Suppose an individual knows that the prices of a particular color TV have a uniform distribution between \(\$ 300\) and \(\$ 400\). The individual sets out to obtain price quotes by phone. a. Calculate the expected minimum price paid if this individual calls \(n\) stores for price quotes. b. Show that the expected price paid declines with \(n\), but at a diminishing rate. c. Suppose phone calls cost \(\$ 2\) in terms of time and effort. How many calls should this individual make in order to maximize his or her gain from search?

For a uniform distribution between \(a\) (lower limit) and \(b\) (upper limit), the probability density function (pdf) is given by: \(f(x) = \frac{1}{(b-a)}\) for \(a \leq x \leq b\) Here, lower limit (a) = \(300\) and upper limit (b) = \(400\) The expected minimum price, when calling 'n' stores, will be the minimum of 'n' random variables with the same uniform distribution. We will denote this as \(E(M_n)\), where \(M_n\) is the minimum value of the 'n' uniform random variables. To find \(E(M_n)\), we will first calculate the cumulative distribution function (CDF) of \(M_n\) and then differentiate it to get the pdf of \(M_n\). Finally, we'll calculate the expected value.

The CDF of \(M_n\), denoted as \(F_{M_n}(x)\), represents the probability that the minimum price is less than or equal to x. In our case, x falls between \(300\) and \(400\), and the individual calls 'n' stores. \(F_{M_n}(x) = P(M_n \leq x) = 1 - P(M_n > x)\) Since \(M_n\) is the minimum value of the 'n\( independent uniform random variables, the probability that \)M_n > x$ is equal to the product of the probabilities that all 'n' random variables are greater than 'x': \(P(M_n > x) = P(X_1 > x) * P(X_2 > x) * ... * P(X_n > x)\) For the uniform distribution mentioned previously, we know that: \(P(X_i > x) = \frac{b-x}{b-a}\) for \(a \leq x \leq b\) Therefore, \(P(M_n > x) = \left(\frac{b-x}{b-a}\right)^n\) And thus, \(F_{M_n}(x) = 1 - \left(\frac{b-x}{b-a}\right)^n\)

To find the pdf of \(M_n\), denoted as \(f_{M_n}(x)\), we need to differentiate the CDF \(F_{M_n}(x)\) with respect to \(x\): \(f_{M_n}(x) = \frac{d}{dx} F_{M_n}(x) = \frac{d}{dx} \left[ 1 - \left(\frac{b-x}{b-a}\right)^n \right]\) \(f_{M_n}(x) = n\left(\frac{1}{b-a}\right)\left(\frac{b-x}{b-a}\right)^{n-1}\)

Finally, we calculate the expected minimum price paid after calling 'n' stores, \(E(M_n)\): \(E(M_n) = \int_{a}^{b} x f_{M_n}(x) dx\) \(E(M_n) = \int_{300}^{400} x \cdot n\left(\frac{1}{b-a}\right)\left(\frac{b-x}{b-a}\right)^{n-1} dx\) After solving this integral, we get: \(E(M_n) = \frac{300n + 400}{n+1}\)

To show that the expected price declines with 'n' at a diminishing rate, we can find the first and second derivatives of \(E(M_n)\) with respect to 'n'. First derivative: \(\frac{dE(M_n)}{dn} = - \frac{100}{(n+1)^2} < 0\) Since the first derivative is negative, it indicates that \(E(M_n)\) is decreasing with respect to 'n'. Second derivative: \(\frac{d^2E(M_n)}{dn^2} = \frac{200}{(n+1)^3} > 0\) Since the second derivative is positive, it indicates that the rate of decrease of \(E(M_n)\) is diminishing with respect to 'n'.

Let's denote the gain from search as 'G' and the cost per call as 'c'. The gain from search depends on the expected minimum price paid and the cost of calling 'n' stores: \(G = E(M_n) - cn\) In this situation, the cost per call (c) is $2. We can use this formula to determine the optimal number of calls by finding the maximum value of 'G' with a relation to 'n'. This can be done by setting the first derivative of 'G' with respect to 'n' to 0: \(\frac{dG}{dn} = - \frac{100}{(n+1)^2} - 2 = 0\) Solving for 'n', we get: \(n \approx 4.6\) Since the number of calls must be a whole number, and the function is concave, we can conclude that the individual should call 5 stores to maximize his or her gain from search.