By P. A. Martin

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**Additional resources for Modeles en Mecanique Statistique des Processus Irreversibles**

**Example text**

The sample space is C = {(i, j) : 1 ≤ i, j ≤ 6}. Because the dice are fair, P [{(i, j)}] = 1/36. The random variable X is X(i, j) = i + j. The space of X is D = {2, . . , 12}. By enumeration, the pmf of X is given by Range value x 2 3 4 5 6 7 8 9 10 11 12 Probability pX (x) 1 36 2 36 3 36 4 36 5 36 6 36 5 36 4 36 3 36 2 36 1 36 To illustrate the computation of probabilities concerning X, suppose B1 = {x : x = 7, 11} and B2 = {x : x = 2, 3, 12}. Then, using the values of pX (x) given in the table, PX (B1 ) = pX (x) = 2 8 6 + = 36 36 36 pX (x) = 2 1 4 1 + + = .

If x ∈ S, then pX (x) is equal to the size of the discontinuity of FX at x. If x ∈ S then P [X = x] = 0 and, hence, FX is continuous at this x. 2. A lot, consisting of 100 fuses, is inspected by the following procedure. Five of these fuses are chosen at random and tested; if all ﬁve “blow” at the correct amperage, the lot is accepted. 31931. More generally, let the random variable X be the number of defective fuses among the ﬁve that are inspected. The pmf of X is given by pX (x) = 80 (20 x )(5−x) 100 (5) 0 for x = 0, 1, 2, 3, 4, 5 elsewhere.

100, zero elsewhere. (b) Compute P (X ≤ 50). 40 Probability and Distributions (c) Show that the cdf of X is F (x) = [x]([x] + 1)/10100, for 1 ≤ x ≤ 100, where [x] is the greatest integer in x. 10. 1. 11. Let X be a random variable with space D. For D ⊂ D, recall that the probability induced by X is PX (D) = P [{c : X(c) ∈ D}]. Show that PX (D) is a probability by showing the following: (a) PX (D) = 1. (b) PX (D) ≥ 0. (c) For a sequence of sets {Dn } in D, show that {c : X(c) ∈ ∪n Dn } = ∪n {c : X(c) ∈ Dn }.