:42:18.083Z
Regio
<blockquote>
  <p>I wonder if it possible to compare risks between two lotteries without giving the utility function.</p>
</blockquote>

<p>What do you mean by this?  Do you mean that we don't know the utility function?  Or that we assume it is linear, e.g., <span class="math-container">$U(x) = ax + b$</span>?  If the latter case, then you're not really avoiding a utility-based formulation of risk preferences... you're just assuming risk-neutrality.</p>

<blockquote>
  <p>Can we say that one lottery is riskier than the other?</p>
</blockquote>

<p>As Giskard notes, this is a weird question to ask because there are many possible definitions of risk, and it requires you to supply a lot of extra information that may not have a robust backing. You give up a lot by ignoring utility functions in discussions about risk preferences - including the ability to evaluate lotteries in terms of preference, rather than "riskiness", which is about as close to a standard for comparing risks as we have.</p>

CC BY-SA 4.0
2020-01-30T18:24:27.447Z
2020-01-30T18:24:27.447Z
heh
<p>I understand the concepts of risk-aversion, risk neutrality and risk-attraction. I wonder if it possible to compare risks between two lotteries without giving the utility function. For instance,  Let X = (10, 20, 30) and consider two lotteries L1 = (1/3, 1/3, 1/3) and L2 = (5/12,1/6,5/12). Can we say that one lottery is riskier than the other?</p>

CC BY-SA 4.0
2020-01-30T17:53:24.697Z
You can say whatever you want, but what do you want riskiness to measure? Will variance do as a measure?
Giskard
CC BY-SA 4.0
2020-01-30T17:56:33.723Z
does it mean that there is no standard definition to compare the concept "risk"?
Alex Wang
CC BY-SA 4.0
2020-01-30T18:06:03.490Z
There are several.
Giskard
<microeconomics><uncertainty>
Can we compare risks of lotteries?
Alex Wang
# Can we compare risks of lotteries?

(Asked by: username_0 on 2020-01-30)

I understand the concepts of risk-aversion, risk neutrality and risk-attraction. I wonder if it possible to compare risks between two lotteries without giving the utility function. For instance, Let X = (10, 20, 30) and consider two lotteries L1 = (1/3, 1/3, 1/3) and L2 = (5/12,1/6,5/12). Can we say that one lottery is riskier than the other?

**Tags:** microeconomics, uncertainty
> **Comments:**
> **username_1**: You can say whatever you want, but what do you want riskiness to measure? Will variance do as a measure?  
> **username_0**: does it mean that there is no standard definition to compare the concept "risk"?  
> **username_1**: There are several.  

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## 3 answers

### Answer by username_2 on 2020-01-30. Score: 4

You might be looking for the concept of [stochastic dominance](https://en.wikipedia.org/wiki/Stochastic_dominance):

Roughly speaking, first order stochastic dominance of distribution A over distribution B, means that for every possible outcome gamble A pays out weakly more and in at least one state it pays out strictly more. Roughly speaking, second order stochastic dominance means that distribution A has the same risk as distribution B but a higher mean (but not necessarily weakly-higher in every state of the world. Alternatively, you preserve the mean of distribution B and you "scrunch" the probability mass to create distribution A. "All risk-averse expected-utility maximizers (that is, those with increasing and concave utility functions) prefer a second-order stochastically dominant gamble to a dominated one."

The seminal paper is *Rules for Ordering Uncertain Prospects* ([Hadar and Russell (1969)](https://www.jstor.org/stable/1811090?seq=2#metadata_info_tab_contents))


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### Answer by username_3 on 2020-01-31. Score: 3

Risk is a subjective measure. Perhaps climbing a tree is very risky for me, while you think it is a child's game and entails no risk. Intuitively this is why there are many ways to define risk, for example, the variance is a common one.

The variance has limitations of course, for example, if the distribution is very concentrated in two different outcomes (bimodal) the variance becomes harder to interpret since it will be a large number, even though the lottery yields one of the two outcomes with very high probability.

The other approach suggested by @BKay is that of stochastic dominance. Let me expand on it.

This approach takes seriously the fact that people might disagree in their risk preferences, but if lottery A is preferred to lottery B by any agent that prefers to have more money rather than less, we say that A first-order stochastic dominates (FSD) B (beware there are other equivalent definitions out there, but I find this one very intuitive), so without specifying a utility function you can say that A is better than B. Do you want to say that A is less risky than B? - maybe, maybe not. For many pairs of lotteries, however, you will not be able to say that A FSD B, because the requirement is strong: every money maximizer should prefer A to B.

If you are willing to be less strict you can ask, will at least every risk-averse agent prefer A to B? If the answer is yes, we say that A second-order stochastically dominates (SSD) B. Here every agent that dislikes risk prefers A to B regardless of the specific shape of their utility. Do you now want to say that A is less risky than B? It would not be crazy to say so, because even if A has a higher variance than B, since A SSD B, it must be that in average A gives a better payoff. So it is less risky not in terms of having less uncertainty about what the outcome will be, but it is less risky in that it is likely to give a better outcome than the other lottery.

In your example, $L\_1 \succsim\_{SSD} L\_2$ because they have the same expected value, but L2 has a higher variance - These two properties are sufficient conditions for SSD. Any risk-averse agent will prefer $L\_1$ to $L\_2$.

One of the limitations of stochastic dominance is that it cannot possibly meassure how much more "risky" a lottery is, it simply can say which one is more and which one is less. Another very important limitation is that there are still a lot of pairs of lotteries where some risk-averse agents will prefer one of them and the others will prefer the other one, so you cannot say which one is riskier using this definition.


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### Answer by username_4 on 2020-01-30. Score: 0

> I wonder if it possible to compare risks between two lotteries without giving the utility function.

What do you mean by this? Do you mean that we don't know the utility function? Or that we assume it is linear, e.g., $U(x) = ax + b$? If the latter case, then you're not really avoiding a utility-based formulation of risk preferences... you're just assuming risk-neutrality.

> Can we say that one lottery is riskier than the other?

As Giskard notes, this is a weird question to ask because there are many possible definitions of risk, and it requires you to supply a lot of extra information that may not have a robust backing. You give up a lot by ignoring utility functions in discussions about risk preferences - including the ability to evaluate lotteries in terms of preference, rather than "riskiness", which is about as close to a standard for comparing risks as we have.


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