4.2 Continuous Models

Transcription

1 Ismor Fischer, 8//8 Stat 54 / Continuous Models Horseshoe Crab (Limulus polyphemus) Not true crabs, but closely related to spiders and scorpions. Living fossils eisted since Carboniferous Period, 35 mya. Found primarily on Atlantic coast, with the highest concentration in Delaware Bay, where males and the much larger females congregate in large numbers on the beaches for mating, and subsequent egg-laying. Pharmaceutical (and many other scientific) contributions! Blue hemolymph (due to copper-based hemocyanin molecule) contains amebocytes, which produce a clotting agent that reacts with endotoins found in the outer membrane of Gram-negative bacteria. Several East Coast companies have developed the Limulus Amebocyte Lysate (LAL) assay, used to detect bacterial contamination of drugs and medical implant devices, etc. Equal amounts of LAL reagent and test solution are mied together, incubated at 37 C for one hour, then checked to see if gelling has occurred. Simple, fast, cheap, sensitive, uses very small amounts, and does not harm the animals probably. (Currently, a moratorium eists on their harvesting, while population studies are ongoing ) Photo courtesy of Bill Hall, bhall@udel.edu. Used with permission.

2 Ismor Fischer, 8//8 Stat 54 / 4-4 Continuous Random Variable: X = Length (inches) of adult horseshoe crabs Sample Sample n = 5; lengths measured to nearest inch n = ; lengths measured to nearest ½ inch e.g., in [, 6), 6 in [6, ), 9 in [, 4) e.g., 8 in [, 4), 4 in [4, 6), etc Eamples: P(6 X < ) =.4 P(6 X < ) =.6 +. =.8 In the limit as n, the population distribution of X can be characterized by a continuous density curve, and formally described by a density function f(). Males are smaller, on average f() Total Area = f () d = Females are larger, on average X a b 4 b Thus, P(a X < b) = f() d = area under the density curve from a to b. a

3 Ismor Fischer, 8//8 Stat 54 / 4-5 Definition: f() is a probability density function for the continuous random variable X if, for all, f() AND f() d =. The cumulative distribution function (cdf) is defined as, for all, F() = P(X ) = f(t) dt. Therefore, F increases monotonically and continuously from to. Furthermore, P(a X b) = f() d = F(b) F(a). FTC!!!! a b f(t) dt f() Total Area = f () d = X F() The cumulative probability that X is less than or equal to some value i.e., P(X ) is characterized by: () the area under the graph of f up to, or () the height of the graph of F at. But note: f() NO LONGER corresponds to the probability P(X = ) [which =, since X is here continuous], as it does for discrete X. X

4 Ismor Fischer, 8//8 Stat 54 / 4-6 Eample : Uniform density This is the trivial constant function over some fied interval [a, b]. That is, f ( ) = b a for a b (and f ( ) = otherwise). Clearly, the two criteria for being a valid density function are met: it is non-negative, and the (rectangular) area under its graph is equal to its base (b a) height ( / b a), which is indeed. Moreover, for any value of in the interval [a, b], the (rectangular) area under the graph up to is equal to its base ( a) height ( / b a). That is, the cumulative a distribution function (cdf) is given by F( ) =, the graph of which is a straight b a line connecting the left endpoint (a, ) to the right endpoint (b, ). [[Note: Since f ( ) = outside the interval [a, b], the area beneath it contributes nothing to F() there; hence F() = if < a, and F() = if > b. Observe that, indeed, F increases monotonically and continuously from to ; the graphs show f ( ) and F( ) over the interval [, 6], i.e., a =, b = 6. Compare this eample with the discrete version in section 3..]] /5 5 Thus, for eample, the probability P(.6 X 3.8) is equal to the (rectangular) area under f ( ) over that interval, or in terms of F( ), simply equal to the difference between the heights F(3.8) F(.6) = =.56.3 =

