Question

An individual has a nice project of a new product, but needs money for that.
She can only produce if she raises the amount of money required (noted "K")
She has the possibility of raising K if she participates on a given program (let's say, a government program).
The probability of raising K through this money is p.
However, to participate on this program, she needs to incur some costs c < K.

The problem is that there is a probability (1 - p) that she does not raise the money, but only incurs c.

Summing up:

positive payoff = (K - c) with probability p
negative payoff = (-c) with probability (1-p)


In order to try to reduce the costs c, she can ask for some professional help.
The costs for this professional help are sunk (for example, it can be a government agency that provides advice for free to inventors and entrepreneurs). With the help, the costs become c.a with 0 < a < 1.

Therefore:

positive payoff = (K - ca) with probability p
negative payoff = (-ca) with probability (1-p)

I want to calculate:

. What is the minimum p in the first case for the individual to participate on the program?
. What is the minimum p in the second case for the individual to participate on the program?
. What would be the correct notation for EU = p(K-c) + (1-p)c?
. What is the variable I should set as the one I am interested in, i.e., the variable which she will maximize for? For me it was "c", but when I change the setting, the result is the same and it makes no sense!

Answer 1

Expected utility is the sum of the utility of every possible outcome each multiplied by the probability of its occurrence.

Question# a

The minimum of p should be always greater than or equal to the expected utility

Expected utility for the first case E (u) = (k-c) p+ (p-1) c

 P≥ (k-c) p+ (p-1) c

Question #b

Expected utility in the second case E (u) = (k-ca) p+ (p-1) ca

p≥ (k-ca) p+ (p-1) ca

Question# c

First Case;

Negative payoff= (k-c) with probability p

Positive payoff= (-c) with probability (1-p)

Multiplying and summing:

E (u) = p (k-c) + (p-1) c

Second case;

Positive payoff= (k-ca) with probability of p

Negative payoff= (-ca) with probability (1-p)

E (u) = (k-ca) p+ (p-1) ca

Question# d

Expected profit =Revenue-cost

In our case is k-c in the first scenario and (k-ca) in the second scenario

One should always maximize the revenue. The value that should be maximized is k.

One should always maximize the revenue. The value that should be maximized is k.