Chapter 2 - Random Variables

Transcription

1 Chapter - Random Variables In this and the chapters that follow, we denote the real line as R = (- < < ), and the etended real line is denoted as R + R{±}. The etended real line is the real line with ± thrown in. Put simply (and incompletely), a random variable is a function that maps the sample space S into the etended real line. Eample -1: In the die eperiment, we assign to the si outcomes f i the numbers X(f i ) = 10i. Thus, we have X(f 1 ) = 10, X(f ) = 0, X(f 3 ) = 30, X(f 4 ) = 40, X(f 5 ) = 50, X(f 6 ) = 60. For an arbitrary value 0, we must be able to answer questions like what is the probability that random variable X is less than, or equal to, 0? Hence, the set { S : X() 0 } must be an event (i.e., the set must belong to -algebra F ) for every 0 (sometimes, the algebraic, non-random variable 0 is said to be a realization of the random variable X). This leads to the more formal definition. Definition: Given a probability space (S, F, P), a random variable X() is a function + X: S R. (-1) That is, random variable X is a function that maps sample space S into the etended real line R +. In addition, random variable X must satisfy the two criteria discussed below. 1) Recall the Borel -algebra B of subsets of R that was discussed in Chapter 1 (see Eample 1-9). For each B B, we must have 1 X (B) S : X( ) B F, BB. (-) A function that satisfies this criteria is said to be measurable. A random variable X must be a measurable function. ) P[ S : X() = ± ] = 0. Random variable X is allowed to take on the values of ± ; Updates at -1

2 however, it must take on the values of ± with a probability of zero. These two conditions hold for most elementary applications. Usually, they are treated as mere technicalities that impose no real limitations on real applications of random variables (usually, they are not given much thought). However, good reasons eist for requiring that random variable X satisfy the conditions 1) and ) listed above. In our eperiment, recall that sample space S describes the set of elementary outcomes. Now, it may happen that we cannot directly observe elementary outcomes S. Instead, we may be forced to use a measuring instrument (i.e., random variable) that would provide us with measurements X(), S. Now, for each R, we need to be able to compute the probability P[- < X() ], because [- < X() is a meaningful, observable event in the contet of our eperiment/measurements. For probability P[- < X() ] to eist, we must have [- < X() as an event; that is, we must have S : X( ) F (-3) for each R. It is possible to show that Conditions (-) and (-3) are equivalent. So, while (-) (or the equivalent (-3)) may be a mere technicality, it is an important technicality. Sigma-Algebra Generated by Random Variable X Suppose that we are given a probability space (S, F, P) and a random variable X as described above. Random variable X induces a -algebra (X) on S. (X) consists of all sets of the form { S : X() B, B B}, where B denotes the -algebra of Borel subsets of R that was discussed in Chapter 1 (see Eample 1-9). Note that (X) F ; we say that (X) is the sub -algebra of F that is generated by random variable X. Probability Space Induced by Random Variable X Suppose that we are given a probability space (S, F, P) and a random variable X as described above. Let B be the Borel -algebra introduced by Eample 1-9. By (-), for each B B, we have { S : X() B} F, so P[{ S : X() B}] is well defined. Updates at -

3 This allows us to use random variable X to define (R +, B, P), a probability space induced by X. Probability measure P is defined as follows: for each B B, we define P(B) P[{ S : X() B]. We say that P induces probability measure P. Distribution and Density Functions The distribution function of the random variable X() is the function F() P[X( ) ] P[ S : X( ) ], (-4) where <. Eample -: Consider the coin tossing eperiment with P[heads] = p and P[tails] = q 1 - p. Define the random variable X(head) = 1 X(tail) = 0. If 1, then both X(head) = 1 and X(tail) = 0 so that F() 1 for 1. If 0 1, then X(head) = 1 and X(tail) = 0 so that F() P [X ] q for 0 < 1 Finally, if 0, then both X(head) = 1 and X(tail) = 0 so that F() P [X ] 0 for < 0. Updates at -3

4 1 F() See Figure -1 for a graph of F(). Properties of Distribution Functions First, some standard notation: q Figure -1: Distribution function for X(head) =1, X(tail) = 0 random variable. 1 F( ) limit F(r) and F( ) limit F(r). r r Some properties of distribution functions are listed below. Claim #1: F(+) = 1 and F(-) = 0. (-5) Proof: F(+ ) = limit P[X ] = P[ S ]=1 and F(- ) = limit P[X ] = P[{ }]=0. Claim #: The distribution function is a non-decreasing function of. If 1, then F( 1 ) F( ). Proof: 1 implies that {X() 1 } {X() }. But this means that P[{X() 1 }] P[{X() }] and F( 1 ) F( ). Claim #3: P[X > ] = 1 - F(). (-6) Proof: {X() 1 } and {X() 1 } are mutually eclusive. Also, {X() 1 }{X() 1 } = Updates at -4