5 Ismor Fischer, 8//8 Stat 54 / 4-7 Eample : Power density (A special case of the Beta density: β = ) For any fied p >, let p f ( ) = p for < <. (Else, f ( ) =.) This is a valid density function, since f() and f() d = p p d = p =. The corresponding cdf is therefore F() = p f(t) dt = pt dt = on [, ]. (And, as above, F() = if <, and F() = if >.) Again observe that F indeed increases monotonically and continuously from to, regardless of f ; see graphs for p =, 3, 3. (Note: p = corresponds to the uniform density on [, ].) t p = p

6 Ismor Fischer, 8//8 Stat 54 / 4-8 Eample 3: Cauchy density The function f ( ) = π + for it satisfies the two criteria above: f() AND therefore F() = f(t) dt = < <+ is a legitimate density function, since t dt = π + f() d =. (Verify it!) The cdf is arctan + for < < +. π Thus, for instance, P( X ) = F() F() = π ( ) ( ) Eample 4: Eponential density π 4 + π + = 4. For any a > fied, f() = a e a for (and = for < ) is a valid density function, since it satisfies the two criteria. (Details are left as an eercise.) The corresponding cdf is given by F() = f(t) dt = The case a = is shown below. a e at dt = a e, for (and = otherwise). e e Thus, for instance, P(X ) = F() = e =.8647, and P(.5 X ) = F() F(.5) = ( e ) ( e.5 ) = =.47.

7 Ismor Fischer, 8//8 Stat 54 / 4-9 Eercise: (Another special case of the Beta density.) Sketch the graph of f () = 6 ( ) for (and = elsewhere); show that it is a valid density function. Find the cdf F (), and sketch its graph. Calculate P(¼ X ¾). e Eercise: Sketch the graph of f ( ) = ( e + ) that it is a valid density function. Find the cdf Find the quartiles. Calculate P( X ). for F( ) < <+, and show, and sketch its graph. If X is a continuous numerical random variable with density function f(), then the population mean is given by the first moment + μ = E[X] = f() d and the population variance is given by the second moment about the mean + or equivalently, σ = E[(X μ) ] = ( μ) f() d, σ = E[X ] μ = f() d μ. (Compare these continuous formulas with those for discrete X.) + Thus, for the eponential density, μ = The calculation of σ is left as an eercise. a e a d =, via integration by parts. a Eercise: Sketch the graph of f () = π for < (and elsewhere); show that it is a valid density function. Find the cdf F (), and sketch its graph. Calculate P(X ½), and find the mean. Eercise: What are the mean and variance of the power density? Eercise: What is the mean of the Cauchy density? Faites attention! Ce n est pas aussi facile qu il apparaît... Augustin-Louis Cauchy

8 Ismor Fischer, 8//8 Stat 54 / 4- Eample: Crawling Ants and Jumping Fleas Consider two insects on a (si-inch) ruler: a flea, who makes only discrete integer jumps (X), and an ant, who crawls along continuously and can stop anywhere (Y).. Let the discrete random variable X = length jumped (,,, 3, 4, 5, or 6 inches) by the flea. Suppose that the flea is tired, so is less likely to make a large jump than a small (or no) jump, according to the following probability distribution (or mass) function f() = P(X = ), and corresponding probability histogram. f() Probability Table f() = P(X = ) /8 6/8 5/8 3 4/8 4 3/8 5 /8 6 /8 X The total probability is P( X 6) =, as it should be. P(3 X 6) = 4/8 + 3/8 + /8 + /8 = /8 P( X < 3) = 7/8 + 6/8 + 5/8 = 8/8, or Not = P(3 X 6) = /8 = 8/8 equal! P( X 3) = 8/8 + 4/8 = /8, because P(X = 3) = 4/8 Eercise: Confirm that the flea jumps a mean length of μ = inches. Eercise: Sketch a graph of the cumulative distribution function F() = P(X ), similar to that of. in these notes.

