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H OMEWORK 1.3 I) Show that the following properties of a Cumulative Distribution Function (CDF) follow from the definition of the CDF and the properties of probability: (1) (2) (3) (4) (5) F (−∞) = 0 and F (+∞) = 1. F is a non-decreasing function, i.e. if x1 ≤ x2 , then F (x1 ) ≤ F (x2 ). If F (x0 ) = 0, then F (x) = 0 for every x ≤ x0 . P {X > x} = 1 − F (x). P ({x1 < X ≤ x2 }) = F (x2 ) − F (x1 ). Proof. Property 1: By definition, FX (a) = P (A) where A = {X ≤ a}. Then F (= +∞) = P (A), with A = {X ≤ +∞}. This event is the certain event, since A = {X ≤ +∞} = {X ∈ R} = S. Therefore F (= +∞) = P (S) = 1. Similarly, Then F (= −∞) = P (A), with A = {X ≤ −∞}. This event is the impossible event, RV X is a real variable, so F (= −∞) = P (∅) = 0. Proof. Property 2: This is a consequence of the fact that probability is a nondecreasingf function, i.e. if A ⊆ B then P (A) ≤ P (B). Indeed, recall that if A ⊆ B then the set B can be partitioned into B = A + B ∩ Ā , so, using the additivity property of probability, P (B) = P (A) + P B ∩ Ā ≥ P (A). If x1 ≤ x2 , then the corresponding events A = {X ≤ x1 } and B = {X ≤ x2 } have the property that A ⊂ B. Therefore P ({X ≤ x1 }) ≤ P ({X ≤ x2 }), which by the definition of a CDF means F (x1 ) ≤ F (x2 ). Proof. Property 3: This property is related to the nondecreasing property of the CDF: going backwards, to the left, toward −∞, the CDF can only decrease or stay constant. If F (x0 ) = 0, to the left of x0 it can only stay constant (=0) since it cannot take negative values. For x ≤ x0 the events A = {X ≤ x} and B = {X ≤ x0 } have the property that A ⊂ B, therefore 0 ≤ F (x) = P ({X ≤ x}) ≤ P ({X ≤ x0 }) = F (x0 ) = 0 therefore F (x) = 0. Proof. Property 4: This is a consequence of the properties of probability of complementary events: recall that P Ā = 1 − P (A). SetA = {X ≤ x}; then its complement (the set of all elementary events which are not in A) is Ā = {X > x}, i.e. {X ≤ x} + {X > x} = S = R. As above, P Ā = 1 − P (A), but with from the definition of the CDF, P Ā = P ({X > x}) = 1 − P ({X ≤ x}) = 1 − F (x). 1 Proof. Property 5: This property shows how to use the CDF to calculate the probability that RV X returns a value in a given interval [x1 , x2 ]. The CDF is defined on semi-infinite intervals of the form (−∞, x1 ] and (−∞, x2 ]. Note that the events A = {X ∈ (−∞, x1 ]}, B = {X ∈ (−∞, x2 ]} and C = {X ∈ (x1 , x2 ]} are in the relationship B =A+C since A and C are disjoint and B = A ∪ C. Therefore P (B) = P (A) + P (C), or P (C) = P (B) − P (A). Hence the property. II) The number xu for which the CDF F (x) has the value F (xu ) = u is called the udF is even, that is f (−x) = f (x), percentile of F . Show that if the derivative f (x) = dx then (1) F (−x) = 1 − F (x) (2) x1−u = −xu . (3) Draw a sketch of the CDF and its derivative. Explain on the plot the above relations. dF , one can write dx Z x Z F (x) = f (s) ds, and, F (−x) = Proof. 1) Since f (x) = −∞ −x f (s) ds. −∞ Changing the variable in the last integral to t = −s, with s = −t; ds = −dt; s = −x 7→ t = x; and using the fact that f (−t) = f (t) yields Z −x Z x Z ∞ Z f (s) ds = f (−t)(−dt) = f (−t) dt = −∞ ∞ x s = −∞ 7→ t = ∞; ∞ f (t) dt = F (∞) − F (x) = 1 − F (x). x 2)Using the above definition of the percentile, F (x1−u ) = 1 − u and F (xu ) = u. Obviously, this means that F (x1−u ) = 1 − F (xu ). From point 1), this implies that x1−u = −xu . 3) A sketch is shown below 2 III) Show that if X is an RV that can take values only in a finite interval, a ≤ X ≤ b, then F (x) = 1 for every x ≥ b and F (x) = 0 for every x ≤ a. Proof. If a ≤ X ≤ b, from the definition of a CDF, F (a) = 0 and F (b) = 1. Since F is nondecreasing with values in the interval [0, 1], from x ≥ b follows that 1 = F (b) ≤ F (x) ≤ 1, therefore F (x) = 1 for any x ≥ b. A similar reasoning works for the second statement: if x ≤ a, 0 ≤ F (x) ≤ F (a) = 0. IV) Show that if X and Y are two RVs such that X ≤ Y at every trial, then FY (a) ≤ FX (a) for every a ∈ R. Proof. From the hypothesis, X and Y are two RVs such that X ≤ Y at every trial, i.e. for every elementary event, X(ζ) ≤ Y (ζ). Let AX be the set of elementary events ζ such that the observed value X(ζ) ≤ a. In set algebra notation AX = {ζ : X(ζ) ≤ a} . Denote similarly, AY = {ζ : Y (ζ) ≤ a}. By the definition of CDF, FX (a) = P (AX ) ; FY (a) = P (AY ) . If AY ⊆ AX , then using the “monotonicity” property of probability, P (AY ) ≤ P (AX ), and consequently FY (a) ≤ FX (a) . Indeed, let ζ ∈ AY . This means that Y (ζ) ≤ a, but also since X(ζ) ≤ Y (ζ) ≤ a, so ζ also belongs to AX , ζ ∈ AX . V) RV X takes only positive values. Show that if the conditional probability P ({s < X ≤ s + s1 } |{X > s}) = P ({X ≤ s1 }) for any s, s1 ≥ 0, then the Cumulative Distribution Function (CDF) FX (t) = 1 − e−ct , where c is a constant. 3 Proof. Denote A = {s < X ≤ s + s1 }, B = {X > s} and C = {X ≤ s1 }; with the definition of conditional probability, the relationship can be written P (A ∩ B) = P (C) . P (B) Note also that A ⊂= B, which means that A ∩ B = A = {s < X ≤ s + s1 }, so in fact the relation is P (A) = P (C) . P (B) Denote by FX (t) the CDF of RV X. For the sake of clarity, let’s rename s1 = ∆s. Then P (A |B ) = P (A) = F (s + ∆s) − F (s); P (B) = 1 − F (s); P (C) = F (∆s). The relation becomes F (s + ∆s) − F (s) = F (∆s). 1 − F (s) valid for any s, ∆s ≥ 0. For a fixed s, let ∆s → ds. Then dF ds, F (∆s) → F (ds) = F (0) + ds 0 dF F (s + ∆s) − F (s) → F (s + ds) − F (s) = ds. ds Note also that F (0) = 0.(Why?) The relation then becomes 1 dF dF ds = ds 1 − F (s) ds ds 0 or equivalently F0 dF . = c, where c = 1−F ds 0 This is a differential equation; integrating from s = 0: Z F Z F dF d(1 − F ) dF = c ds, = cs; − = cs. 1−F 1−F 0 1−F 0 Z F − d log(1 − F ) = cs; log (1 − F (s)) = −cs, so 0 F (s) = 1 − e−cs . 4

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