Solutions to assignment 1
1.
Harvey loves Sabrina. Sabrina loves Fred. Jennifer loves Fred. Fred loves himself.
xLy means x loves y
S = {Harvey, Sabrina, Fred, Jennifer}
1.1
Based on this information define the relation L on the set S as a set of ordered
pairs.
L = {(h,s), (s,f), (j,f), (f,f)}
1.2
Is L symmetric? Why?
A relation R defined on a set S is symmetric iff (x,yS)(if xRy then yRx).
The relation L is not symmetric because hLs but not sLh (there are more examples that
show that L is not symmetric)
1.3
Is L reflexive? Why?
A relation R defined on a set S is reflexive iff (xS) xRx
L is not reflexive because (h,h)L, i.e. hLh is not true.
1.4
Is L transitive? Why?
A relation R defined on a set S is transitive iff (x,y,zS)(if xRy and yRz, then xRz).
L is not transitive because hLs and sLf, but not hLf.
1.5
Is L total? Why?
A relation R defined on a set S is total iff (x,yS)(at least one of the following is true:
xRy or yRx).
L is not total since neither hLf nor fLh are true
1.6
If L is not reflexive or symmetric, which other ordered pairs must be included in
L to make it reflexive and symmetric?
Reflexive: (h,h), (s,s), (j,j).
Symmetric: (s,h), (f,s), (f,j)
2.
An agent assigns the following utilities to the four options: u(o1) = 2, u(o2) = 3, u(o3) = 1,
u(o4) = 2
2.1 Define the agent’s strict preference relation and the indifference relation on the set
O = (o1, o2, o3, o4} as a set of ordered pairs based on the given utility function u.
2.2 Provide a different score function which generates the same preference relation
 = {(o1, o3), (o2, o1), (o2, o3), (o2, o4), (o4, o3)}
u(o1) = 20, u(o2) = 30, u(o3) = 10, u(o4) = 20
3.
If a weak preference relation  is rational, the corresponding strict preference relation 
is transitive. Prove.
Proof 1
Suppose that  is not transitive. Then there are three options a, b, c such that
u(a)<u(b)<u(c) and u(a)u(c). It is not possible to find numbers for which this holds
(since ‘>’ is transitive). Therefore there is no utility function that assigns numbers to the
given options. According to Theorem 2 from our first class, this means that the
corresponding weak preference relation  is not rational. But  is rational by assumption.
Therefore  is transitive (since the opposite assumption leads to a contradiction). QED
Proof 2
According to Theorem 2 from our first class, since  is rational, there is a utility function
that assigns a number to each option from the set of options on which  is defined. For
the corresponding strict preference relation it holds that for any options x and y, x  y iff
u(x)>u(y). Thus ‘’ defined on the set of options has the same properties as the relation
‘>’ defined on the set of values of the utility function. Since ‘>’ is transitive, ‘’ must
also be transitive. QED
4.
Suppose that this matrix describes the choices you are facing. Like everybody, you prefer
to pay less, to have fewer roommates, and be close to U of A.
Dimensions
Options
Strathcona
Apt.
McKernan
Apt.
Northside Apt.
rent
roommates
distance to U of
A
$400
2
5 min
$400
3
10 min
$500
4
20 min
4.1 Define your strict preference relation d and your weak preference relation d for
some dimension d (your choice, specify which dimension you choose).
roommates: SA r McA r NA
4.2 Do any options strongly Pareto-dominate any other options? If so, which
dominate which? If not, explain why.
An option strongly Pareto-dominates another option if it is strongly preferred for any
dimension.
SA. strongly Pareto-dominates NA and McA strongly Pareto-dominates NA
4.3 Do any options weakly Pareto-dominate any other options? If so, which dominate
which? If not, explain why.
An option weakly Pareto-dominates another option if it is weakly preferred for any
dimension and strongly preferred for at least one dimension..
SA. weakly Pareto-dominates NA and McA weakly Pareto-dominates NA
SA weakly Pareto-dominates McA