9 Ismor Fischer, 8//8 Stat 54 / 4-. Let the continuous random variable Y = length crawled (any value in the interval [, 6] inches) by the ant. Suppose that the ant is tired, so is less likely to crawl a long distance than a short (or no) distance, according to the following probability density function f(y), and its corresponding graph, the probability density curve. (Assume that f = outside of the given interval.) f(y) = 6 y 8, y 6 / Y The total probability is P( Y 6) = ½ (6)(/3) =, as it should be. P(3 Y 6) = ½ (3)(/6) = /4 (Could also use calculus.) Equal! P( Y < 3) = P(3 Y 6) = /4 = 3/4 P( Y 3) = 3/4 also, because P(Y = 3) = Why? Eercise: Confirm that the ant crawls a mean length of μ = inches. Eercise: Find the cumulative distribution function F(y), and sketch its graph.

10 Ismor Fischer, 8//8 Stat 54 / 4- An etremely important bell-shaped continuous population distribution Normal Distribution (a.k.a. Gaussian Distribution): X ~ N(μ, σ) f() = π σ μ σ ² e, < < + π = e =.788 Total Area = f () d = Standard Deviation σ Johann Carl Friedrich Gauss ( ) Left tail Right tail Mean μ X Eamples: Total Area = f () d = small σ large σ μ = 98.6 X = Body Temp ( F) μ = X = IQ score (discrete!)

11 Ismor Fischer, 8//8 Stat 54 / 4-3 Eample: Two eams are given in a statistics course, both resulting in class scores that are normally distributed. The first eam distribution has a mean of 8.7 and a standard deviation of 3.5 points. The second eam distribution has a mean of 8.8 and a standard deviation of 4.5 points. Carla receives a score of 87 on the first eam, and a score of 9 on the second eam. Which of her two eam scores represents the better effort, relative to the rest of the class? X ~ N(8.7, 3.5) X ~ N(8.8, 4.5) σ = 3.5 σ = 4.5 μ = 8.7 = 87 μ = 8.8 = 9 Z-score Transformation X ~ N(μ, σ) Z = X μ ~ N(, ) σ Standard Normal Distribution The Z-score tells how many standard deviations σ the X-score lies from the mean μ. -score = 87 z-score = higher relative score N(, ) -score = 9 = z-score = 4.5 = Z

12 Ismor Fischer, 8//8 Stat 54 / 4-4 Eample: X = Age (years) of UW-Madison third-year undergraduate population Assume: X ~ N(,.5), i.e., X is normally distributed with mean μ = yrs, and s.d. σ =.5 yrs. How do we check this? And what do we do if it s not true, or we can t tell? Later... σ =.5 μ = Suppose that an individual from this population is randomly selected. Then P(X < ) =.5 (via symmetry) P(X < 9) = P 9 Z <.5 = P(Z <.8) =.9 (via table or software) X ~ N(,.5) Z ~ N(, ) = 9 X.8 Z Therefore P(9 X < ) =.5.9 =.88 Likewise, P(9 X < 9.5) = =.37 P(9 X < 9.5) =.36.9 =.8 P(9 X < 9.5) =.3.9 =. P(9 X < 9.5) =..9 =. P(X = 9. ) =, since X is continuous! 9

13 Ismor Fischer, 8//8 Stat 54 / 4-5 Two Related Questions. Given X ~ N(μ, σ). What is the probability that a randomly selected individual from the population falls within one standard deviation (i.e., ±σ) of the mean μ? Within two standard deviations (±σ)? Within three (±3σ)? Solution: We solve this by transforming to the tabulated standard normal distribution Z ~ N(, ), via the formula Z = X μ, i.e., X = μ + Zσ. σ P(μ σ X μ + σ) = P( Z + ) =.687 P(Z +) P(Z ) = =.687 μ σ μ μ + σ X P(μ σ X μ + σ) = P( Z + ) =.9545 P(Z +) P(Z ) = =.9545 μ σ μ σ μ μ + σ μ + σ X Likewise, P(μ 3σ X μ + 3σ) = P( 3 Z +3) = These values can be used as an informal check to see if sample-generated data derive from a population that is normally distributed. For if so, then 68%, or approimately /3, of the data should lie within standard deviation of the mean; approimately 95% should lie within standard deviations of the mean, etc. Other quantiles can be checked similarly. Superior methods also eist See my homepage to view a ball drop computer simulation of the normal distribution: (requires Java)

