Applications of Series to Probability

Applications of Series to Probability We will begin by considering a specific situation involving coin tossing that leads, in a natural way, to an infinite series. Along the way, we will introduce some of the terminology that appears in books on probability. Suppose you had a fair coin and decided to toss it until you obtained a head (H). If you counted the number of tosses required, you might let X denote the number of tosses. Thus, with T representing tails, TTH would give X = 3, H would give X =€1, TTTTH would give X = 5. In probabilistic language, the variable X is called a random variable, i.e., a variable whose value is determined by chance. Now, since the coin is fair, the probabilities of the three sequences above are 1/8, 1/2 and 1/25 = 1/32, respectively. If we use P to denote probability, we could say P (X = 3) = 1 8, P (X = 1) = 1 2, P (X = 5 ) = 1 32 . More generally then, P ( X = x ) = 1 x for x = 1, 2, K. Note carefully the standard 2 convention of using the lower case equivalent of the name of the random variable to represent a value of the random variable. Now consider the infinite series ∞ ∞ x =1 x=1 ∑ P (X = x ) = ∑ (1 2 ) . If we believe that sooner x or later, the coin must land head up, then we should believe that the value of this sum is 1. In fact, it is, since the series is a geometric series with first term and ratio 1/2, and the sum of an infinite geometric series is given by L < 1. Special Focus: Calculus Each random variable has an associated distribution function, F. Here we write F( x) + P (X ≤ x). Even though F ( x) is defined for all real values of x, we will only consider values of x that are positive integers. With this understanding, we see that F (x ) = P (X ≤ x ) = x x k=1 k=1 ∑ P(X = k) = ∑ x 1  1− (1 2)  x   = 1− (1 2) , using the (1 2) = ⋅  2  1−1 2  k standard formula for the sum of a finite geometric series, i.e., € n a + ar + ar + ... + ar = ∑ ar = 2 n k a (1 − r n +1 ) k =0 1− r if r ≠ 1 . Note that the distribution function for the random variable is really just the xth ∞ partial sum for the infinite series (1 2 ) , and this shows that partial sums have ∑ k =1 k a useful probabilistic interpretation. In fact, the distribution function is even more useful than it might seem at first since it is clear that if a and b are integers with 1 ≤ a ≤ b, then P (a < X ≤ b) = F ( b) – F (a). Thus, for example, the probability that more than three but no more than seven tosses are needed to obtain a head is given by P (3 < X ≤ 7) = F (7) – F (3) + (1 – 1/27) – (1 – 1/23) = 1/23 – 1/27 = 15/128. We are about to introduce another probabilistic concept but need to make some preliminary remarks first. Suppose that we obtained N values of the random variable X. We let V be the set of these values, say, V = {x1, x2, …, xN}. If N were large, we would believe that the average of the values in V provided a representative “average value” for the random variable X. To find the average of the values in V we would compute x 1 + x 2 + ... + x N . N There is a more convenient way to compute this value, however. If we let n( x) x + x 2 + ... + x N 1 ∞ = ∑ x ⋅ n ( x ) . (1) denote the number of x1 in V that equal x, then 1 N N x =1 To see why (1) is true, let’s look at an example. Suppose V = {2 ,4, 2, 3, 1, 6, 1, 2, 2, 3}. Then 2 + 4 + 2 + 3 + 1 + 6 + 1 + 2 + 2 + 3 = 1 ∙ 2 + 2 ∙ 4 + 3 ∙ 2 + 4 ∙ 1 + 5 ∙ 0 + 6 ∙ 1 10 10 = 1∙n(1)+2∙n(2)+3∙n(3)+ 4∙n(4)+5 ∙n(5)+ 6∙( n6) 10 Applications of Series to Probability ∞ n (x ) 1 ∞ To see why (1) is useful, note that ∑ x ⋅ n (x ) = ∑ x ⋅ . Now if N is large, N x =1 N x =1 n (x ) then will be very close to P (X = x) so we will get a very good approximation N ∞ to the average value of X if we use ∑ x ⋅ P (X = x ) , and we take this sum to be the x =1 definition of the expected value or mean of the random variable X. We denote the expected value of X by E(X) and think of it as the average of a large number of values of X, a “long-run” average, if you like. ∞ In the current case, E (X ) = ∑ x (1 2 ) , another infinite series, and we would like x x =1 to find the value of this series. We will use a geometric method first and then use a method that exploits some important properties of power series. Consider the following arrangement in which the bottom row contains the terms ∞ in the infinite geometric series ∑ (1 2 ) , the next row up contains all of these terms x x =1 except 1/2, etc. ... 1/16 ... 1/8 1/16 ... 1/4 1/8 1/16 ... 1/2 1/4 1/8 1/16 ... If we sum the entries in the bottom row, we get 1; if we sum the entries in the next row up, we get 1/2; in the next row up, 1/4; etc. Thus, the sum of all the entries is 1 + 1 2 + 1 4 + ... , which is 2. On the other hand, we note that exactly one of the entries is 1/2, exactly two are 1/4, exactly three are 1/8, etc. Thus, the sum of the entries can ∞ also be expressed as ∑ x (1 2 ) x , and therefore we have x =1 ∞ E (X ) = ∑ x (1 2 ) = 2 . x x =1 CAUTIONARY NOTE: In general, evaluating infinite series by rearranging the terms in the series can lead to invalid results. However, in this case, the series in question is absolutely convergent, and rearrangement is allowed. Special Focus: Calculus As an alternative to the argument above, consider ∞ 1 f (x) = = 1 + x + x 2 + ... = ∑ x k , which is valid for |x| < 1, using the formula for 1− x k =0 the sum of an infinite geometric sequence. Now, if we differentiate both sides, we obtain 1 ∞ 2 (1 − x ) = f ′ (x ) = ∑ kx k −1 k =1 x so that = xf ′(x ) = 2 (1− x ) ∞ ∑ kx . k k=1 12 k If we let x = 1/2 here, we get ∑ k (1 2 ) = 2 = 2 , confirming the previous 1 − 1 2) k =1 ( computation. € It might seem strange that the average value of the random variable X is 2, since X can take on arbitrarily large values. However, it does so with small probability. Remember that X is 1 or 2 with probability 3/4 = 1/2 +1/4, and 1 ∞  5 1 1 2  P (X > 4) = = = . k   1 16 2 k= 5 1−   2 ∞ ∑ The following exercise generalizes everything we have done so far. € Exercise I Suppose a coin lands head up with probability p where 0 < p < 1. Suppose that this coin is tossed repeatedly until a head appears and that Xp denotes the number of tosses required. Let q = 1 – p. The random variable Xp is said to be geometric with parameter p. (Note that in the example we just considered, p = 1/2 and the random variable is X1/2.) (a) Convince yourself that P(Xp = x)=qx–1p for x = 1, 2, 3, .... ∞ (b) Show that ( ) ∞ ∑ P X p = x = ∑ q x −1 p = 1 . (Hint: This is an easy application of x =1 x =1 the formula for the sum of an infinite geometric series.) (c) Show that the distribution function of Xp is 1 – qx for x = 1, 2, 3, .... (d) Use both of the methods given above to show that E(XP) = 1/p, and then convince yourself that it is reasonable that if p is close to 0, the average number of tosses needed to get a head should be large (because it is very hard to get a head), and that if p is close to 1, the average number of tosses should be very close to 1 (because we will typically get a head on the first toss). Applications of Series to Probability Next we extend the ideas in Exercise I. Using