5 S. Hence, P[{X 1 }] + P[{X 1 }] = P[S] = 1. Claim #4: Function F() may have jump discontinuities. It can be shown that a jump is the only type of discontinuity that is possible for distribution F() (and, F() may have a countable number of jumps, at most). F() must be right continuous; that is, we must have F( + ) = F(). At a jump, take F to be the larger value; see Figure -. Claim #5: P[ 1 X ] = F( ) - F( 1 ) (-7) Proof: {X() 1 } and { 1 X() } are mutually eclusive. Also, {X() } = {X() 1 } { 1 X() }. Hence, P[X( ) ] P[X( ) 1] + P [ 1< X( ) ] and P[ 1 X() ] = F( ) - F( 1 ). Claim #6: P[X ] = F( ) - F( ). (-8) Proof: P[ - X ] = F() - F( - ). Now, take limit as to obtain the desired result. Claim #7: P[ 1 X ] F( ) - F( 1 ). (-9) Proof: { 1 X } { 1 X }{X 1 } so that P[ X ] = ( F( ) F( ) )+( F( ) F( ) )= F( ) F( ). 1 F( 0 ) 0 Figure -: Distributions are right continuous. Updates at -5

6 F () 1 Figure -3: Distribution function for a discrete random variable. Continuous Random Variables Random variable X is of continuous type if F () is continuous. In this case, P[X = ] = 0; the probability is zero that X takes on a given value. Discrete Random Variables Random variable X is of discrete type if F () is piece-wise constant. The distribution should look like a staircase. Denote by i the points where F () is discontinuous. Then F ( i ) F ( i ) = P[X = i ] = p i. See Figure -3. Mied Random Variables Random variable X is said to be of mied type if F () is discontinuous but not a staircase. Density Function The derivative df () f () (-10) d is called the density function of the random variable X. Suppose F has a jump discontinuity at a point 0. Then f() contains the term Updates at -6

7 Distribution Function F () } k Jump Discontinuity at 0 0 Density Function f () Delta Function at 0 k( - 0 ) Figure -4: "Jumps" in distribution function causes delta function in density function. The distribution jumps by the value k at = 0. 0 F( 0 ) F( 0 ) () 0 F() 0 F( 0 ) () 0. (-11) See Figure -4. Suppose that X is of a discrete type taking values i, i I. The density can be written as f () P[X ] (), i i ii where I is an inde set. Figures -5 and -6 illustrate the distribution and density, respectively, of a discrete random variable. Properties of f The monotonicity of F implies that f () 0 for all. Furthermore, we have df () f () F () f ( )d d (-1) Updates at -7

8 1 Distribution Function F () k 1 UVW k q k 3 } k Figure -5: Distribution function for a discrete random variable. Density Function f () k 1 k k 3 k Figure -6: Density function for a discrete random variable. P[ 1 < X ] = F ( ) - F ( 1 ) = f ( )d. (-13) 1 If X is of continuous type, then F () = F ( - ), and P[ 1 X ] = F ( ) - F ( 1 ) = f ( )d 1. For continuous random variables P[ X + ] f () for small. Normal/Gaussian Random Variables Let,, and, > 0, be constants. Then f () g ep[ ] (-14) 1 1 f() ep[ ] 1.0 F() 0.5 F() 1 ep[ ½ ]d Figure -7: Density and distribution functions for a Gaussian random variable with = 0 and = 1. Updates at -8

9 is a Gaussian density function with parameters and. These parameters have special meanings, as discussed below. The notation N(;) is used to indicate a Gaussian random variable with parameters and. Figure -7 illustrates Gaussian density and distribution functions. Random variable X is said to be Gaussian if its distribution function is given by 1 ( ) F() P [X ] ep d (-15) for given, - < <, and, > 0. Numerical values for F() can be determined with the aid of a table. To accomplish this, make the change of variable y = ( - )/ in (-15) and obtain ( )/ 1 y F() ep dy G. (-16) Function G() is tabulated in many reference books, and it is built in to many popular computer math packages (i.e., Matlab, Mathcad, etc.). Uniform Random variable X is uniform between 1 and if its density is constant on the interval [ 1, ] and zero elsewhere. Figure -8 illustrates the distribution and density of a uniform f () F () Figure -8: Density and distribution functions for a uniform random variable. Updates at -9

10 random variable. Binomial Random variable X has a binomial distribution of order n with parameter p if it takes the integer values 0, 1,..., n with probabilities n k nk P [X k] p q, 0 k n, 0 p 1, p q 1. (-17) k Both n and p are known parameters where p + q = 1, and n n!. (-18) k k!(n k)! We say that binomial random variable X is B(n,p). The Binomial density function is n f () p q (k) k n k nk, (-19) k0 and the Binomial distribution is m n k nk F() pq, k0k m m 1n 1, n. (-0) Note that m depends on. Updates at -10