14 Ismor Fischer, 8//8 Stat 54 / 4-6. Given X ~ N(μ, σ). What symmetric interval about the mean μ contains 9% of the population distribution? 95%? 99%? General formulation? Solution: Again, we can answer this question for the standard normal distribution Z ~ N(, ), and transform back to X ~ N(μ, σ), via the formula Z = X μ σ, i.e., X = μ + Zσ. The value z.5 =.645 satisfies.5 z.5 = = z z.5 = = z z.5 = = z.5 Z Z Z P( z.5 Z z.5 ) =.9, or equivalently, P(Z z.5 ) = P(Z z.5 ) =.5. Hence, the required interval is μ.645σ X μ +.645σ. The value z.5 =.96 satisfies P( z.5 Z z.5 ) =.95, or equivalently, P(Z z.5 ) = P(Z z.5 ) =.5. Hence, the required interval is μ.96σ X μ +.96σ. The value z.5 =.575 satisfies P( z.5 Z z.5 ) =.99, or equivalently, P(Z z.5 ) = P(Z z.5 ) =.5. Hence, the required interval is μ.575σ X μ +.575σ. In general Def: The critical value z α/ satisfies P( z α/ Z z α/ ) = α, α/ α α/ or equivalently, the tail probabilities P(Z z α/ ) = P(Z z α/ ) = α/. Hence, the required interval satisfies z α/ z α/ Z P(μ z α/ σ X μ + z α/ σ) = α.

15 Ismor Fischer, 8//8 Stat 54 / 4-7 Normal Approimation to the Binomial Distribution (continuous) (discrete) Eample: Suppose that it is estimated that % (i.e., π =.) of a certain population has diabetes. Out of n = randomly selected individuals, what is the probability that (a) eactly X = are diabetics? X = 5? X =? X = 5? X = 3? Assuming that the occurrence of diabetes is independent among the individuals in the population, we have X ~ Bin(,.). Thus, the values of P(X = ) are calculated in the following probability table and histogram. P(X = ) = (.) (.8) (.8) 9 = (.8) 85 =.486 (.8) 8 = (.8) 75 = (.)3 (.8) 7 =.59 X ~ Bin(,.) μ = (b) X are diabetics? X 5? X? X 5? X 3? Method : Directly sum the eact binomial probabilities to obtain P(X ). For instance, the cumulative probability P(X ) = (.) (.8) + (.) (.8) 99 + (.) (.8) (.) 3 (.8) (.) 4 (.8) (.) 5 (.8) (.) 6 (.8) (.) 7 (.8) (.) 8 (.8) (.) 9 (.8) 9 + (.8) 9 =.57

16 Ismor Fischer, 8//8 Stat 54 / 4-8 Method : Despite the skew, X ~ N(μ, σ), approimately (a consequence of the Central Limit Theorem, 5.), with mean μ = nπ, and standard deviation σ = nπ ( π). Hence, becomes Z = Z = X μ ~ N(, ) σ X nπ nπ ( π) ~ N(, ). In this eample, μ = nπ = ()(.) =, and σ = nπ ( π) = (.)(.8) = 4. So, approimately, X ~ N(, 4); thus Z = X 4 ~ N(, ). X N(, 4) μ = For instance, P(X ) P Z 4 = P(Z.5) =.6. The following table compares the two methods for finding P(X ). Binomial Normal Normal (eact) (approimation) (with correction) Comment: The normal approimation to the binomial generally works well, provided nπ 5 and n( π) 5. A modification eists, which adjusts for the difference between the discrete and continuous distributions: Z = X nπ ±.5 nπ ( π) ~ N(, ) where the continuity correction factor is equal to +.5 for P(X ), and.5 for P(X ). In this eample, the corrected formula becomes Z = X ~ N(, ).

