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**Question regarding stochastical dominance**

About slide 743:

In the Example the x-axis is “Negative cost” does this mean -4 would be a payoff of +4 then? Meaning the “further left” the better?

Or is stochastical dominance independent of the acutal value in this case and is only about the size of the integral?

Intuitively I would have thought S1 is better than S2, since it has more density in the less negative area than S2.

Thank you in advance.

No, the opposite: Cost is bad, the lower the cost, the better. Hence it makes sense to define our utility function (which is supposed to be of the form „the larger U, the better“, so that we can specify our goal as *maximizing* U!) as the *negative cost*. The larger the negative cost, the better

Depending on what you mean by „actual value“. If you mean the actual cost, then the whole point is that we don’t *know* the actual cost. We only know a *range* of possible costs via some probability distribution, which is why we deal with integrals in the first place.

I agree with this interpretation

I should add, that slide 743 is missing an explanation in the notes. I recommend Russell&Norveig section 6.4 (page 622f in the 3rd edition), where those two diagrams seem to be inspired by. In which case the left graphs would be probability density functions and the graphs on the right would be the cumulative distributions for those (i.e. the graphs on the right are of the integrals of the graphs on the left, which is why they approach 1 at around x=-2.5). Hence the left side shows the functions to integrate according to the definition of stochastical dominance, and the graphs on the right show those integrals to demonstrate that clearly S1 dominates S2. The graphs on the right thus tell you (for each value x for U) the probability that the *actual* value will be *at most* x. Since the probability is lower for S1 than for S2 everywhere, and since the domains of both functions range over *negative* values only, the fact that the graph for S1 is smaller than the graph for S2 everywhere implies that the actual utility is *larger* for S1 than for S2 everywhere; hence S1 dominates.

Thank you for that explanation.

I guess I just misunderstood which distribution is dominating.

The first two questions were just me trying to find ways to explain what was completely against my intuition…