Mixture of Discrete and Continuous Random Variables

Transcription

1 Mixture of Discrete and Continuous Random Variables What does the CDF F X (x) look like when X is discrete vs when it s continuous? A r.v. could have a continuous component and a discrete component. Ex 1 & 2 from MixedRandomVariables.pdf. 1

2 Example 1: Consider a r.v. X with cdf F (x) = x < x 3 x < 2 1 x 2 The support of X is [, 2] = A 1 A 2 where A 1 = [, 2) a, and A 2 = {2}. The distribution of X has different expressions over the two regions: (continuous portion) pdf on A 1 with f(x) = 1/3. (discrete portion) pmf on A 2, with p(2) = 1/3. When computing expectations, we use pmf or pdf, in each region. a It doesn t matter if we write A 1 = (, 2). 2

3 Example 2 Suppose a > b. The difference between the two sets, (, b] and (, a], is (b, a]. So F (a) F (b) = P(X a) P(X b) = P ( X (, a] ) P ( X (, b] ) = P ( X (b, a] ) = P(b < X a). 3

4 How to use CDF to compute those probabilities? Note 1) P(b < X a) = F (a) F (b); 2) if a point x is in the continuous portion, then we can use and > (or, and <) interchangeably. P(1 < X < 1.5) = P(1 < X 1.5) = F (1.5) F (1), where the first equality is due to the fact that 1.5 is in the continuous portion. P(1 X < 1.5) = P(1 X 1.5) = P(X = 1) + P(1 < X 1.5) =.5 + F (1.5) F (1). P(1 X 1.5) = P(X = 1) + P(1 < X 1.5). 4

5 Example: An insurance policy reimburses a loss up to a benefit limit of C and has a deductible of d. Suppose the policyholder s loss, X, follows Ex(1/5). Let Y denote the benefit paid under the insurance policy. Find the distribution of Y. x < d Y = Benefit Paid = x d d x < C + d C x C + d Discrete portion of Y: p() = d 1 5 e x/5 dx = 1 e d/5, p(c) = C+d 1 5 e x/5 dx = e (C+d)/5. Continuous portion of Y: f Y (y) = 1 5 e (y+d)/5, < y < C. 5

6 Distributions of Two Random Variables Major concepts (chap 2): Joint pdf/pmf Marginal pdf/pmf Conditional pdf/pmf, conditional expectations 6

7 Let X and Y be discrete random variables. The joint pmf p(x, y) is defined by and p(x, y) = P(X = x, Y = y), P((X, Y ) A) = p(x, y). (x,y) A The marginal pmfs of X and of Y are given by p X (x) = all y p(x, y), p Y (y) = all x p(x, y). Check that p X (x) and p Y (y) are legit pmfs. 7

8 Example 1 (2.1.1 on p.74) X / Y a) Find P(X + Y = 2). p(1, 1) + p(, 2) + p(2, ) = 2 8. b) Find P(X < Y ). c) Find the marginal probability distributions p X (x) of X and p Y (y) of Y. For example, the marginal pmf of X is given by the row sums of the table p X () = = 1 4, p X(1) = = 1 4, p X(2) =

9 Let X and Y be continuous random variables. Then f(x, y) is the joint pdf for X and Y if for any two-dimensional set A P((X, Y ) A) = f(x, y)dxdy. In particular, f(x, y)dxdy = 1. The marginal pdfs of X and of Y are given by A f X (x) = f(x, y)dy, f Y (y) = f(x, y)dx. Check that f X (x) is a legit pdf: apparently f X (x). and f X (x)dx = f(x, y)dydx = 1. 9

10 Let X and Y be random variables. Then their joint cdf is defined by F (x, y) = P(X x, Y y). For continuous rvs, f(x, y) = 2 F (x, y). x y Let X and Y be random variables and g some real valued function, i.e., g : R 2 R. E[g(X, Y )] = g(x, y) p(x, y) all x all y E[g(X, Y )] = g(x, y)f(x, y)dxdy. R 2 1

