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September 11

Expected Value and Magnitude of Utility

I've noticed an interesting pattern when changing the order of magnitude in an expected value problem. I'll use the most recent example I've stumbled across. Let probabilities Π={0.41, 0.49,0.57,0.65,0.72}, negative utilities Φ={29,30,31,32,33}, and positive utilities Ψ={165,215,275,335,415}. In this case and in all cases, all variables are positively correlated. Now, let Φ'=100Φ and let Ψ'=0.01Ψ. Finally, let Ζ=ΦΠ+Ψ(1-Π), Α=Φ'Π+Ψ(1-Π), Β=ΦΠ+Ψ'(1-Π), and Γ=Φ'Π+Ψ'(1-Π), where Ζ, Α, Β, and Γ are all expected value equations. It is surprising to me that Ζ is a parabolic function, where MAX(Ζ)=φ₃π₃+ψ₃(1-π₃), but Α, Β, and Γ are linear, where the function is maximized as the with the fifth element of their respective vectors.

Why is it that changing the order of magnitude of the utility vectors changes a quadratic function into a linear function? I'm not a mathematician, so please ask if I'm not making sense with my notation or if I'm not using the correct vocabulary. Thank you Wikipedians! 47.187.81.103 (talk) 16:15, 11 September 2020 (UTC)[reply]

You may have made a calculation error. I compute:

ζ 3 = φ 3 π 3 +ψ 3 (1-π 3) = 31\times0.57 + 275\times0.43 = 17.67 + 156.75 = 135.92

,

ζ 5 = φ 5 π 5 +ψ 5 (1-π 5) = 33\times0.72 + 415\times0.28 = 23.76 + 298.80 = 139.96

.

But perhaps the error is in the values

π i

,

φ i

and

ψ i

you gave above. When plotted against

i

, all give an almost straight line – for

φ i

even perfectly straight –, which means that all can be approximated very well by a linear function. Now the product of two linear functions is a quadratic function, and a sum of quadratic functions is again quadratic, so

ζ i

is very close to being quadratic (giving a parabola when plotted), but so are

α i

,

β i

and

γ i

. However, the parabola for

ζ

is different: it is like a hill, having a maximum somewhere; the other three are valleys with a low point. For a valley, the maximal value is always at one of the extremes of the range of

π

; for a hill, it can be anywhere. When the leading coefficient

a

in the quadratic polynomial

ai 2 + bi + c

is positive, its graph is a valley; when it is negative, we get a hill. The reason that

ζ

is the odd man out here, is that for this case the trend of the positive utilities is dominant, about a factor 60 larger than the trend of the negatives. --Lambiam 21:14, 11 September 2020 (UTC)[reply]

Oh! Well of course the negative utilities should be negative... Φ={-29,-30,-31,-32,-33}. I understand that the quadratic term in a quadratic function changes the shape of the parabola, but what I do not understand is why the parabolas for the four functions are different. Sorry to be obtuse. 47.187.81.103 (talk) 16:08, 12 September 2020 (UTC)[reply]

I see. I thought you called the

Φ

utilities "negative" because they are the utilities that kick in in case of the "negative" – that is, less favorable – outcome of a random variable with a Bernoulli distribution – basically a toss of a biased coin, The sign of the

Φ

values does not change the fact that all four graphs are (almost) parabolas, but now they all have become hills, with a peak. Instead of plotting

i

on the x-axis, we may as well use the value of

π

, which I interpret as the single parameter of the Bernoulli distribution of which it is also the mean. Interpreted as a probability, it should of course satisfy

0 \leq π \leq 1

, which gives only a short segment of the parabola, but the function for the parabola allows us to plot the graph for values outside that range. All parabolas have the same shape in the same sense that all ciircles have the same shape: each can be changed into any other by a similarity transformation (for which only translation and scaling are needed). As the absolute magnitude of the ratio between the trends of the negative and positive utilities increases, the peak becomes more narrow, but it also shifts farther and farther to the left, with the values of

π

for which the maximum is obtained deep into the negative part of the x-axis. The farther away you are from the vertex of the parabola, the less its curvature, so the small segment between

0.41

and

0.72

may seem linear, in the same way the Earth seems locally flat. --Lambiam 19:46, 12 September 2020 (UTC)[reply]