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Additional resources for 40 Puzzles and Problems in Probability and Mathematical Statistics (Problem Books in Mathematics)

Example text

On average the area of her rectangle equals AP aula = E[U1 · U2 ] = E[U1 ] · E[U2 ] = E2 [U] because U1 and U2 are independent and have the same distribution as U. Therefore, AP eter − AP aula = E[U2 ] − E2 [U] = Var[U] ≥ 0 that is, AP eter ≥ AP aula . On average, the area of Peter’s square is larger than that of Paula’s rectangle, even though all lengths and widths of all rectangles (a square is a rectangle) are generated by realizations of the same generic rv, U. 9 Maximize Your Gain a. As stated in the Hints, in order to win a large amount, one would like to choose a rather large value of c.

2n − 1, 2n). Of these n pairs we may select k to associate exactly one female player with each of them; the number of ways to select these k pairs from a total of n pairs is nk . However, within each of the k pairs chosen, the respective female may be placed either at the odd or at the even rank of that pair, yielding a total of 2k · nk different ways to distribute k female players among 2n ranks such that each of the n pairs has at most one female. , p(k, n) = 2k · 2n k n k = 2n−k n 2n n · 2k Note that the formula also covers the trivial cases of k = 0, 1.

Let us call the chosen value of c the “strategy” of the player and denote its expected gain as G(c). Thus, with the strategy c, the gain will be zero with probability P(U ≤ c) = F (c), and it will be equal to c with probability P(U > c) = 1 − F (c). Putting these two cases together, we get G(c) = 0 · F (c) + c · [1 − F (c)] = c · [1 − F (c)] 20 Amount won, G(c) 15 10 5 0 0 50 100 150 200 Number chosen, c Fig. 5. The relation between the expected gain, G(c), and the number, c, that was chosen. The example assumes that U has an exponential distribution f (t) = λ exp(−λ t) with mean 1/λ = 50.

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