11 Poisson with Random variable X is Poisson with parameter a > 0 if it takes on integer values 0, 1,... k a a [X k] e, k 0, 1,, 3,... k! P (-1) The density and distribution of a Poisson random variable are given by k0 k a a f () e (k) (-) k! m k a a F () e m m 1, m 0, 1,.... k! k0 (-3) Rayleigh The random variable X is Rayleigh distributed with real-valued parameter, > 0, if it is described by the density f () ep 1, 0 (-4) 0, 0. See Figure -9 for a depiction of a Rayleigh density function. The distribution function for a Rayleigh random variable is Updates at -11

12 1 ep 1 1 f( ) ep 1 ( ) ej U L NM O QP f( ) ep U ( ) / Figure -9: Rayleigh density function. u F() e du, 0 u / Figure -10: Eponential density function. (-5) 0, 0 To evaluate (-5), use the change of variable y = u /, dy = (u/ )du to obtain / y / F() e dy 1e, 0 0 (-6) 0, 0 as the distribution function for a Rayleigh random variable. Eponential The random variable X is eponentially distributed with real-valued parameter, > 0, if it is described by the density f () ep, 0 0, 0. (-7) See Figure -10 for a depiction of an eponential density function. Updates at -1

13 The distribution function for an eponential random variable is y 0 F() e dy 1e, 0 (-8) 0, 0. Conditional Distribution Let M denote an event for which P[M] 0. Assuming the occurrence of event M, the conditional distribution F(M) of random variable X is defined as [X,M] F(M) P[X M] P P[M]. (-9) Note that F(M) = 1 and F(-M) = 0. Furthermore, the conditional distribution has all of the properties of an "ordinary" distribution function. For eample, [1 X,M] P[1 X M] F(M) F(1M) P P[M]. (-30) Conditional Density The conditional density f(m) is defined as d [ X M] f (M) F(M) limit d 0 P. (-31) The conditional density has all of the properties of an "ordinary" density function. Eample -3: Determine the conditional distribution F(M) of random variable X(f i ) = 10i of the fair-die eperiment where M={f, f 4, f 6 } is the event "even" has occurred. First, note that X must take on values in the set {10, 0, 30, 40, 50, 60}. Hence, if 60, then [X ] is the Updates at -13

14 certain event and [X, M] = M. Because of this, P[X,M] P[M] F(M) 1, 60 P[M] P[M]. If 40 < 60, then [X, M] = [f, f 4 ], and P[X,M] P[f,f 4] /6 F(M), P[M] P[f,f,f ] 3/ If 0 < 40, then [X, M] = [f ], and P[X,M] P[f ] 1/6 F(M), P[M] P[f,f,f ] 3/6 4 6 Finally, if < 0, then [X, M] = [], and P[X,M] P[{ ] 0 F(M), 0. P[M] P[f,f,f ] 3/6 4 6 Conditional Distribution When Event M is Defined in Terms of X If M is an event that can be epressed in terms of the random variable X, then F(M) can be determined from the "ordinary" distribution F (). Below, we give several eamples. As a first eample, consider M = [X a], and find both F(M) = F(X a) = P[X X a] and f(m). Note that P[X X a] F(X a) P[XX a] P[X a]. Hence, if a, we have [X, X a] = [X a] and Updates at -14

15 P[X a] F(X a) 1, a. P[X a] If a, then [X, X a] = [X ] and P[X ] F () F(X a), a. P[X a] F (a) At = a, F(X a) would jump 1 - F X (a - )/F X (a) if F X (a - ) F X (a). The conditional density is f (), a F(a) d d F () F (a ) f(x a) F(X a) 1 (a). d d F (a) F (a) 0, a As a second eample, consider M = [b < X a] so that P[X b X a] F(b X a). P[b X a] Since [X, b < X a] = [b < X a], a [b < X ], b < < a = { }, b, we have Updates at -15

16 F(b < X a) 1, a F() F(b), F(a) F(b) b a 0, b. F(b < X a) is continuous at = b. At = a, F(b < X a) jumps 1 - {F X (a ) - F X (b)}/{f X (a) - F X (b)}, a value that is zero if F X (a ) F X (a). The corresponding conditional density is d f(b Xa) F(b Xa) d f (), b a F(a) F(b) F(a ) F(b) 1 (a). F(a) F(b) 0, otherwise Eample -4: Find f(x - ), where X is N(). First, note that F() P[X] G d 1 f () G g, d where G is the distribution, and g is the density, for the case = 0 and = 1. Also, note that the inequality X implies X so that Updates at -16