the coin from the exercise, toss the coin until it lands head up twice and let B denote the number of tosses that are required. If B=b, then since exactly two of the tosses are H’s, the remaining tosses are T’s and the first H can appear on any one of the first b–1 tosses, we have P(B = b) = (b–1)qb–2p2. Note that, much as in part (b) of Exercise I, we have ∞ ∞ ∞ b=2 b=2 b =1 ∑ (b − 1)qb − 2 p 2 = p 2 ∑ (b − 1)qb − 2 = p 2 ∑ bqb −1 = p 2 . However, since q = 1 − p, that P (B = b ) = (b − 1)q We have b−2 ∞ ∞ b=2 b=2 p2 (1 − q )2 d  ∞ b 1  p2 2 d  q = p = ∑ dq  b = 0  dq  1 − q  (1 − q )2 = 1 and therefore ∞ ∑ P (B = b ) = 1 , so our claim b=2 p is consistent. Now we will find the expected value of B. 2 E (B ) = ∑ bP (B = b ) = ∑ b (b − 1)q b−2 2  1  d2  ∞ b  2 2 d p =p q =p . = p2 2 ∑ 2   € dq  b = 0  dq  1 − q  (1 − q )3 2 2 But this last expression equals 2/p, and this should really not come as much of a surprise. Note that on average we need 1/p tosses to get the first head and then another 1/p tosses to get the second head, or 2/p tosses in all. The random variable B has a name as well; it is called a negative binomial random variable with parameters n = 2 and p. Before moving on to another “famous” random variable, we need to extend the definition of expected value to a more general setting. Suppose that X is a random variable that takes on the values {x1, x2, ...}. Then the expected value of X, i.e., E(X), ∞ is defined as ∑ x P (X = x ), or more simply ∑ xP (X = x ) , where the second sum is i =1 i i understood to be taken over all of the values assumed by X. An example should help clarify this idea. Example 1 Suppose X takes on the values 1, 3, 5,... with probabilities 1/2, 1/4, 1/8, ... respectively, i.e., P ( X = 2 k − 1) = 1 k for k = 1, 2,.... Then the expected value of X is 2 given by ∞ 1 1 1 1 E ( X ) = 1 ⋅ + 3 ⋅ + 5 ⋅ + ... = ∑ ( 2 k − 1) ⋅ k . 2 4 8 2 k =1 We can find the sum of this series using ideas like those used above, but there is an easier way. Note that if Y is an arbitrary random variable and Z = a + bY where a and b are constants with b, not zero, then since Z = a + bY if and only if Y = y, we have E (Z ) = ∑ (a + by )P (Z = a + by ) = ∑ (a + by )P (Y = y ) = ∑ aP (Y = y ) + ∑ byP (Y = y ) . ( ) (( )) Special Focus: Calculus ) ( (∑ yP (Y = y ))), ( The final expression above can be written as a∑ P (Y = y ) + b and we see that the value of this expression is a + bE (Y ) since ∑ P (Y = y ) = 1 as the probability that Y takes on some value is 1 and, by definition of expected value, ∑ yP (Y = y ) = E (Y ). Thus, we have shown that E (a + bY ) = a + bE (Y ) . This is a very useful result. In this case, using the notation of Exercise I, we see that the random variable X in this example can be written as 2X1/2 –1, and then using the formula we have just established E(X) = 2E(X1/2)–1 = 2 ∙ 2 – 1 = 3. It also needs to be noted that not every random variable has an expected value. Here is a simple example that illustrates this fact. Example 2 1 for x = 1, 2, 3,.... We need to check that the sum of x ( x + 1) the P(X = x) equals 1 to be sure that we have a well-defined random variable. To this n ∞ 1 1 end, consider ∑ , the nth partial sum of the infinite series ∑ . Since x =1 x(x + 1) x =1 x(x + 1) Let X = x with probability 1 1 1 = − , the partial sum “telescopes” as follows: x (x + 1) x x + 1 n 1 1 1 1 1 1 1 1 ∑ x ( x + 1) =  1 − 2  +  2 − 3  + ... +  n − n + 1 = 1 − n + 1 . x =1 Since the limit of this last expression is 1 as n goes to infinity, the infinite series 1 ∑ x(x + 1) does have a value