Answer a): The intuition is exactly the same here. The higher the 𝑎, the higher is the
probability that the signal matches the true value, thereby making it more “precise”.
Similarly, the lower the 𝑎, the lower is the signal a good depiction of the true value.
Answer b): Pr(𝑠 = 6) = Pr(𝑠 = 6|𝑣 = 6) Pr(𝑣 = 6) + Pr(𝑠 = 6|𝑣 =
2) Pr(𝑣 = 2) =
2−𝑎
3
and Pr(𝑠 = 2) = Pr(𝑠 = 2|𝑣 = 6) Pr(𝑣 = 6) +
Pr(𝑠 = 2|𝑣 = 2) Pr(𝑣 = 2) =
1+𝑎
3
1
. Comparing the two, 𝑎 < 2 it is more likely that
the consumer receives a high signal of 6, i.e. there should be sufficient noise. Notice
1
2
that as 𝑎 → 1, Pr(𝑠 = 6) = 3 and Pr(𝑠 = 2) = 3. Therefore, the more precise the
signal, the closer it is to the distribution of the true values, i.e. if 𝑎 = 1, the consumer
can be sure that whatever signal she receives will also be her true value.
Answer c): As discussed in the lecture, it might not be possible for the seller to
provide no information (or provide only noise) about the product through
advertisement. For example, even though the advert might not talk about the features
and characteristics of the product, it might show the way it looks. In this way,
consumer will still learn something about her match value from viewing the advert.
This assumption merely says that it is not possible to hide all information about the
product with probability 1.
Answer d): As an illustration, Pr(𝑣 = 6|𝑠 = 6; 𝑎) =
𝑎
2−𝑎
1−𝑎
1+𝑎
Pr(𝑠
= 6|𝑣 = 6; 𝑎 ) Pr(𝑣=6)
Pr(𝑠=6)
=
2𝑎
. Similarly, Pr(𝑣 = 2|𝑠 = 2; 𝑎) = 1+𝑎 , Pr(𝑣 = 6|𝑠 = 2; 𝑎) =
, and Pr(𝑣 = 2|𝑠 = 6; 𝑎) =
2−2𝑎
2−𝑎
. To find out the lower bound, consider first if the
buyer receives a signal of 6. We want to find an 𝑎 such that Pr(𝑣 = 6|𝑠 = 6; 𝑎) ≥
2
Pr(𝑣 = 2|𝑠 = 6; 𝑎). This gives 𝑎 ≥ 3. Similarly, if the buyer receives a signal of 2,
1
then for her to think it is more likely that her valuation is also 2, 𝑎 ≥ 3. The two
lower bounds are different because given that the prior favors a value of 2, it is easier
to convince the buyer that her valuation is also 2 even by choosing a lower value of 𝑎.
However, the signal should be sufficiently accurate for the buyer to think that her
valuation is 6 even when she receives a signal of 6.
Answer e): 𝑝𝑠=6 = 6 Pr(𝑣 = 6|𝑠 = 6; 𝑎) + 2 Pr(𝑣 = 2|𝑠 = 6; 𝑎) =
𝑝𝑠=2 = 6 Pr(𝑣 = 6|𝑠 = 2; 𝑎) + 2 Pr(𝑣 = 2|𝑠 = 2; 𝑎) =
4+2𝑎
2−𝑎
6 − 2𝑎
1+𝑎
1
Answer f): Since 𝑎 > 2, we have that 𝑝𝑠=6 >𝑝𝑠=2 .
1, if 0 ≤ 𝑝 ≤ 𝑝𝑠=2
2−𝑎
𝑄 = {Pr(𝑠 = 6) =
, if 𝑝𝑠=2 < 𝑝 ≤ 𝑝𝑠=6
3
0, 𝑖𝑓 𝑝 > 𝑝𝑠=6
Note that if the price is lower than the lower WTP (𝑝𝑠=2), then the consumer will buy
independent of the signal she receives. However, if the price is higher than the lower
WTP, she would only buy it if she receives a signal of 6. Hence the demand function.
Yes, the demand function traces the usual negative relationship between price and
quantity.
Answer g): The only two relevant prices to look at are 𝑝𝑠=2 and 𝑝𝑠=6 .
6 − 2𝑎
if 𝑝 = 𝑝𝑠=2
1+𝑎
Π={
2 − 𝑎 4 + 2𝑎
.
if 𝑝 = 𝑝𝑠=6
3
2−𝑎
Answer h): The monopolist charges the higher price if and only if
2−𝑎 4+2𝑎
3
.
2−𝑎
≥
6−2𝑎
1+𝑎
.
Cancelling 2 − 𝑎 from the numerator and the denominator in the LHS and solving for
the threshold level we get 𝑎̅ = 1. For 𝑎 > 1, profits are higher from charging a higher
price. The reverse is true when 𝑎 < 1. Therefore, given the limits on the accuracy
level, the firm finds it profitable to charge a lower price and sell the product with
probability 1.
Answer i): The function that you should therefore plot is given by Π(a) =
6−2𝑎
1+𝑎
. This
is a decreasing hyperbola in the range of 𝑎 we are considering. Yes, the function is
𝟏
convex and the profit maximizing level of accuracy is 1min 𝑎 ≈ 𝟐.
2
<𝑎≤1
Answer j): Note first that given the limits on the range of accuracy level it is more
likely that consumer receives a lower signal (i.e. a signal of 2). Moreover, the
accuracy level would have to be sufficiently high for the consumer to believe that her
signal of 6 is trustworthy. Thus the monopolist prefers to set to the lowest possible 𝑎,
sell the product with probability 1 and charge a lower price, rather than setting a
higher 𝑎 and hoping for the buyer to receive a higher signal.