17 P[ X] F ( ) F ( ) G( ) G( ) G( ) 1. By the previous eample, we have F( X ) 1, G G( ), G( ) 1 0,, a conditional distribution that is a continuous function of. Differentiate this to obtain 1 1 ep 1 g f( X ), G( ) 1 G( ) 1 0, otherwise, a conditional density that contains no delta functions. Distributions/Densities Epressed In Terms of Conditional Distributions/Densities Let X be any random variable and define B = [X ]. Let A 1, A,..., A n be a partition of sample space S. That is, n A i S and Ai A j { for i j. (-3) i=1 From the discrete form of the Theorem of Total Probability discussed in Chapter 1, we have P[B] = P[B A 1 ]P[A 1 ] + P[B A ]P[A ] P[B A n ]P[A n ]. Now, with B = [X ], this becomes Updates at -17

18 P[X ] = P[X A 1 ]P[A 1 ] + P[X A ]P[A ] P[X A n ]P[A n ]. (-33) Hence, F () = F(A ) P[A ] + F(A ) P[A ] F(A ) P[A ] 1 1 n n f () = f(a ) P[A ] + f(a ) P[A ] f(a ) P[A ]. 1 1 n n (-34) Total Probability Continuous Form The probability of an event can be conditioned on an event defined in terms of a random variable. For eample, let A be any event, and let X be any random variable. Then we can write P[AX ] P[X A] [A] F( A) [A] [A X ] P P P. P[X ] P[X ] F () (-35) As a second eample, we derive a formula for P[A 1 < X ]. Now, the conditional distribution F(A) has the same properties as an "ordinary" distribution. That is, we can write P[ 1 < X A] = F( A) - F( 1 A) so that P[1 X P[A 1X ] P[A] P[1 X F( F(1 P[A]. F( F( 1 (-36) In general, the formulation P[AX ] P[AX ] P[X ] (-37) Updates at -18

19 may result in an indeterminant 0/0 form. Instead, we should write F( A) F(A) P[AX ] limit P[A X ] limit P[A] F() F() [F( A) F(A)]/ limit P(A), [F ( ) F ()]/ + 0 (-38) which yields f(a) P[AX ] P [A]. (-39) f () Now, multiply both sides of this last result by f and integrate to obtain [AX ]f ()d [A] f (A)d P P. (-40) But, the area under f(a) is unity. Hence, we obtain [A] [AX ]f ()d P P, (-41) the continuous version of the Total Probability Theorem. In Chapter 1, we gave a finite dimensional version of this theorem. Compare (-41) with the result of Theorem 1-1; conceptually, they are similar. P[AX = ] is the probability of A given that X =. It is a function of. Equation (-41) tells us to average this function over all possible values of X to find P[A]. For the Total Probability Theorem, the discrete and continuous cases can be related by the following simple, intuitive argument. We start with the discrete version (see Theorem 1.1, Updates at -19

20 Chapter 1 of class notes). For small > 0 and integer i, < i <, define the event X ( +1) : X( ) ( +1) Ai i i S i i. Note that the A i, < i <, form a partition of the sample space. Let A be an arbitrary event. By Theorem 1-1 of Chapter 1, the discrete version of the Total Probability Theorem, we can write P P P i A AAi Ai PA i X ( i+1) Pi X ( i+1) i P AX i f () i, i where i is any point in the infinitesimal interval (i, (i + 1)]. In the limit as approaches zero, the infinitesimal becomes d, and the sum becomes an integral so that A P P A X= f ()d, the continuous version of the Theorem of Total Probability. Eample -5 (Theorem of Total Probability) Suppose we have a large set S c of distinguishable coins. We do not know the probability of heads for any of these coins. Suppose we randomly select a coin from S c (all coins are equally likely to be selected). Its probability of heads will be estimated by a method that uses eperimental data obtained from tossing the selected coin. This method is developed in this and Updates at -0

21 the net eample. For each coin in S c, we model its probability of heads as a continuous random variable p, 0 p 1, described by known density f p (p). Density f p (p) should peak at or near p = ½ and decrease as p deviates from ½. After all, most coins are well balanced and nearly fair. Also, note the requirements p 1 p p P p p p f (p)dp. (-4) p 1 f 0 p (p)dp1 P. (-43) Realistically, it is likely that we do not know f p (p) eactly. We may make a good guess ; on the other hand, we could choose a uniform density for f p (p), a choice that implies considerable ignorance. As shown in Eample -6, given a specific coin S c, it may be possible to use eperimental data and estimate its probability of heads without knowledge of f p (p). Consider a combined, or joint, eperiment of randomly selecting a coin from S c (the first eperiment) and then tossing it once (the second eperiment, with sample space S = [h, t], is independent of the first eperiment). For this combined eperiment, the relevant product sample space is (,h): (,t): S S S S. (-44) c c c For the combined eperiment, the event we get a heads or heads occurs is H (,h): S c. (-45) Updates at -1