of 1 as required. Now what about the expected value of x =1 ∞ ∞ ∞ x 1 =∑ . X ? Consider E (X ) = ∑ xP (X = x ) = ∑ x =1 x =1 x(x + 1) x =1 x + 1 But we recognize this last series as the harmonic series with its first term missing, and we know that this series diverges, or diverges to positive infinity. We would say then either that X has an infinite expected value or that the expected value of X does not exist. As a final example before going on to our next (and last) famous random variable, we note that random variables can assume negative values. When they do, there may be some ambiguity about the existence of their expected values. The following example illustrates this. ∞ Applications of Series to Probability Example 3 Suppose that X = 1, 2, 3, –4,... with P ( X = ( −1) k + 1 k ) = 1 for k = 1, 2,.... Then, k ( k + 1) 1 = 1 so the random variable is well defined. k =1 k (k + 1) ∞ 1 1 1 1 1 Moreover, E ( X ) = ∑ xP ( X = x ) = ∑ ( −1) k + 1 k = − + − + ... , and this k ( k + 1) 2 3 4 5 k =1 series obviously satisfies the hypotheses of the alternating series test and hence converges. Thus, X does have a finite expected value, at least for the moment. In fact, 1 1 1 1 1 1 1 since − + − + ... = 1 −  1 − + − + ... , and the expression in parentheses is   2 3 4 5 2 3 4 the “alternating harmonic series,” we see that E(X) = 1 – ln(2). ∞ as in Example 2, ∑ P (X = x ) = ∑ However, many mathematicians would claim that E(X) does not exist, since for them, the existence of the expected value requires that ∑ x P (X = x ) converge, i.e., that the series that defines the expected value converges absolutely. And, of course, we have already seen that this does not happen in this case. Thus, for these mathematicians, the expected value of X does not exist. We introduce another justly famous random variable. If λ is a positive real λ x e− λ number, define the random variable, X, by P ( X = x ) = for x = 0, 1, 2, 3,.... This x! is a Poisson random variable, named after the French mathematician Siméon-Denis Poisson (1781–1840). We say that it has parameter λ. Before exploring some of the λ x e− λ = 1. x! x=0 ∞ properties of X, let us check that it is well defined, i.e., that ∞ ∑ P (X = x ) = ∑ x=0 λx ∑ x! = eλ , x=0 which is valid for all real λ. This familiar fact is usually established by considering the Maclaurin Series of eλ, showing first that the series converges for all real λ (using the ratio test, perhaps) and then showing that the series converges to eλ for all real λ (using one of the forms of Taylor’s theorem, perhaps). An alternative derivation involving separable differential equations uses the following ideas: the derivative ∞ ∞ ∞ ∞ ∞ tx xt x−1 xt x−1 t x−1 tx with respect to t of the series ∑ is = = = so x! x! x! (x −1)! x = 0 x! ∞ In fact, this is an immediate consequence of the well-known identity ∑ x= 0 ∑ x=1 ∑ ∑ x= 0 x=1 the function y = y(t) satisfies y´ = y, and since y(0) = 1. y (t) = e . A key step is this t derivation is the observation that € x/x! = x/(x(x – 1)!) = 1/x(x – 1)! for x >0. Special Focus: Calculus Let’s find the expected value of X. Using, once again, x/x! = 1/(x –1)!, we have ∞ ∞ ∞ λ x e− λ λx λ x −1 λx −λ −λ −λ E (X ) = ∑ x =e ∑ = λe ∑ = λe ∑ = λe− λ eλ = λ . x! x=0 x =1 (x − 1)! x =1 (x − 1)! x = 0 x! ∞ Thus, the parameter λ is actually the expected value of the random variable. Poisson Random Variables Poisson random variables are useful in a number of settings. They generally model the number of occurrences of a “rare” event in time intervals of fixed length, for example, the number of extinctions of marine invertebrate families in a time interval of fixed length. As a second example, the so-called binomial probabilities can be approximated by Poisson probabilities. If p is a probability and n is positive integer, n! is a k!