11 The mgf of a random vector (X, Y ) M XY (t 1, t 2 ) = E(e t 1X+t 2 Y ), if it exists for t 1 < h, t 2 < h. M X (t) = M XY (t, ) M Y (t) = M XY (, t). 11

12 Example 2: Consider two random variables X and Y with the mgf M(t 1, t 2 ) =.1 +.2e t1 +.3e 2t 2 +.4e t 1+t 2. (1) Find the joint pmf p(x, y). Recall that M(t 1, t 2 ) = E(e t 1X+t 2 Y ) = all (x,y) p(x, y)ex t 1+y t 2. Write (1) as.1e ()t 1+()t 2 +.2e (1)t1+()t 2 +.3e ()t 1+2t 2 +.4e (1)t 1+(1)t 2. So X / Y

13 Example 3 (2.1.5 & on p.8-81) Let X and Y have the pdf f(x, y) = 8xy, < x < y < 1;, elsewhere. The very first step before doing any calculation: sketch the support. a) Verify that f(x, y) is a legitimate pdf. 1. f(x, y) for all (x, y) R f(x, y)dxdy = 1 this is because 1 y 8xydxdy = = 1 1 ( y 8y 4y 3 dy ) xdx dy where y xdx = x2 2 y = y2 2 = y 4 1 = 1. 13

14 b) Find P(X + Y <.5) and P(2X Y ). P(X + Y <.5) = = = = (.5 x x (.5 x 4x x ) 8xydy dx ) 2ydy dx [ 4x (.5 x) 2 x 2] dx 4x(.25 x)dx =.25 (x 4x 2 )dx = P(2X Y ) = =.5 1 ( 2x ( y x y/2 ) 8xy dy dx + ) 8xy dx dy 1.5 ( 1 x ) 8xy dy dx, OR 14

15 c) Find the marginal pdf for X. f X (x) = 1 f(x, y)dy = 1 x 8xy dy = 4x 4x 2, < x < 1. d) Find the marginal pdf for Y. f X (x) = 1 f(x, y)dx = y 8xy dx = 4y 3, < y < 1. e) Find E(XY 2 ), E(Y ), E(7XY 2 + 5Y ) (see p.8). 15

16 f) Let Z = X/Y. Find the distribution of Z. First find the support of Z < x < y < 1 = < z < 1. Then use the CDF approach F Z (z) = P(X/Y z) = P(X zy ) = 1 = z 2. ( zy ) 8xy dx dy = 1 4y 3 z 2 dy The pdf is given by f Z (z) = 2z, < z < 1. 16

17 Go through the examples from JointDistributions.pdf by yourself. 17

18 Independent Random Variables Random variables X and Y are independent if for all (x, y) p(x, y) = p X (x) p Y (y), (discrete) f(x, y) = f X (x) f Y (y), (continuous). If X and Y independent, 1. F (x, y) = F X (x) F Y (y) 2. M(t 1, t 2 ) = M X (t 1 ) M Y (t 2 ) 3. E[g(X)h(Y )] = Eg(X) Eh(Y ), specially Var(X + Y ) = Var(X) + Var(Y ) (see the proof on the next slide). 18

19 E(X + Y ) = EX + EY = µ X + µ Y Var(X + Y ) = E ( X + Y (µ X + µ Y ) ) 2 ( ) 2 = E X µx + Y µ Y [ ] = E (X µ X ) 2 + (Y µ Y ) 2 + 2(X µ X )(Y µ Y ) = E(X µ X ) 2 + E(Y µ Y ) 2 + 2E(X µ X )(Y µ Y ) = Var(X) + Var(Y ) + 2E(X µ X ) E(Y µ Y ) = Var(X) + Var(Y ) + 19

20 Example 1 (revisit): Are X and Y independent? (NO) Sign that they are dependent: some entries in pmf table are zero. Example 3 (revisit): Are X and Y independent? (NO) Sign that they are dependent: support is not rectangle. More Examples from Independence and Covariance.pdf 2