22 The conditional probability of H, given that the tossed coin has (Probability of Head) = p (i.e., p p), is simply PH p p P[(,h): S c] p p p. (-46) The Theorem of Total Probability can be used to epress P[H] in terms of known f p (p). Simply use (-46) and the Theorem of Total Probability (-41) to write 1 1 p p f 0 p(p)dp pf 0 p(p)dp PH P H. (-47) On the right hand side of (-47), the integral is called the epected value, or ensemble average, of the random variable p. Equation (-47) is used in Eample -6 below to compute an estimate for the probability of heads for any coin selected from S c. Bayes Theorem - Continuous Form From (-39) we get [AX ] f(a) P f (). (-48) P[A] Now, use (-41) and (-48) to write f(a) P[AX ] f (), (-49) P[AX ]f ( )d a result known as the continuous form of Bayes Theorem. Often, f X () is called the a-priori density for random variable X. And, f(a) is called the a-posteriori density conditioned on the observed event A. In an application, we might cook Updates at -

23 up a density f X () that (crudely) describes a random quantity (i.e., variable) X of interest. To improve our characterization of X, we note the occurrence of a related event A and compute f(a) to better characterize X. The value of that maimizes f(a) is called the maimum a-posteriori (MAP) estimate of X. MAP estimation is used in statistical signal processing and many other problems where one must estimate a quantity from observations of related random quantities. Given event A, the value m is the MAP estimate for X if f(m f( ), m. Eample -6: (Bayes Theorem & MAP estimate of probability of heads for Eample -5) Find the MAP estimate of the probability of heads in the coin selection and tossing eperiment that was described in Eample -5. First, recall that the probability of heads is modeled as a random variable p with density f p (p) (which we may have to guess). We call f p (p) the a- priori ("before the coin toss eperiment") density of p. Suppose we toss the coin n times and get k heads. We want to use these eperimental results with our a-priori density f p (p) to compute the conditional density f (pk heads, in a specific order, in n tosses of a selected coin) f (p. (-50) Observed Event A This density is called a-posteriori density ("after the coin toss eperiment") of random variable p given the eperimentally observed event A [k heads, in a specific order, in n tosses of a selected coin]. (-51) Updates at -3

24 f p (p) The a-posteriori density f(pa) may give us a good idea (better than the a-priori density ) of the probability of heads for the randomly selected coin that was tossed. Conceptually, think of f(pa) as a density that results from using eperimental data/observation A to update f p (p). In fact, given that A occurred, the a-posteriori probability that p is between p 1 and p is p p f(pa)dp. (-5) 1 Finally, the value of p that maimizes f(pa) is the maimum a-posteriori estimate (MAP estimate) of p for the selected coin. To find this a-posteriori density, recall that p is defined on the sample space S c. The eperiment of tossing the randomly selected coin n times is defined on the sample space (i.e., product space) S c S n, where S = [h, t]. The elements of S c S n have the form, t h t h where S and t h t h S. n outcomes c n outcomes n Now, given that p = p, the conditional probability of event A is k n k P[k heads in specific order in n tosses of specific coin p = p] = p (1 p) Observed Event A. (-53) Substitute this into the continuous form of Bayes rule (-49) to obtain k Observed Event A f (pa) f (pk heads, in a specific order, in n tosses of a selected coin ) nk p p (1 p) f (p), (1 ) f ( )d 1 k n k 0 p (-54) Updates at -4

25 a result known as the a-posteriori density of p given A. In (-54), the quantity is a dummy variable of integration. Suppose that the a-priori density f p (p) is smooth and slowly changing around the value p = k/n (indicating a lot of uncertainty in the value of p). Then, for large values of n, the numerator p k (1-p) n-k f (p) and the a-posteriori density f(pa) have a sharp peak at p = k/n, p indicating little uncertainty in the value of p. When f(pa) is peaked at p = k/n, the MAP estimate of the probability of heads (for the selected coin) is the value p = k/n. Eample -7: Continuing Eamples -5 and -6, for > 0, we use the a-priori density 1 (p1/) ep f p(p), 0p1,, (-55) 1 1 G where G() is a zero mean, unit variance Gaussian distribution function (verify that there is unit area under f p (p)). Also, use numerical integration to evaluate the denominator of (-54). For =.3, n = 10 and k = 3, a-posteriori density f(pa) was computed and the results are plotted on Figure -11. For =.3, n = 50 and k = 15, the calculation was performed a second time, and the results appear on Figure -11. For both, the MAP estimate of p is near.3, since this is were the plots of f(p A) peak. Epectation Let X denote a random variable with density f (). The epected value of X is defined as E[X] f ()d. (-56) Also, E[X] is known as the mean, or average value, of X. If f is symmetrical about some value, then is the epected value of X. For eample, Updates at -5

