STAT 430/510: Lecture 10

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1 STAT 430/510: Lecture 10 James Piette June 9, 2010

2 Updates HW2 is due today! Pick up your HW1 s up in stat dept. There is a box located right when you enter that is labeled "Stat 430 HW1". It ll be out for the next week or so. Given that so many people take accounting, I m changing my hours to 2:30 to 3:30 on MTTh. No more Wednesdays. If you do not have webcafe and want to see your grades as I post them, then get an account at Wharton computing s office.

3 Formalization Def: We say that X is a continuous random variable if there exists a nonnegative f, defined on all x (, ), such that for any set B of real numbers (i.e. subset of (, )), P(X B) = f (x)dx B The function f is called the probability density function of the random variable X. This is synonymous to the pmf for discrete random variables. The cumulative distribution function for a continuous r.v. is given by x F(X) = P(X x) = f (s)ds

4 Properties 1 1 = P(X (, )) = f (x)dx 2 P(a X b) = b a f (x)dx 3 P(X = a) = a a f (x)dx = 0 4 P(X < a) = P(X a) = a f (x)dx

5 Properties Another way of thinking about f (x), our pdf, is the following: P ( x ɛ 2 X x + ɛ ) x+ ɛ 2 = f (x)dx ɛ f (x) 2 a ɛ 2 where ɛ is small. In otherwords, the probability that our r.v. X is contained in that a tiny interval of length ɛ around the point x is close to ɛ f (x).

6 Example 1 Suppose the amount of time (in hours) that a computer functions before breaking down is a continuous r.v. with pdf given by: {λe x 100 if x 0 f (x) = 0 if else Question: What is the probability that a computer will function between 50 and 150 hours before breaking down? Solution: We need to determine what λ is. What property can we use about pdf s to help us? Thus, λ = = f (x)dx = λ 0 e x 100 = λ(100)e x = 100λ

7 Example 1 (cont.) Now, we can calculate the probability that the computer will function between 50 and 150 hours before breaking down: P(50 X 150) = = e x e x 100 dx = e 1 2 e Question: What is the probability that a computer will function for fewer than 100 hours?

8 Example 1 (cont.) Solution: We already know λ for this problem, so the only question is how can we formalize the probability of interest? F(100) = P(X 100) = e x 100 dx = e x = 1 e

9 Relationship between PDF and CDF Going off of the definition we presented before about the cdf of a continuous r.v., it is obvious that we can easily get the cdf from the pdf: F(x) = P(X (, x]) = x f (x)dx We can also do the antithesis of that; that is, we can (less easily) get the pdf from the cdf: d F(x) = f (x) dx The density (or pdf) of a continuous r.v. is the derivative of the cdf.

10 Example 2 Suppose X is a continuous r.v. with cdf F X and pdf f X. Question: What is the pdf of Y = 2X? Solution 1: First, we derive the distribution function (i.e. cdf) of Y. F Y (x) = P(Y x) = P(2X x) = P(X x 2 ) ( x ) = F X 2 All that is left is to differentiate: d dx F Y (x) = d ( x ) dx F X = 1 ( x ) 2 2 f x 2

11 Example 2 (cont.) Solution 2: Another way to determine the density of Y is by considering a very small interval... ( ɛf Y (x) P x ɛ 2 Y x ɛ ) ( 2 = P x ɛ 2 2X x ɛ ) ( 2 x = P 2 ɛ 4 X x 2 4) ɛ ɛ ( x ) 2 f X 2 Dividing by ɛ gets us to the same answer as before.

