EE 178/278A Probabilistic Systems Analysis. Spring 2014 Tse/Hussami Lecture 11. A Brief Introduction to Continuous Probability

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1 EE 178/278A Probabiistic Systems Anaysis Spring 2014 Tse/Hussami Lecture 11 A Brief Introduction to Continuous Probabiity Up to now we have focused excusivey on discrete probabiity spaces Ω, where the number of outcomes ω Ω is either finite or countaby infinite (such as the integers). As a consequence we have ony been abe to tak about discrete random variabes, which take on ony a finite (or countaby infinite) number of vaues. But in rea ife many quantities that we wish to mode probabiisticay are continuous-vaued; exampes incude the position of a partice in a box, the time at which a certain incident happens, or the direction of trave of a meteorite. In this ecture, we discuss how to extend the concepts we ve seen in the discrete setting to this continuous setting. As we sha see, everything transates in a natura way once we have set up the right framework. The framework invoves some eementary cacuus. Continuous uniform probabiity spaces Suppose we spin a whee of fortune and record the position of the pointer on the outer circumference of the whee. Assuming that the circumference is of ength and that the whee is unbiased, the position is presumaby equay ikey to take on any vaue in the rea interva [0,]. How do we mode this experiment using a probabiity space? Consider for a moment the (amost) anaogous discrete setting, where the pointer can stop ony at a finite number m of positions distributed eveny around the whee. (If m is very arge, then presumaby this is in some sense simiar to the continuous setting.) Then we woud mode this situation using the discrete sampe space Ω = {0, m, 2 (m 1) m,..., m }, with uniform probabiities P(ω) = 1 m for each ω Ω. In the continuous word, however, we get into troube if we try the same approach. If we et ω range over a rea numbers in [0,], what vaue shoud we assign to each P(ω)? By uniformity this probabiity shoud be the same for a ω, but then if we assign to it any positive vaue, the sum of a probabiities P(ω) for ω Ω wi be! Thus P(ω) must be zero for a ω Ω. But if a of our outcomes have probabiity zero, then we are unabe to assign meaningfu probabiities to any events! To rescue this situation, consider instead any non-empty interva [a, b] [0, ]. Can we assign a non-zero probabiity vaue to this interva? Since the tota probabiity assigned to [0,] must be 1, and since we want our probabiity to be uniform, the natura assignment of probabiity to the interva [a,b] is ength of [a,b] P([a,b]) = ength of [0,] = b a. (1) In other words, the probabiity of an interva is proportiona to its ength. Note that intervas are subsets of the sampe space Ω and are therefore events. So in continuous probabiity, we are assigning probabiities to certain basic events, in contrast to discrete probabiity, where we assigned probabiity to points in the sampe space. But what about probabiities of other events? Actuay, by specifying the probabiity of intervas we have aso specified the probabiity of any event E which can be written EE 178/278A, Spring 2014, Lecture 11 1

2 Figure 1: The event E as a subset of the sampe space Ω. as the disjoint union of (a finite or countaby infinite number of) intervas, E = i E i. For then we can write P(E) = i P(E i ), in anaogous fashion to the discrete case. Thus for exampe the probabiity that the pointer ends up in the first or third quadrants of the whee is /4 + /4 = 1 2. For a practica purposes, such events are a we reay need. 1 An exampe: Buffon s neede Here is a simpe appication of continuous probabiity to the anaysis of a cassica procedure for estimating the vaue of π known as Buffon s neede, after its 18th century inventor Georges-Louis Lecerc, Comte de Buffon. Here we are given a neede of ength, and a board rued with horizonta ines at distance apart. The experiment consists of throwing the neede randomy onto the board and observing whether or not it crosses one of the ines. We sha see beow that (assuming a perfecty random throw) the probabiity of this event is exacty 2/π. This means that, if we perform the experiment many times and record the proportion of throws on which the neede crosses a ine, then the Law of Large Numbers (Lecture Note 10) tes us that we wi get a good estimate of the quantity 2/π, and therefore aso of π; and we can use Chebyshev s inequaity as in the other estimation probems we considered in that same Lecture Note to determine how many throws we need in order to achieve specified accuracy and confidence. To anayze the experiment, we first need to specify the probabiity space. Note that the position where the neede ands is competey specified by two numbers: the vertica distance y between the midpoint of the neede and the cosest horizonta ine, and the ange θ between the neede and the vertica. The vertica distance y ranges between 0 and /2, whie θ ranges between π/2 and π/2. Thus, the sampe space is the rectange Ω = [ π/2,π/2] [0,/2]. Note that, compared to the whee-of-fortune exampe, the sampe space is two-dimensiona rather than one-dimensiona. But ike the whee-of-fortune exampe, the sampe space is aso continuous. Now et E denote the event that the neede crosses a ine. It is a subset of the sampe space Ω. We need to identify this subset expicity. By eementary geometry the vertica distance of the endpoint of the neede from its midpoint is 2 cosθ, so the neede wi cross the ine if and ony if y 2 cosθ. The event E is sketched in Figure 1. 1 A forma treatment of which events can be assigned a we-defined probabiity requires a discussion of measure theory, which is beyond the scope of this course. EE 178/278A, Spring 2014, Lecture 11 2