(n − k )! so-called binomial coefficient. Computing one of these binomial probabilities is quite a then C ( n,k)p k (1 – p)n–k is a binomial probability where C (n, k ) = chore if n is large. However, it turns out that λ k e− λ n− k C (n, k ) p k (1 − p ) ≈ if λ = np and n is large and p is small . k! This is not too hard to prove in general, but we will prove only one case, i.e., k = 1, in which case we need to show that −λ λ 1e − λ n −1 1 np (1 − p ) ≈ = λe if λ = np and n is large and p is small . 1! We will establish this by showing that the limit of the left-hand side, as n goes to infinity, is the right-hand side. We assume that λ = np so that p = λ/n. np (1 − p ) 1 n −1 λ  = λ 1 −   n n −1 = λ  λ 1 −   n λ   1 −  n 1 n . As n goes to infinity, the numerator of the fraction goes to λe–λ, and the denominator goes to 1. Thus, we have the result we need. (To find the limit of the n r  numerator we used the well-known result lim  1 +  = er with r = –λ. You can use n→∞  n L’Hopital’s Rule to prove this limit theorem.) To see how good an approximation we have, consider n = 25 and p = 0.05, and note that np1(1 – p)n–1 = 0.364986 while λe–λ = 0.358131. We finish this unit with an interesting result. Suppose that for i = 1, 2, X i is a Poisson random variable with parameter λi We say that X1 and X2 are independent Applications of Series to Probability random variables if P(X1 = x1 and X2 = x2) = P(X1 = x1)P(X2 = x2). (Independence is a crucial property in probability theory and is a much more natural idea than it might seem from this definition.) Consider the “new” random variable S = X1 + X2. Let’s see if we can compute P(S = s) where s is a nonnegative integer. We see that s s s λ1x e− λ1 λ2 s − x e− λ2 . P (S = s ) = ∑ P (X1 = x and X2 = s − x ) = ∑ P (X1 = x )P (X 2 = s − x ) = ∑ x! s − x ! ( ) x=0 x=0 x=0 But λ1x e− λ1 λ2 s − x e− λ2 e− λ1 e− λ2 ∑ x! (s − x )! = s! x=0 s s! e− (λ1 + λ2 ) s x s− x λ λ = C (s, x )λ1x λ2 s − x , ∑ x!(s − x ) 1 2 ∑ s! x = 0 x=0 s and we (hopefully) recognize this last sum as the result of applying the binomial theorem to (λ1 + λ2)S. Then we have proved that λ1 + λ2 ) e− (λ1 + λ2 ) ( e− (λ1 + λ2 ) s P (S = s ) = . But this implies that S is a Poisson (λ1 + λ2 ) = s! s! s random variable with parameter λ1 + λ2. This is certainly an interesting result. The following exercise, which concludes this unit, gives another interesting result of the same type. Exercise II Suppose that X1 and X2 are independent geometric random variables with the same parameter p. So, P(Xi = x) = pqx–1 where q = 1 – p. Let S = X1 + X2. (a) Show that P(S = s) = (s – 1)p2qs – 2 for s = 2, 3,... . (b) Compare the probabilities given in part (a) with those associated with the negative binomial distribution that was introduced right after Exercise I and conclude that S is a negative binomial random variable with parameters n = 2 and p. (c) Convince yourself that the sum of two independent geometric random variables, each with parameter p, should be a negative binomial random variable with parameters n = 2 and p. (Hint: If we toss a coin until we get two heads (H’s), we could let X1 denote the number of tosses required to get the first head and X2 denote the number of tosses required to get the second head. Then S = X1 + X2 is the total number of tosses required to get two H’s.)