STAT 430/510: Lecture 16

Transcription

1 STAT 430/510: Lecture 16 James Piette June 24, 2010

2 Updates HW4 is up on my website. It is due next Mon. (June 28th). Starting today back at section 6.7 and will begin Ch. 7.

3 Joint Distribution of Functions of Cont. R.V. s Let X 1 and X 2 be jointly cont. r.v. s with joint pdf f X1,X 2. Suppose there are two r.v. s, Y 1 and Y 2, such that Y 1 = g 1 (X 1, X 2 ) and Y 2 = g 2 (X 1, X 2 ) for some functions g 1 and g 2. There are two assumptions: 1 Equations y 1 = g 1 (x 1, x 2 ) and y 2 = g 2 (x 1, x 2 ) can be uniquely solved for x 1 and x 2 in terms of y 1 and y 2, with solutions given by x 1 = h 1 (y 1, y 2 ) and x 2 = h 2 (y 1, y 2 ). 2 The functions g 1 and g 2 have cont. partial derivatives at all points (x 1, x 2 ) and are s.t. the 2 2 determinant: g 1 J(x 1, x 2 ) = x 1 g 2 x 2 g 1 g 1 x 1 x 2 0

4 Joint Distribution of Functions of Cont. R.V. s (cont.) Under the previous two assumptions, the r.v. s Y 1 and Y 2 are jointly cont. with joint density given by f Y1 Y 2 (y 1, y 2 ) = f X1 X 2 (x 1, x 2 ) J(x 1, x 2 ) 1 where x 1 = h 1 (y 1, y 2 ), x 2 = h 2 (y 1, y 2 ).

5 Example 1 Let X 1, X 2, X 3 be independent standard normal r.v. s. Question: If Y 1 = X 1 + X 2 + X 3, Y 2 = X 1 X 2 and Y 3 = X 1 X 3, then what is the joint density of Y 1, Y 2, Y 3? Solution: Need to verify the assumptions. The first assumption... X 1 = Y 1 + Y 2 + Y 3 3 X 2 = Y 1 2Y 2 + Y 3 3 X 3 = Y 1 + Y 2 2Y 3 3 And the second assumption J(x 1, x 2 ) = = 3

6 Example 1 (cont.) Remember that... f X1 X 2 X 3 (x 1, x 2, x 3 ) = 1 3 e 1 x i 2 /2 (2π) 3/2 Now, all we need to do is plug in the appropriate values into the formulas: f Y1 Y 2 Y 3 (y 1, y 2, y 3 ) = 1 ( 3 f y1 + y 2 + y 3 X 1 X 2 X 3, y 1 2y 2 + y 3, y 1 + y = 3(2π) 3/2 e Q(y 1,y 2,y 3 )/2 where... Q(y 1, y 2, y 3 ) = ( ) y1 + y 2 + y 2 ( ) 3 y1 2y 2 + y 2 ( 3 y1 + y = y y y y 2y 3 3

7 Recall: Expectation The expected value of a discrete r.v. X with pmf p(x) is given by E(X) = xp(x) x The expected value of a continuous r.v. X with pdf f (x) is given by E(X) = xf (x)dx If P(a X b) = 1, then a E(X) b.

8 Proposition If X and Y have a joint pmf p(x, y), then E(g(X, Y )) = g(x, y)p(x, y) y x If X and Y have a joint pdf f (x, y), then E(g(X, Y )) = g(x, y)f (x, y)dxdy

9 Example 2 An accident occurs at point X that is uniformly dist. on a road of length L. At the time of the accident, an ambulance is at location Y that is also uniformly dist. on the road. Question: Assuming X and Y are independent, what is the expected distance between the ambulance and point of the accident? Solution: In this scenario, we are looking to calculate the expectation of the r.v.... X Y because we want distance between the two.