3 Since we are assuming a competey random throw, probabiity of the event E is: P(E) = area of E area of Ω. This is the two-dimensiona generaization of equation (1) in the whee-of-fortune exampe, where the probabiity of anding in an interva is proportiona to the ength of the interva. The area of the whoe sampe space is π/2. The area of E is: π/2 [ ] π/2 cosθ dθ = π/2 2 2 sinθ =. π/2 Hence, P(E) = π/2 = 2 π. This is exacty what we caimed at the beginning of the section! Continuous random variabes Reca that in the discrete setting we typicay work with random variabes and their distributions, rather than directy with probabiity spaces and events. This is even more so in continuous probabiity, since numerica quantities are amost aways invoved. In the whee-of-fortune exampe, the position X of the pointer is a random variabe. In the Buffon neede exampe, the vertica distance Y and the ange Θ are random variabes. In fact, they are a continuous random variabes. These random variabes are a reativey simpe, in the sense that they are a uniformy distributed on the range of vaues they can take on. (Because of this simpicity, we didn t even need to worry about the random variabes expicity when cacuating probabiities in these exampes, and instead reason directy with the sampe space.) But more compicated random variabes do not have a uniform distribution. How, precisey, shoud we define the distribution of a genera continuous random variabe? In the discrete case the distribution of a r.v. X is described by specifying, for each possibe vaue a, the probabiity P(X = a). But for the r.v. X corresponding to the position of the pointer, we have P(X = a) = 0 for every a, so we run into the same probem as we encountered above in defining the probabiity space. The resoution is essentiay the same: instead of specifying P(X = a), we instead specify P(a < X b) for a intervas [a,b]. 2 To do this formay, we need to introduce the concept of a probabiity density function (sometimes referred to just as a density, or a pdf ). Definition 11.1 (Density): A probabiity density function for a random variabe X is a function f : R R satisfying b P(a < X b) = f (x)dx for a a b. (2) a Let s examine this definition. Note that the definite integra is just the area under the curve f between the vaues a and b (Figure 2(a)). Thus f pays a simiar roe to the histogram we sometimes draw to picture the distribution of a discrete random variabe. In order for the definition to make sense, f must obey certain properties. Some of these are technica in nature, which basicay just ensure that the integra is aways we defined; we sha not dwe on this issue 2 Note that it does not matter whether or not we incude the endpoints a,b; since P(X = a) = P(X = b) = 0, we have P(a < X < b) = P(a < X b) = P(a < X < b). EE 178/278A, Spring 2014, Lecture 11 3

4 Figure 2: (a) The area under the density curve between a and b is the probabiity that the random variabe ies in that range. (b) For sma δ, the area under the curve between x and x + δ can be we approximated by the area of the rectange of height f (x) and width δ. here since a the densities that we wi meet wi be we behaved. What about some more basic properties of f? First, it must be the case that f is a non-negative function; for if f took on negative vaues we coud find an interva in which the integra is negative, so we woud have a negative probabiity for some event! Second, since the r.v. X must take on some vaue everywhere in the space, we must have f (x)dx = P( < X < ) = 1. (3) In other words, the tota area under the curve f must be 1. A caveat is in order here. Foowing the histogram anaogy above, it is tempting to think of f (x) as a probabiity. However, f (x) doesn t itsef correspond to the probabiity of anything! For one thing, there is no requirement that f (x) be bounded by 1 (and indeed, we sha see exampes of densities in which f (x) is greater than 1 for some x). To connect f (x) with probabiities, we need to ook at a very sma interva [x,x +δ] cose to x. Assuming that the interva [x,x +δ] is so sma that the function f doesn t change much over that interva. we have x+δ P(x < X x + δ) = f (z)dz δ f (x). (4) x This approximation is iustrated in Figure 2(b). Equivaenty, f (x) P(x < X x + δ). δ The approximation in (5) becomes more accurate as δ becomes sma. Hence, more formay, we can reate density and probabiity by taking imits: P(x < X x + δ) f (x) = im. (6) δ 0 δ Thus we can interpret f (x) as the probabiity per unit ength in the vicinity of x. Note that whie the equation (2) aows us to compute probabiities given the probabiity density function, the equation (6) aows us to compute the probabiity density function given probabiities. Both reationships are usefu in probems. (5) EE 178/278A, Spring 2014, Lecture 11 4

