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There are 3 possible payoffs - \$4, \$9 and \$36. The utility function for these payoffs is $\sqrt x $.

I have to find all the lotteries preferable over getting $9 with probability 1 in a probability simplex. The answer is the shaded area in the probability simplex below.

Probability simplex

I am not able to find the 3D equation of line required for it. But I have approached the problem as follows:

Any point on line BC gives expected utility of $ 6p_{3} + 3(1-p_{3}) = 3p_{3} + 3 > 3 $, so all points on BC are to be included.

All points on AB are eliminated as expected utility on them is less than 3.

Only those points on AC are included which give the expected utility greater than 3 i.e.

$$ 2p_{1} + 6(1-p_{1}) = 6 - 4p_{1} >3 \implies p_{1} < \dfrac{3}{4} $$

From here how do we arrive at point D ?

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3 Answers 3

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In your last inequality, point $A$ is where $p_1=1$ and point $C$ is where $p_1=0$. So the condition $p_1<\frac34$ is captured by the line segment $\overline{CD}$.

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The question asks how to find the point D.

For the strategy below we find the vector AC, scale it down to AD then set the vector's tail at A so that the tip is at D. Finally, adding the vector to A gives us point D.

The vector from point A to point C is found as point C minus point A so is (-1,0,1). You want to travel a quarter of the way down this vector to get to point D .

Thus point D is at the tip of the vector (-1/4, 0, 1/4) when the vector starts at A.

so D = (1, 0, 0) + (-1/4, 0, 1/4) = the point at (3/4,0, 1/4)

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You can find point D on line AC with the following explanation -

Key here is here is to find a point where expected utility from a lottery on line AC equals the utility derived from $9 . Proceeding with this idea , equate 2p1 + (1-p1).6 = 3 and you will find p1= 3/4 (point D).

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