I think that the problem is with taking the expectation over 1effect, instead of multiplying the expected values mean(1effect)â mean(cost).
I agree! Itâs the division thatâs not linear, not the multiplication. How do you think we could make it clearer in the post?
Edit: on more thought, Iâm not sure I understand your point.mean(1/effect)â mean(cost) gives the âwrongâ result (e.g. if effect can be 0 with non 0 probability, mean(1/effect) will be +â)
People are asking the question âHow much money do you have to donate to get an expected value of 1 unit of goodâ.
I think the question is:
How can I do as much good as possible with C units of cost?
This corresponds to the problem of maximising E(U(C)), where U(c) is the utility achieved (via a certain intervention) for the cost c (which must not exceed C). If the budget C is small enough (thinking at the margin):
U(C) = Uâ(0)*C, where Uâ(c) is the derivate of U with respect to cost.
Assuming Uâ(0) and C are independent, mean(âeffectâ/ââcostâ) equals mean(âeffectâ)/âmean(âcostâ):
mean(âeffectâ)/âmean(âcostâ) = E(U(C))/âE(C) = E(Uâ(0))*E(C)/âE(C) (assuming independence between Uâ(0) and C) = E(Uâ(0)).
So, it seems that, regardless of the metric we choose, we should maximise E(Uâ(0)), i.e. the expected marginal cost-effectiveness. However, Uâ(0) and C will not be independent for large C, so I think it is better to maximise mean(âeffectâ/ââcostâ).
I think Michael Dickens explains this better than me in a more recent comment. The point is that we usually care about effect/âcost rather than the other way around (although Iâd love to understand more clearly why and when exactly thatâs true). In your example, you have implicitly computed this and then compare it to Guesstimateâs model computing cost/âeffect.
How do you think we could make it clearer in the post?
I think your example should focus on the mean of the ratio between the effect and cost, not on the mean of the ratio between the cost and effect. The latter is a bad metric because:
A very small âcostâ/ââeffectâ could correspond to interventions that are either quite bad or good (since âcostâ/ââ = 0). This means small numerical errors could lead to large differences in mean(âcostâ/ââeffectâ), which is bad.
When changing from negative values of âcostâ/ââeffectâ to positive ones, the goodness of the intervention increases (changing from harmful to beneficial). However, for negative and positive values, a higher âcostâ/ââeffectâ corresponds to a worse intervention.
The metric âeffectâ/ââcostâ has good properties:
A higher value always implies a better intervention (at least in theory).
I agree! Itâs the division thatâs not linear, not the multiplication.How do you think we could make it clearer in the post?
Edit: on more thought, Iâm not sure I understand your point.mean(1/effect)â mean(cost) gives the âwrongâ result (e.g. if effect can be 0 with non 0 probability, mean(1/effect) will be +â)
Thank you so much for the post! I might communicate it as:
People are asking the question âHow much money do you have to donate to get an expected value of 1 unit of goodâ Which could be formulated as:
E(good(x))=1
where x is the amount you donate and good(x) is the amount of utility you get out of it.
In most cases, this is linear, so: good(x)=goodcostâx. And E(goodcostx)=1.
Solving for x in this case gets x=E(goodcost)â1, but the mistake is to solve it and get x=E(costgood).
Please correct me if this is a bad way to formulate the problem! Canât wait to see your future work as well
nice explanation :)
I think the question is:
How can I do as much good as possible with C units of cost?
This corresponds to the problem of maximising E(U(C)), where U(c) is the utility achieved (via a certain intervention) for the cost c (which must not exceed C). If the budget C is small enough (thinking at the margin):
U(C) = Uâ(0)*C, where Uâ(c) is the derivate of U with respect to cost.
Assuming Uâ(0) and C are independent, mean(âeffectâ/ââcostâ) equals mean(âeffectâ)/âmean(âcostâ):
mean(âeffectâ/ââcostâ) = E(U(C)/âC) = E(Uâ(0)*C/âC) = E(Uâ(0)).
mean(âeffectâ)/âmean(âcostâ) = E(U(C))/âE(C) = E(Uâ(0))*E(C)/âE(C) (assuming independence between Uâ(0) and C) = E(Uâ(0)).
So, it seems that, regardless of the metric we choose, we should maximise E(Uâ(0)), i.e. the expected marginal cost-effectiveness. However, Uâ(0) and C will not be independent for large C, so I think it is better to maximise mean(âeffectâ/ââcostâ).
Is that a typo on your final bullet point? Should say mean(âeffectâ)/âmean(âcostâ).
Hi Stan,
Yes, it was a typo, which I have now corrected. Thanks for catching it! I have also added one extra sentence at the end:
Thanks for commenting here, and thanks again for your initial feedback!
I donât really have anything planned in this area, what would you be excited to see?
I think Michael Dickens explains this better than me in a more recent comment. The point is that we usually care about effect/âcost rather than the other way around (although Iâd love to understand more clearly why and when exactly thatâs true). In your example, you have implicitly computed this and then compare it to Guesstimateâs model computing cost/âeffect.
Replied to Michael Dickens, curious about your thoughts! We should definitely add a section on this
I think your example should focus on the mean of the ratio between the effect and cost, not on the mean of the ratio between the cost and effect. The latter is a bad metric because:
A very small âcostâ/ââeffectâ could correspond to interventions that are either quite bad or good (since âcostâ/ââ = 0). This means small numerical errors could lead to large differences in mean(âcostâ/ââeffectâ), which is bad.
When changing from negative values of âcostâ/ââeffectâ to positive ones, the goodness of the intervention increases (changing from harmful to beneficial). However, for negative and positive values, a higher âcostâ/ââeffectâ corresponds to a worse intervention.
The metric âeffectâ/ââcostâ has good properties:
A higher value always implies a better intervention (at least in theory).
A null value correspond to neutrality.