10 Example 2 (cont.) First, we need the joint density function of X and Y : f (x, y) = 1 L 2, 0 < x < L, 0 < y < L From Proposition 7.2.1,... E( X Y ) = 1 L 2 L L 0 0 x y dydx

11 Example 2 (cont.) Let s evaluate the first integral by splitting it up: L 0 x y dy = x 0 (x y)dy + L x (y x)dy = x L2 2 x 2 x(l x) 2 = L2 2 + x 2 xl Therefore, E( X Y ) = 1 L ( ) L 2 L x 2 xl dx = L 3 0

12 Properties Let X and Y be continuous r.v. s If X Y, then E(X) E(Y ). E( n i=1 X i) = n i=1 E(X i). E(aX + by ) = ae(x) + be(y ).

13 Sample Mean Let X 1,..., X n be i.i.d r.v. s having distribution F and expected value µ. Such a sequence of r.v. s is said to constitute a sample from the distribution F. The quantity X = n i=1 X i n is called the sample mean.

14 Sample Mean (cont.) The expectation of X is [ n ] E( X) X i = E n i=1 [ n ] = 1 n E X i i=1 = 1 n E(X i ) n i=1 = µ since E(X i ) µ

15 Expectation of Binomial R.V. Let X be a binomial r.v. with parameters (n, p). Recalling that a binomial represents the number of successes in n independent trials when each trail has probability p of being a success, we have that X = X 1 + X X n where X i = { 1 if the ith trial is a success 0 if the ith trial is a failure Hence, X i is a Bernoulli r.v. with expectation... E(X i ) = 0(1 p) + 1(p) = p. So... E(X) = E(X 1 ) + E(X 2 ) E(X n ) = np

16 Expectation of Negative Binomial R.V. Let X represent the number of trials needed to get a total of r successes, where the probability of each success is p. Then X is a negative binomial r.v. with parameters (r, p). Once again, X can be represented by X = X 1 + X X r where X i is the number of trials required after after the (i 1)th success until a total of i successes is obtained. Thus, X i is a geometric r.v. with parameter p. So... E(X) = E(X 1 ) + E(X 2 ) E(X r ) = r p

17 Example 3 Suppose there are N different types of coupons and each time one obtains a coupon, it is equally likely to be any one of the N types. Question: What is the expected number of coupons needed to amass before you have a complete set of at least one of each type? Solution: Let X denote the number of coupons collected before complete set is attained. Then, X = X 0 + X X N 1 where X i = number of additional coupons needed to be obtained after i distinct types have been collected in order to obtain another distinct type. How is X i distributed? Geometric with parameter N i, with expectation... E(X i ) = N N i. N

18 Example 3 (cont.) This implies that E(X) = X 0 + X X N 1 = 1 + N N 1 + N N N [ 1 = N N ] N

19 Example 4 Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of all the others. Question: If each hunter independently hits his target with probability p, then what is the expected number of ducks that escape unhurt when a flock of 10 flies overhead? Solution: Let X denote the number of ducks that escape. Then, we can represent X as X = X 1 + X X 10 where X i is an indicator (remember back?) such that { 1 if the ith duck escapes unhurt X i = 0 otherwise

20 Example 4 (cont.) Like with any indicator, its expected value is... E(X i ) = P(X i = 1) = (1 p 10 )10 Therefore, E(X) = E(X 1 ) + E(X 2 ) E(X 10 ) = 10(1 p 10 )10

21 Motivation Expected value describes the average of the r.v. s. Variance desrcibes the variation of the r.v. s. Covariance describes the relationship between two r.v. s (i.e. how the interact).

22 Formalization Def: The covariance between two r.v. s X and Y is defined as Cov(X, Y ) = E[(X E(X))(Y E(Y ))] An alternative form of covariance is Cov(X, Y ) = E[XY ] E[X]E[Y ] Remembering back, Cov comes up when we look at X and Y not independent and... Var(X + Y ) = Var(X) + Var(Y ) + 2Cov(X, Y )

23 Let s cover self-test problems We ve now finished Ch. 6 and sections , plus started 7.4. Remember, I ve posted HW4 up on my website and it is due on Mon. June 28th.