ECE270: Handout 11Pairs of RVs (Part I)Outline:1. notation of pairs of RVs,2. a general rule to find probabilities of events concerning two RVs,3. joint CDF of two RVs and its properties,4. joint pmf of two discrete RVs,5. joint pdf of two continuous RVs.F Notion of Pairs of RVs• Many random experiments involve more than one RV. In some experiments more than onequantity is measured, e.g., the voltage at different points in a circuit at some specific time. Sofar, we have learned how to calculate probabilities of events involving a single RV in isolation.We want to extend the concepts we learned to two RVs.• For two RVs X and Y we discuss the joint CDF FX,Y(x, y) and the marginal CDFs FX(x)and FY(y). We also we consider two important cases when i) X and Y are both discrete andwe discuss joint pmf pX,Y(x, y) = P (X = x, Y = y) and marginal pmfs pX(x) and pY(y), ii)X and Y are both continuous and we discuss joint pdf fX,Y(x, y) and marginal pdfs fX(x)and fY(y). We use the joint CDF, joint pmf, and joint pdf to calculate the probabilities ofevents that involve the joint behavior of two RVs.• The notion of a RV as a mapping from sample space S to the real line R can be generalized tothe case of two RVs: we are interested in a two dimensional vector function X = (X, Y ) thatmaps a point ξ ∈ S in the sample space to a point X(ξ) = (X(ξ), Y (ξ)) ∈ R2in the real plane.F A General Rule to Find Probabilities of Events Concerning Two RVs• To find the probability that the pair (X, Y ) ∈ B (where B ⊂ R2) we need to find the set ofoutcomes A, A ⊂ S that are mapped to B, i.e., the set A = {ξ : (X(ξ), Y (ξ)) ∈ B}.• If event A occurs then (X(ξ), Y (ξ)) ∈ B ⇒ event B occurs.• If event B occurs then (X(ξ), Y (ξ)) ∈ B implies ξ ∈ A ⇒ event A occurs.• We have P ((X, Y ) ∈ B) = P (A) = P ({ξ : (X(ξ), Y (ξ)) ∈ B}). We refer to A and B asequivalent events.EXAMPLE 1Let a random experiment consist of selecting a student’s name from an urn. Let ξ denote theoutcome of this experiment and define the following two RVs: H(ξ) is the height of studentξ in cm and W (ξ) is the weight of student ξ in pound. Let B = {H ≤ 183, W ≤ 160}. Theequivalent event A = {ξ : H(ξ) ≤ 183, W (ξ) ≤ 160} and P (B) = P (A).EXAMPLE 2Let the random experiment be selecting a point inside a unit circle and ξ be the outcomeof this experiment. We define the following two RVs: Θ(ξ) is the angle of the point andR(ξ) is the distance of the point from the center. We can be interested in P (B) whereB = {π/5 < Θ ≤ 3π/4, 0.2 < R ≤ 0.8}. The equivalent event A = {ξ : π/5 < Θ(ξ) ≤3π/4, 0.2 < R(ξ) ≤ 0.8} and P (B) = P (A).F Joint CDF of Two RVs• Recall that by defining the CDF FX(x) = P (X ≤ x) we were able to find the probabilitiesof other events in terms of the CDF, e.g., P (a < X ≤ b) = FX(b) − FX(a). We extent theconcept of CDF for two RVs.• The joint CDF of two RVs X and Y is defined as, regardless of X and Y being discrete orcontinuous, the probability of the event B = {X ≤ x, Y ≤ y}:FX,Y(x, y) = P (B) = P (X ≤ x, Y ≤ y) for − ∞ < x < ∞ − ∞ < y < ∞• The event B = {X ≤ x, Y ≤ y} and its probability vary as x, y are varied. Hence,FX,Y(x, y) is a function of the variables x and y.• In terms of the underlying random experiment FX,Y(x, y) = P ({ξ : X(ξ) ≤ x, Y (ξ) ≤ y}| {z }equivalent event A).F Properties of the Joint CDF• The three axioms of probability and the corresponding properties imply that the joint CDFhas the following properties:(i) 0 ≤ FX,Y(x, y) ≤ 1(ii) FX,Y(x, −∞) = 0 FX,Y(−∞, y) = 0 FX,Y(∞, ∞) = 1(iii) FX(x) = limy→∞FX,Y(x, y) = FX,Y(x, ∞) FY(y) = limx→∞FX,Y(x, y) = FX,Y(∞, y)We obtain the marginal CDFs by removing the constraint on one of the RVs. Note thatthe knowledge of the marginal CDF’s is insufficient to specify the joint CDF.(iv) FX,Y(x, y) is a nondecreasing function of x and y, that is, if x1≤ x2and y1≤ y2then FX,Y(x1, y1) ≤ FX,Y(x2, y2).(v) FX,Y(x, y) is continuous from the “north” and from the “east”:limx→a+FX,Y(x, y) = FX,Y(a, y) limy→b+FX,Y(x, y) = FX,Y(x, b)Page 2(vi) The probability of the rectangle {x1< x ≤ x2, y1< y ≤ y2} is given by:{x1< X ≤ x2, y1< Y ≤ y2} = FX,Y(x2, y2)−FX,Y(x2, y1)−FX,Y(x1, y2)+FX,Y(x1, y1)EXAMPLE 3The joint CDF of the vector RV X = (X, Y ) is given by:FX,Y(x, y) =½(1 − e−αx)(1 − e−βy) x ≥ 0, y ≥ 00 elsewherea) Find the marginal CDF’s.b) Find the probability of the event A where A = {X ≤ 1, Y ≤ 1}.c) Find the probability of the event B where B = {1 < X ≤ 2, 2 < Y ≤ 5}.d) Find the probability of the event C where C = {X > 3, Y > 4}.F Joint pmf of Two Discrete RVs• Let the pair (X, Y ) take values from a countable set Sx,y= {(xj, yk), j = 1, 2, ..., k = 1, 2, ...}.The joint pmf of (X, Y ) is defined as:pX,Y(xj, yk) = P (X = xj, Y = yk) for (xj, yk) ∈ Sx,ywhereXall jXall kpX,Y(xj, yk) =Xall jXall kP (X = xj, Y = yk) = 1• We can obtain the join CDF FX,Y(x, y) from the joint pmf pX,Y(xj, yk):FX,Y(x, y) =X{j:xj≤x}X{k:yk≤y}pX,Y(xj, yk)• Let B ⊂ Sx,ybe an event. The probability of event B is P (B) =PP(xj,yk)∈BpX,Y(xj, yk).• We obtain the marginal pmfs by removing the constraint on one of the RVs.pX(xj) = P (X = xj) =Xall kpX,Y(xj, yk)similarly,pY(yk) = P (Y = yk) =Xall jpX,Y(xj, yk)Note that the knowledge of the marginal pmf’s is insufficient to specify the joint pmf.Page 3EXAMPLE 4Let N be the length of a randomly selected message in bytes. Suppose the messages are brokeninto packets of fixed length M bytes. Let Q be the number of full packets in a message andR be the number of bytes left over. Suppose N is a Geometric RV with parameter 1 − p, i.e.,P (N = k) = (1 − p)pk, and range Sn= {0, 1, 2, 3, 4, ...}.a) What is Srand Sq?b) Find the joint pmf of R and Q, i.e., P (Q = q, R = r).c) Find the marginal pmf’s.d) Find the probability that a message is at least 5M bytes, i.e., P (Q ≥ 5).e) Find the probability that the last packet is less than half, i.e., P (R …