There is an example of using the probability described in Morgan and Manning (1983). Authors claim that it is possible to solve with a pocket calculator, however, I managed to replicate only one result. Apparently, I am doing something wrong but I cannot figure out what. Here is a problem:
- Construction firms price discriminate by offering two prices: 50,000 pounds with a probability of 0.2 and 60,000 pounds with a probability of 0.8.
- Present value of a complete construction is 80,000.
- Discount factor is 1.01 per period
- A customer can choose how to search - asking for quotes sequentially or simultaneously.
- Each quote costs 1,000 pounds.
- If a customer prefers to ask all four firms to send her quotes in the first period, then the value of such a search will be:
$$V = -4,000 + \frac{1}{1.01}\mathbb{E}\left(\frac{1}{1.01}80,000 - p_4^{min}\right)$$
The result in the paper is $$V = 20,863.29$$.
- If a customer prefers to ask sequentially, then in the first period she will pay 1,000 for one quote. She will accept 50,000 but will reject $60,000$ and will pay another 1,000 to get the next quote. This type of search yields the value:
$$V = -1,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{\frac{1}{1.01}80,000 - p_1^{min},-1,000+\mathbb{E}\left(\frac{1}{1.01}80,000 - p_2^{min}|p_1^{min}\right)\right\}\right)$$
The result in the paper is $$V = 20,623.70$$
- Or a customer prefers to ask sequentially different amounts of offers. In the first period, she asks for three quotes, if all free quotes will return without a price of 50,000 pounds, then she will request another four quotes. The value of this type of search is:
$$V = -3,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{\frac{1}{1.01}80,000 - p_3^{min},-4,000+\frac{1}{1.01}\mathbb{E}\left(\frac{1}{1.01}80,000 - p_7^{min}|p_3^{min}\right)\right\}\right)$$
The result in the paper is $$V = 21,688.58$$.
My solutions:
Case 1 (managed to get the same result as in the paper): $$V = -4,000 + \frac{1}{1.01}\mathbb{E}\left(\frac{1}{1.01}80,000 - p_4^{min}\right)$$
First of all, I need to find the probability to get the cheapest price. By the complement method, such probability will be calculated as follows:
The probability of none of the four firms offering 50,000 is $$0.8^4 = 0.4096$$
Therefore, the probability at least one firm offers 50,000 is $$(1 - 0.4096) = 0.5904$$
Therefore, the expected value of $$p_4^{min} = 60,000 * 0.4096 + 50,000 * 0.5904$$ $$p_4^{min} = 54,096$$
Now we have all information to solve Case 1: $$V = -4,000 + \frac{1}{1.01}\mathbb{E}\left(79,207.9208 - 54,096\right)$$ $$V = -4,000 + \frac{1}{1.01}25,111.9208$$ $$V = -4,000 + 24,863.2879$$ $$V = 20,863.2879$$
Case 2 (my result does not match the paper):
$$V = -1,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{\frac{1}{1.01}80,000 - p_1^{min},-1,000+\frac{1}{1.01}\mathbb{E}\left(\frac{1}{1.01}80,000 - p_2^{min}|p_1^{min}\right)\right\}\right)$$
To solve this case, I need to identify a probability to obtain the cheapest price in two rounds:
The probability of a firm doesn't offer a price of $50,000$ in the first round is 0.8.
The probability that after 2 rounds the consumer didn't receive a price of 50,000 is $$0.8 * 0.8 = 0.64$$
The probability that after 2 rounds the consumer saw at least one price of 50,000 is $$1 - 0.64 = 0.36$$
Therefore, the expected values of $$p_1^{min} = 60,000 * 0.8 + 50,000 * 0.2$$ or $$p_1^{min} = 58,000$$ and $$p_2^{min} = 60,000 * 0.64 + 50,000 * 0.36$$ or $$p_2^{min} = 56,400$$
Now we have all information to solve Case 2:
$$V = -1,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{79,207.9208 - 58,000,-1,000+\frac{1}{1.01}\mathbb{E}\left(79,207.9208 - 56,400\right)\right\}\right)$$
$$V = -1,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{79,207.9208 - 58,000,-1,000+\frac{1}{1.01}\mathbb{E}\left(79,207.9208 - 56,400\right)\right\}\right)$$
$$V = -1,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{21,207.9208, 21,582.0998\right\}\right)$$
$$V = -1,000 + 21,368.4156$$
$$V = 20,368.4156$$
Case 3 (My result doesn't match the paper):
$$V = -3,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{\frac{1}{1.01}80,000 - p_3^{min},-4,000+\mathbb{E}\left(\frac{1}{1.01}80,000 - p_7^{min}|p_3^{min}\right)\right\}\right)$$
Firstly, the probability of none of the three firms offering 50,000 in the first round is $$0.8^3 = 0.512$$
The probability of one firm offering 50,000 in the first round is $$1 - 0.512 = 0.488$$
The probability of none firms offering 50,000 in both rounds is $$0.8^7 = 0.2097$$
The probability of one firm offering 50,000 in 2 rounds is $$1 - 0.2097 = 0.7903$$
Therefore, $$p_3^{min} = 60,000 * 0.512 + 50,000 * 0.488$$ or $$p_3^{min} = 55,120$$ and $$p_7^{min} = 60,000 * 0.2097 + 50,000 * 0.7903$$ or $$p_7^{min} = 52,097$$
Plugging it into the given equation:
$$V = -3,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{\frac{1}{1.01}80,000 - 55,120,-4,000+\frac{1}{1.01}\mathbb{E}\left(\frac{1}{1.01}80,000 - 52,097\right)\right\}\right)$$
$$V = -3,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{24,087.9208,-4,000+\frac{1}{1.01}\mathbb{E}\left(79,207.9208 - 52,097\right)\right\}\right)$$
$$V = -3,000 + \frac{1}{1.01}\mathbb{E}\left(max\left\{24,087.9208, 22,842.4958\right\}\right)$$
$$V = -3,000 + 23,849.4265$$
$$V = 20,849.4265$$