In particular if x is a continuous type random

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Unformatted text preview: nition, the limit of the sum of the first n terms as n → ∞. Therefore, n P {X < c} = lim n→∞ P (Gk ) = lim P {G1 ∪ G2 ∪ · · · ∪ Gn } = lim P {X ≤ cn } = lim FX (cn ). k=1 n→∞ n→∞ n→∞ That is, P {X < c} = FX (c−). The second conclusion of the proposition follows directly from the first: P {X = c} = P {X ≤ c} − P {X < c} = FX (c) − FX (c−) = FX (c). Example 3.1.3 Let X have the CDF shown in Figure 3.2. (a) Determine all values of u such that P {X = u} > 0. (b) Find P {X ≤ 0}. (c) Find P {X < 0}. 1 0.5 0 −1 Figure 3.2: An example of a CDF. Solution: (a) The CDF has only one jump, namely a jump of size 0.5 at u = 0. Thus, P {X = 0} = 0.5, and there are no values of u with u = 0 such that P {X = u} > 0. (b) P {X ≤ 0} = FX (0) = 1. (c) P {X < 0} = FX (−0) = 0.5. Example 3.1.4 Let X have the CDF shown in Figure 3.3. Find the numerical values of the following quantities: (a) P {X ≤ 1}, (b) P {X ≤ 10}, (c) P {X ≥ 10}, (d) P{X=10}, (e) P {|X − 5| ≤ 0.1}....
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