PS 3
In general a lottery may have s possible independent outcomes, each outcome occurs with some probability πs . To
each outcome corresponds in general a dierent payo. Furthermore, the expected utility depends on my initial
wealth W (because the slope of the utility function is dierent for diernt levels of wealth) and the cost of the
ticket.
EU (W +L) = π1 u (W − ticket cost + payof f1 )+π2 u (W − ticket cost + payof f2 )+...+πs u (W − ticket cost + payof fs )
Usually you will have to deal with only two outcomes (I win or I lose), so the problem restricts to
EU (W + L) = πu (W − ticket cost + payof f1 ) + (1 − π) u (W − ticket cost + payof f2 )
The expected utility in general is dierent from the expected value, which is the average payo I get from participating to the lottery. Therefore I have
EV (W + L) = π (W − ticket cost + payof f1 ) + (1 − π) (W − ticket cost + payof f2 )
EV (L) = π (−ticket cost + payof f1 ) + (1 − π) (−ticket cost + payof f2 )
Note that I can dene the expected value of the lottery EV (L), while it doesn't make any sense to dene EU (L)
because the expected utility I get from the lottery is conditional to my initial level of wealth (make sure you
understand this point)!
Problem 1
1. I need to confront EU (accept) = EU (W + L) with EU (not accept) = U (W ), π = 21 .
EU (accept) = 0.5ln (14, 000 + 400) + 0.5ln (14, 000 − 6256)
EU (not accept) = ln (14, 000)
2. ... π = 0.9
EU (accept) = 0.9ln (14, 000 + 400) + 0.1ln (14, 000 − 6256)
EU (not accept) = ln (14, 000)
3. Professor denes in his notes the certainty equivalent of the lottery CE (L) as:
EU (W + L) = u (W − CE (L))
Therefore we have
EU (accept) = ln (14, 000 + CE (L))
taking the exponential
eEU (accept) = 14, 000 + CE (L)
CE (L) = eEU (accept) − 14, 000
You can also think (for me is more intuitive) in terms of CE (W + L) dened as
EU (W + L) = u (CE (W + L))
we have
EU (accept) = ln (CE (W + L))
CE (W + L) = eEU (accept)
You can think of CE (W ) as of W itself. Therefore, you can heuristically say that
W + CE (L) = eEU (accept)
CE (L) = eEU (accept) − 14, 000
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Problem 2
1. We follow the hint and we have simply to solve the following system
(
(
u (x0 ) − bx0 = a
u (x1 ) = a + bx1
(
u (x0 ) − bx0 = a
0)
b = u(xx11)−u(x
−x0
u (x0 ) = a + bx0
u (x1 ) = a + bx1
(
u (x0 ) − bx0 = a
⇒
u (x1 ) = u (x0 ) − bx0 + bx1
(
0)
a∗ = u (x0 ) − u(xx11)−u(x
x0
−x0
⇒
u(x
)−u(x
)
1
0
b∗ = x1 −x0
Then we follow the hint once again and we get
u (x0 ) −
u (x1 ) − u (x0 )
x1 − x0
x0 +
u (x1 ) − u (x0 )
x1 − x0
EV (L)
but EV (L) = πx1 + (1 − π) x0 , hence
u (x0 ) −
u (x1 ) − u (x0 )
u (x1 ) − u (x0 )
u (x1 ) − u (x0 )
x0 +
πx1 +
x0 −
πx0
x1 − x0
x1 − x0
x1 − x0
u (x1 ) − u (x0 )
π (x1 − x0 ) = u (x0 ) + π (u (x1 ) − u (x0 ))
u (x0 ) +
x1 − x0
u (x1 ) − u (x0 )
x1 − x0
= πu (x1 ) + (1 − π) u (x0 ) ≡ EU (L)
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