5 Figure 3: The density function of the whee-of-fortune r.v. X. Now et s go back and put our whee-of-fortune r.v. X into this framework. What shoud be the density of X? We, we want X to have non-zero probabiity ony on the interva [0,], so we shoud certainy have f (x) = 0 for x < 0 and for x >. Within the interva [0,] we want the distribution of X to be uniform, which means we shoud take f (x) = c for 0 x. What shoud be the vaue of c? This is determined by the requirement (3) that the tota area under f is 1. The area under the above curve is f (x)dx = 0 cdx = c, so we must take c = 1. Summarizing, then, the density of the uniform distribution on [0,] is given by 0 for x < 0; f (x) = 1/ for 0 x ; 0 for x >. This is potted in Figure 3. Note that f (x) can certainy be greater than 1, depending on the vaue of. Another Exampe: Suppose you throw a dart and it ands uniformy at random on a target which is a disk of unit radius. What is the probabiity density function of the distance of the dart from the center of the disk? Let X be the distance of the dart from the center of the disk. We first cacuate the probabiity that X is between x and x + δ. If x is negative or greater than or equa to 1, this probabiity is zero, so we focus on the case that x is between 0 and 1. The event in question is that the dart ands in the ring (annuus) shown in Figure 4. Since the dart ands uniformy at random on the disk, the probabiity of the event is just the ratio of the area of the ring and the area of the disk. Hence, P(x < X x + δ) = π[(x + δ)2 x 2 ] π(1) 2 (7) = x 2 + 2δx + δ 2 x 2 = 2δx δ 2. Using equation (6), we can now compute the probabiity density function of X: P(x < X x + δ) 2δx δ 2 f (x) = im = f (x) = im = 2x. δ 0 δ δ 0 δ Summarizing, we have 0 for x 0; f (x) = 2x for 0 x < 1; 0 for x 1. EE 178/278A, Spring 2014, Lecture 11 5

6 Figure 4: The sampe space is the disk of unit radius. The event of interest is the ring. Figure 5: (a) The probabiity density function and (b) the cumuative distribution function of the distance X from the target center. It is potted in Figure 5(a). Note that athough the dart ands uniformy inside the target, the distance X from the center is not uniformy distributed in the range from 0 to 1. This is because an ring farther away from the center has a arger area than an ring coser to the center with the same width δ. Hence the probabiity of anding in the ring farther away from the center is arger. Cumuative Distribution Function Let us re-interpret equation (7) in the dart throwing exampe above. In words, we are saying: P(x < X x + δ) = area of ring area of target = (area of disk of radius x + δ) (area of disk of radius x) area of target = area of disk of radius x + δ area of disk of radius x area of target area of target = P(X x + δ) P(X x). EE 178/278A, Spring 2014, Lecture 11 6

7 This ast equaity can be understood directy as foows. The event A that X x + δ (dart ands inside disk of radius x + δ) can be decomposed as a union of two events: 1) the event B that X x (dart ands inside disk of radius x), and 2) the event C that x < X x + δ (dart ands inside ring). The two events are disjoint. (See Figure 4.) Hence, P(A) = P(B) + P(C) or P(x < X x + δ) = P(X x + δ) P(X x), (8) which is exacty the same as above. Ceary, the reasoning eading to (8) has nothing much to do the particuars of this exampe but in fact (8) hods true for any random variabe X. A we needed are the facts that A = B C and B and C are disjoint events, and the facts are true in genera. Substituting (8) into (6), we obtain: P(X x + δ) P(X x) f (x) = im. δ 0 δ What does this equation remind you of? To make things even more expicit, et us define the cumuative distribution function F(x) = P(X x). (9) Then we have: F(x + δ) F(x) f (x) = im = d F(x). (10) δ 0 δ dx The function F has a name: it is caed the cumuative distribution function of the random variabe X (sometimes abbreviated as cdf). It is caed cumuative because at each vaue x, F(x) is the cumuative probabiity up to x. Note that the cumuative distribution function and the probabiity density function of a random variabe contains exacty the same information. Given the cumuative distribution function F, one can differentiate to get the probabiity density function f. Given the probabiity density function f, one can integrate to get the cumuative distribution function: F(x) = x f (a)da. So stricty speaking, one does not need to introduce the concept of cumuative distribution function at a. However, for many probems, the cumuative distribution function is easier to compute first and from that one can then compute the probabiity density function. Summarizing, we have: Definition 11.2 (Cumuative Distribution Function): The cumuative distribution function for a random variabe X is a function F : R R defined to be: F(x) = P(X x). (11) Its reationship with the probabiity density function f of X is given by f (x) = d dx F(x), F(x) = x f (a)da. EE 178/278A, Spring 2014, Lecture 11 7

8 The cumuative distribution function satisfies the foowing properties: 1. 0 F(x) 1 2. im x F(x) = 0 3. im x F(x) = 1 EE 178/278A, Spring 2014, Lecture 11 8

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