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.
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