12 Expected Value Thinking back to Chapter 4 when we defined the expected value of a discrete r.v., we defined it as: E(X) = x xp(x = x) Using that same mind set for a continuous r.v., we note that for small dx: f (x)dx P(x X x + dx) Thus, we want to "sum" all of these pieces using integration: E(X) = xf (x)dx

13 Expected Value (cont.) Using this definition for the expectation of a continuous r.v., we can extend this to functions of our r.v. in a similar fashion to what we did before. E(g(X)) = where g is any real-valued function. g(x)f (x)dx Ex: The expectation of a continuous r.v., X, that is squared can be written as E(X 2 ) = x 2 f (x)dx

14 Example 3 Suppose X has a density given by { 1 0 x 1 f (x) = 0 else Question: What is E(e x )? Solution: E(e X ) = 1 0 e x (1)dx = e x 1 0 = e 1

15 Variance The variance of a continuous r.v. is exactly the same as a discrete r.v. with both the original form of: and the alternative form of: Var(X) = E((X µ) 2 ) Var(X) = E(X 2 ) (E(X)) 2 Also, once again, we represent the standard deviation of a continuous r.v. as SD(X) = Var(X) = (E(X 2 ) (E(X)) 2 )

16 Example 4 Let c be some constant and X be a r.v. with pdf { cx 4 0 < x < 2 f (x) = 0 otherwise Question: What is E(X)? Before we begin answering any questions about X, we need to calculate c, using the same technique as earlier. 1 = 2 0 cx 4 dx = c 5 ( ) = 32c 5 c = 5 32

17 Example 4 (cont.) Solution: Now that we know c, we can find the expectation of X: 2 ( ) 5 E(X) = x 32 x 4 dx 0 2 = 5 x 5 dx 32 0 = = 5 3 Question: What is Var(X)?

18 Example 4 (cont.) Solution: First, we must find E(X 2 ): E(X 2 ) = 2 0 = 5 32 ( ) 5 x 2 32 x 4 dx 2 0 x 6 = = 20 7 Second, we compute (E(X)) 2 and plug it into the alternative form: (E(X)) 2 = ( ) 5 2 = Var(X) = = 5 63

19 Properties The same properties about expectations and variance hold the same with continuous r.v. s. We see that E is a linear operand: E(aX + b) = ae(x) + b a, b R Also, Var is unaffected by a shift change and a scale change results in a squared change to the variance: Var(aX + b) = a 2 Var(X) a, b R

20 Formalization Def: A continuous r.v. X is said to be uniformly distributed on the interval [a, b] if the pdf of X is given by f (x) = { 1 b a 0 else a x b In this scenario, X is also said to be a uniform r.v.. The cdf of a uniform r.v. over the interval [a, b] is 0 x a F (x) = x a b a a < x < b 1 x b

21 Properties Let X be a uniform r.v. on [a, b]. Then: E(X) = a+b 2. E(X 2 ) = a2 +ab+b 2 3. Var(X) = (b a)2 12.

22 Example 5 To be a winner in a certain game, you must be successful in three successive rounds. The game depends on the value of U, a uniform r.v., on (0,1). If U > 0.1, then you are successful in round 1; if U > 0.2, then you are successful in round 2; and if U > 0.3, then you are successful in round 3. Question: What is the probability that you are successful in round 1? Solution: This is pretty straight forward: P(U > 0.1) = dx = = 0.9

23 Example 5 (cont.) Question: Given that you won in round 1, what is the probability that you won in round 2? Solution: Think back to conditional probability. Let s say A = win in round 1 and B = win in round 2, so... P(B A) = P(AB) P(A) P((U > 0.1) (U > 0.2)) = P(U > 0.1) P(U > 0.2) = (since (U > 0.2) (U > 0.1)) P(U > 0.1) = 1dx = 0.1 1dx 0.9 = 8 9

24 Example 5 (cont.) Question: Given that you won in round 1 and 2, what is the probability that you won in round 3? Solution: Carrying on with the previous pattern and letting C = win in round 3, we get: P(C AB) = P(ABC) P(AB) P(U > 0.3) = P(U > 0.2) = = 7 8

25 Example 5 (cont.) Question: What is the probability that you are a winner? Solution: Similar to the key problem in the current homework. What probability are we interested in? P(ABC). Now, applying the multiplication rule, we get P(ABC) = P(C AB)P(B A)P(A) = = 7 10

26 (If time permits), let s look at self-test problem Material covered today goes through 5.3; we will continue on Thurs. talking about 5.3 and move on. HW3 will be up soon, as well as practice midterm and HW2 solutions.

STAT 430/510: Lecture 16

STAT 430/510: Lecture 16 James Piette June 24, 2010 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. Joint Distribution of Functions