17 Ismor Fischer, 8//8 Stat 54 / 4-9 Eercise: Recall the preceding section, where a spontaneous medical condition affects % (i.e., π =.) of the population, and X = number of affected individuals in a random sample of n = 3. Previously, we calculated the probability P(X = ) for =,,, 3. We now ask for the more meaningful cumulative probability P(X ), for =,,, 3, 4,... Rather than summing the eact binomial (or the approimate Poisson) probabilities as in Method above, adopt the technique in Method, both with continuity correction and without. Compare these values with the eact binomial sums. A Word about Probability Zero Events (Much Ado About Nothing?) Eactly what does it mean to say that an event E has zero probability of occurrence, i.e. P(E) =? A common, informal interpretation of this statement is that the event cannot happen and, in many cases, this is indeed true. For eample, if X = Sum of two dice, then X = 4, X = 5.7, and X = 3 all have probability zero because they are impossible outcomes of this eperiment, i.e., they are not in the sample space {, 3, 4, 5, 6, 7, 8, 9,,, }. However, in a formal mathematical sense, this interpretation is too restrictive. For eample, consider the following scenario: Suppose that k people participate in a lottery; each individual holds one ticket with a unique integer from the sample space {,, 3,, k}. The winner is determined by a computer that randomly selects one of these k integers with equal likelihood. Hence, the probability that a randomly selected individual wins is equal to /k. The larger the number k of participants, the smaller the probability /k that any particular person will win. Now, for the sake of argument, suppose that there is an infinite number of participants; a computer randomly selects one integer from the sample space {,, 3, }. The probability that a randomly selected individual wins is therefore less than /k for any k, i.e., arbitrarily small, hence =.* But by design, someone must win the lottery, so probability zero does not necessarily translate into the event cannot happen. So what does it mean? Recall that the formal, classical definition of the probability P(E) of any event E is the #(E occurs) mathematical limiting value of the ratio # trials, as # trials. That is, the fraction of the number of times that the event occurs to the total number of eperimental trials, as the eperiment is repeated indefinitely. If, in principle, this ratio becomes arbitrarily small after sufficiently many trials, then such an everincreasingly rare event E is formally identified with having probability zero (such as, perhaps, the random toss of a coin under ordinary conditions resulting in it landing on edge, rather than on heads or tails). * Similarly, any event consisting of a finite subset of an infinite sample space of possible outcomes (such as the event of randomly selecting a single particular value from a continuous interval), has a mathematical probability of zero.

18 Ismor Fischer, 8//8 Stat 54 / 4-3 Uniform f() = Normal For σ >, f() = Log-Normal b a, π σ Classical Continuous Probability Densities (The t and F distributions will be handled separately.) e μ σ a b Consequently, F() = a b a., < < +. Notes on the Gamma and Beta Functions Def: Γ(α) = α e d Thm: Γ(α) = (α ) Γ(α ); therefore, = (α )!, if α =,, 3, Thm: Γ(/) = π Def: Thm: Β(α, β) = α ( ) β d Β(α, β) = Γ(α) Γ(β) Γ(α + β) For β >, f() = π β e ln α β,. Gamma For α >, β >, f() = β α α e /β, Γ(α). Chi-Squared: For ν =,, f() = ν/ Γ(ν/) ν/ e /,. Eponential f() = β e /β,. Thus, F() = e /β. Weibull For α >, β >, f() = α β β e Thus, F() = β α e β α,.. Beta For α >, β >, f() = Β(α, β) α ( ) β,.