26 7 6 f(pa) for n = 50, k = 15 Density Functions f(pa) for n = 10, k = 3 fp~( p) 1 let X be N() so that p Figure -11: Plot of a-priori and a-posteriori densities. f() 1 ep 1 ( ). (-57) Now, (-57) is symmetrical about, so E[X] =. Suppose that random variable X is discrete and takes on the value of i with P[X = i ] = p i. Then the epected value of X, as defined by (-56), reduces to E[X] ipi. (-58) i Eample -8: Consider a B(n,p) random variable X (that is, X is Binomial with parameters n and p). Using (-58), we can write n n n k nk E[X] k P [X k] k p q. (-59) k0 k0 k Updates at -6

27 To evaluate this result, first consider the well-know, and etremely useful, Binomial epansion n n k n (1). (-60) k0 k With respect to, differentiate (-60); then, multiply the derivative by to obtain n n k n1 k n(1). (-61) k0 k Into (-61), substitute = p/q, and then multiply both sides by q n. This results in n k0 n p k p q npq (1 ) np(p q) k q k nk n1 n1 n1 np. (-6) From this and (-59), we conclude that a B(n,p) random variable has E[X] = np. (-63) We now provide a second, completely different, evaluation of E[X] for a B(n,p) random variable. Define n new random variables Xi 1 th, if i trial is a "success", 1 i n = 0, otherwise. (-64) Updates at -7

28 Note that E[X i] 1p 0 (1 p) p, 1 i n. (-65) B(n,p) random variable X is the number of successes out of n independent trials; it can be written as X X X X. (-66) 1 n With the use of (-65), the epected value of X can be evaluated as E[X] E[X X X ] E[X ] E[X ] E[X ] np, 1 n 1 n (-67) a result equivalent to (-63). Consider random variable X, with mean = E[X], and constant k. Define the new random variable Y X + k. The mean value of Y is k k k. (-68) y E[Y] E[X ] E[X] A random variable, when translated by a constant, has a mean that is translated by the same constant. If instead Y kx, the mean of Y is E[Y] = E[kX] = k. In what follows, we etend (-56) to arbitrary functions of random variable X. Let g() be any function of, and let X be any random variable with density f (). In Chapter 4, we will discuss transformations of the form Y = g(x); this defines the new random variable Y in terms of the old random variable X. We will argue that the epected value of Y can be computed as Updates at -8

29 E[Y] E[g(X)] g()f - ()d. (-69) This brief heads-up note (on what is to come in CH 4) is used net to define certain statistical averages of functions of X. Variance and Standard Deviation The variance of random variable X is Var[X] E[(X ) ] ( ) f ()d. (-70) Almost always, variance is denoted by the symbol. The square root of the variance is called the standard deviation of the random variable, and it is denoted as. Finally, variance is a measure of uncertainty (or dispersion about the mean). The smaller (alternatively, larger) is, the more (alternatively, less) likely it is for the random variable to take on values near its mean. Finally, note that (-70) is an application of the basic result (-69) with g = ( - ). Eample -9: Let X be N() then 1 ( ). (-71) VAR[X] E[(X ) ] ( ) ep d Let y = d = dy so that 1 1 y Var[X] y e dy, (-7) a result obtained by looking up the integral in a table of integrals. Consider random variable X with mean and variance = E[{X } ]. Let k denote an arbitrary constant. Define the new random variable Y kx. The variance of Y is Updates at -9

30 y E[{Y y} ] E[ {X } ] k k. (-73) The variance of kx is simply k time the variance of X. On the other hand, adding constant k to a random variable does not change the variance; that is, VAR[X+k] = VAR[X]. Consider random variable X with mean and variance problem by centering and normalizing X to define the new random variable. Often, we can simplify a X Y. (-74) Note that E[Y] = 0 and VAR[Y] = 1. Moments The n th moment of random variable X is defined as n n mn E[X ] f ()d. (-75) The n th moment about the mean is defined as n n n E[(X ) ] ( ) f ()d. (-76) Note that (-75) and (-76) are basic applications of (-69). Variance in Terms of Second Moment and Mean Note that the variance can be epressed as Var[X] E[(X ) ] E[X X ] E[X ] E[X] E[ ], (-77) where the linearity of the operator E[ ] has been used. Now, constants come out front of Updates at -30

31 epectations, and the epectation of a constant is the constant. Hence, Equation (-77) leads to E[X ] E[ X] E[ ] m E[X] m, (-78) the second moment minus the square of the mean. In what follows, this formula will be used etensively. Eample -10: Let X be N(0,) and find E[X n ]. First, consider the case of n an odd integer n, (-79) n 1 E[X ] ep 1 d 0 since an integral of an odd function over symmetrical limits is zero. Now, consider the case n an even integer. Start with the known tabulated integral 1/ e d, > 0. (-80) Repeated differentiations with respect to yields 1 3/ first d/d : e d / second d/d : e d 1 35 (k1) k d/d : e d. th k (k 1) / Let n = k (remember, this is the case n even) and = 1/ to obtain Updates at -31

32 135 (n1) e d n / (n 1) (n+1) n n (n1). (-81) From this, we conclude that the n th -moment of a zero-mean, Gaussian random variable is n n / n mn E[X ] 1 e d 135 (n 1), n = k (n even) (-8) = 0, n = k-1 (n odd). Eample -11 (Rayleigh Distribution): A random variable X is Rayleigh distributed with parameter if its density function has the form f X() ep, 0 (-83) = 0, < 0. This random variable has many applications in communication theory; for eample, it describes the envelope of a narrowband Gaussian noise process (as described in Chapter 9 of these notes). The n th moment of a Rayleigh random variable can be computed by writing n n 1 / 1 E[X ] e d 0. (-84) We consider two cases, n even and n odd. For the case n odd, we have n+1 even, and (-84) becomes n 1 n 1 / n 1 / E[X ] 1 e d 1 e d. (-85) - - Updates at -3

33 On the right-hand-side of (-85), the bracket contains the (n+1) th moment of a N(0,) random variable (compare (-8) and the right-hand-side of (-85)). Hence, we can write n n1 n E[X ] 135n 135 n /, n = k+1 (n odd). (-86) Now, consider the case n even, n+1 odd. For this case, (-84) becomes n 1 n 1 / n / E[X ] e d e d 0 0. (-87) Substitute y = /, so that dy = (/ )d, in (-87) and obtain n n n y n/ n n/ y n/ n EX y e dy y e dy 0 0, (-88)! (note that n/ is an integer here) where we have used the Gamma function k y (k 1) y e dy k! 0, (-89) k 0 an integer. Hence, for a Rayleigh random variable X, we have determined that n E[X ] 135n /, n odd n n/ n = n, n even.! (-90) In particular, Equation (-90) can be used to obtain the mean and variance Updates at -33

34 E[X] Var[X] E[X ] E[X] ( ), (-91) given that X is a Rayleigh random variable. Eample -1: Let X be Poisson with parameter so that k P [X k] e k!, k 0, and k f() e (k). (-9) k! k0 Show that E[X] = and VAR[X] = Recall that e k0 k. (-93) k! With respect to differentiate (-93) to obtain k1 1 e k k k! k! k0 k1 k (-94) Multiply both sides of this result by e - to obtain k = k e E[X] k! k1. (-95) as claimed. With respect to, differentiate (-94) (obtain the second derivative of (-93)) and obtain Updates at -34

35 k k k 1 1 k! k! k! k1 k1 k1. e k(k1) k k Multiply both sides of this result by e - to obtain k k k e k e k P[X k] k P [X k]. (-96) k! k! k1 k1 k1 k1 Note that (-96) is simply E[X ] E[X]. Finally, a Poisson random variable has a variance given by Var[X] E[X ] E[X] E[X ] E[X], (-97) as claimed. Conditional Mean Let M denote an event. The conditional density f(m) can be used to define the conditional mean E[XM] f (d -. (-98) The conditional mean has many applications, including estimation theory, detection theory, etc. Eample -13: Let X be Gaussian with zero mean and variance (i.e., X is N(0,)). Let M = [X > 0]; find E[X M] = E[XX > 0]. First, we must find the conditional density f( X > 0); from previous work in this chapter, we can write Updates at -35

36 [X,X ] [X,X ] [X,X ] F(X ) P[X ] 1- P[X ] 1 F (0) P P P 1 F (0) F() X F(0) X, 0 = 0, < 0, so that f(x0) f (), 0 0, 0. From (-98) we can write EX X 0 ep / d. 0 Now, set y = /, dy = d/ to obtain y EX X 0 e dy. 0 Tchebycheff Inequality A measure of the concentration of a random variable near its mean is its variance. Consider a random variable X with mean variance and density f X (). The larger, the more "spread-out" the density function, and the more probable it is to find values of X "far" from the mean. Let denote an arbitrary small positive number. The Tchebycheff inequality says that the probability that X is outside ( -, + ) is negligible if is sufficiently small. Theorem (Tchebycheff s Inequality) Consider random variable X with mean and variance. For any > 0, we have Updates at -36

37 X- P. (-99) Proof: Note that { : } ( ) f ()d ( ) f ()d [ f ()d ] { : } P[ X- ]. This leads to the Tchebycheff inequality X- P. (-100) The significance of Tchebycheff's inequality is that it holds for any random variable, and it can be used without eplicit knowledge of f(). However, the bound is very "conservative" (or "loose"), so it may not offer much information in some applications. For eample, consider Gaussian X. Note that X- P X- 3 1 P X- 3 1 P G(3) G( 3) G(3), (-101).007 where G(3) is obtained from a table containing values of the Gaussian integral. However, the Tchebycheff inequality gives the rather "loose" upper bound of Updates at -37

38 P X- 3 1/ 9. (-10) Certainly, inequality (-10) is correct; however, it is a very crude upper bound as can be seen from inspection of (-101). Generalizations of Tchebycheff's Inequality For a given random variable X, suppose that f X () = 0 for < 0. Then, for any > 0, we have P[ X ] (-103) To show (-103), note that E[X] f 0 ()d f ()d f ()d [X ] P, (-104) so that P[ X ] as claimed. Corollary: Let X be an arbitrary random variable and and n an arbitrary real number and positive integer, respectively. The random variable X - n takes on only nonnegative values. Hence n P n n E[ X- ] X- n, (-105) which implies n P E[ X- ] X- n. (-106) Updates at -38

39 The Tchebycheff inequality is a special case with = and n =. Application: System Reliability Often, systems fail in a random manner. For a particular system, we denote t f as the time interval from the moment a system is put into operation until it fails; t f is the time to failure random variable. The distribution F a (t) = P[t f t] is the probability the system fails at, or prior to, time t. Implicit here is the assumption that the system is placed into service at t = 0. Also, we require that F a (t) = 0 for t 0. The quantity R(t) 1F (t) P [t t] (-107) a f is the system reliability. R(t) is the probability the system is functioning at time t > 0. We are interested in simple methods to quantify system reliability. One such measure of system reliability is the mean time before failure MTBF Etf t f 0 a(t)dt, (-108) where f a = df a /dt is the density function that describes random variable t f. Given that a system is functioning at time t, t 0, we are interested in the probability that a system fails at, or prior to, time t, where t t 0. We epress this conditional distribution function as P[tf t,tf t ] P[t < tf t] F a(t) F a(t ) F(t tf t ), t t. (-109) P[t t ] P[t t ] 1F (t ) f f a The conditional density can be obtained by differentiating (-109) to obtain Updates at -39

40 d f a (t) f(tt f t) F(tt f t), t t. (-110) dt 1 F (t ) a F(tt f > t) and f(tt f > t) describe t f conditioned on the event t f > t. The quantity f(tt f > t)dt is, to first-order in dt, the probability that the system fails between t and t + dt given that it was working at t. Eample -14 Suppose that the time to failure random variable t f is eponentially distributed. That is, suppose that t t 1e, t 0 F(t) a 0, t 0 e, t 0 f a (t) 0, t 0, (-111) for some constant > 0. From (-110), we see that t e (tt ) f(ttf t) e f t a(t t), t > t. (-11) e That is, if the system is working at time t, then the probability that it fails between tand t depends only on the positive difference t - t, not on absolute t. The system does not wear out (become more likely to fail) as time progresses! With (-110), we define f(tt f > t) for t > t. However, the function f a (t) (t), (-113) 1 F (t) a Updates at -40

41 known as the conditional failure rate (also known as the hazard rate), is very useful. To firstorder, (t)dt (when this quantity eists) is the probability that a functioning-at-t system will fail between t and t + dt. Eample -15 (Continuation of Eample -14) Assume the system has f a and F a as defined in Eample -14. Substitute (-111) into (-113) to obtain t e (t). (-114) t 1 {1e } That is, the conditional failure rate is the constant. As stated in Eample -14, the system does not wear out as time progresses! If conditional failure rate (t) is a constant, we say that the system is a good as new system. That is, it does not wear out (become more likely to fail) overtime. Eamples -14 and -15 show that if a system s time-to-failure random variable t f is eponentially distributed, then f (t tf t ) f a(t t ), t > t (-115) and (t) = constant, (-116) so the system is a good as new system. The converse is true as well. That is, if (t) is a constant for a system, then the system s time-to-failure random variable t f is eponentially distributed. To argure this, use (-113) to write Updates at -41

42 f (t) [1 F (t)]. (-117) a a But (-117) leads to df dt a F(t). (-118) a Since F a (t) = 0 for t 0, we must have t 1e, t 0 F(t) a 0, t 0, so t f is eponentially distributed. For (t) to be equal to a constant, failures must be truly random in nature. A constant requires that there be no time epochs where failure is more likely (no Year 000 type problems!). Random variable t f is said to ehibit the Markov property, or it is said to be memoryless, if its conditional density obeys (-115). We have established the following theorem. Theorem The following are equivalent 1) A system is a good-as-new system ) = constant 3) f(tt f > t) = f a (t t), t > t 4) t f is eponentially distributed. The previous theorem, and the Markov property, is stated in the contet of system reliability. However, both have far reaching consequences in other areas that have nothing to do with system reliability. Basically, in problems dealing with random arrival times, the time between two successive arrivals is eponentially distributed if Updates at -4

43 1) the arrivals are independent of each other, and ) the arrival time t after any specified fied time t is described by a density function that depends only on the difference t - t (the arrival time random variable obeys the Markov property). In Chapter 9, we will study shot noise (caused by the random arrival of electrons at a semiconductor junction, vacuum tube anode, etc), an application of the above theorem and the Markov property. Updates at -43