Generating and characteristic functions. Generating and Characteristic Functions. Probability generating function. Probability generating function

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1 Generating and characteristic functions Generating and Characteristic Functions September 3, 03 Probability generating function Moment generating function Power series expansion Characteristic function Characteristic function and moments Convolution and unicity Inversion Joint characteristic functions /60 /60 Probability generating function Probability generating function Let X be a nonnegative integer-valued random variable. The probability generating function of X is defined to be G X (s) E(s X )= X k 0 s k P(X = k) I If X takes a finite number of values, the above expression is a finite sum. I Otherwise, it is a series that converges at least for s [, ] and sometimes in a larger interval. If X takes a finite number of values x 0, x,...,x n, G X (s) isa polynomial: G X (s) = nx s k P(X = k) k= =P(X = 0) + P(X = ) s + +P(X = n) s n 3/60 4/60

2 Probability generating function If X takes a countable number of values x 0, x,..., x k,...,then G X (s) = X s k P(X = k) k 0 =P(X = 0) + P(X = ) s + +P(X = k) s k + is a series that converges at least for s apple, because X s k P(X = k) apple X s k P(X = k) apple X P(X = k) = k 0 k 0 k 0 Let X be a Bernoulli random variable, X B(p). P(X = 0) = q, P(X = ) = p We have G X (s) = X s k P(X = k) =q + sp, s R k 0 5/60 6/60 Let X Bin(n, p). Then P(X = k) = G X (s) = X k 0 s k P(X = k) = = nx n p k q n k, k =0,,...,n k nx n s k p k q n k n (sp) k q n k =(q + sp) n, s R k k X Poiss( ). P(X = k) =e Then k, k =0,,... G X (s) = X X s k (s ) k P(X = k) =e k 0 = e e s = e (s ), s R 7/60 8/60

3 Unicity X Geom(p) P(X = k) =q k p, k =,,..., 0 < p < We have G X (s) = X X s k P(X = k) = s k q k k 0 = sp k= X (sq) k = sp sq, s < q k= p If two nonnegative, integer-valued random variables have the same generating function, then they follow the same probability law. Let X and Y be nonnegative integer-valued random variables such that G X (s) =G Y (s). Then P(X = k) =P(Y = k) for all k 0. The result is a special case of the uniqueness theorem for power series 9/60 0 / 60 Example (Convolution) If X and Y are independent random variables and Z = X + Y, then G Z (s) =G X (s) G Y (s) Proof: G Z (s) =E(s Z ) =E(s X +Y )=E(s X s Y )=E(s X )E(s Y )=G X (s)g Y (s) Let X Bin(n, p), Y Bin(m, p) be independent random variables and let Z = X + Y We have G Z (s) =G X (s)g Y (s) =(q + sp) n (q + sp) m =(q + sp) n+m Observe that G Z (s) is the probability generating function of a Bin(n + m, p) random variable. By the unicity theorem, X + Y Bin(n + m, p) / 60 / 60

4 More generally, Let X, X,...,X n be independent, nonnegative, integer-valued random variables and set S n = X + X + + X n. A case of particular importance is: Corollary If, in addition, X, X,...,X n are equidistributed, with common probability generating function G X (s), then Then ny G S (s) = G Xk (s). k= G S (s) =(G X (s)) n. 3 / 60 4 / 60 Example: Negative binomial probability law Example: Negative binomial probability law A biased coin such that P(heads) = p is repeatedly tossed until a total amount of k heads has been obtained. Let X be the number of tosses. Notice that where X = X + X + + X k, X i Geom(p) is the number of tosses between the (i )-th and the i-th head. As X,...,X k are independent and identically distributed we can apply the convolution theorem. Thus, G X (s) =G X (s)g X (s) G Xk (s) =(G X (s)) k sp k =, s < sq q 5 / 60 6 / 60

5 Example: Negative binomial probability law Example: Negative binomial probability law Recall that if R, then the Taylor series expansion about 0 of the function ( + x) is, for x (, ), + x + We can write where ( ) x + + ( + x) = X r 0 ( )...( r + ) r! x r r ( )...( r + ) r r! x r + Consider the series expansion of G X (s): where X G X (s) =(sp) k ( sq) k =(sp) k r=0 k ( sq) r r k k( k ) ( k r + ) r r! k + r =( ) r k 7 / 60 8 / 60 Example: Negative binomial probability law Properties Therefore, G X (s) = X k + r p k q r s k+r = k r=0 X n=k n p k q n k Hence, 8 >< 0, n < k P(X = n) = n >: p k q n k, n = k, k +, k This is the negative binomial probability law. k s n I G X (0) = P(X = 0) I G X () = We have 3 G X () = 4 X s k P(X = k) 5 = X P(X = k) = k 0 k 0 s= 9 / 60 0 / 60

6 Properties Properties Proposition Let R be the radius of covergence of G X (s). IfR>, then E(X )=GX 0 () Indeed, GX 0 (s) = d X s k P(X = k) ds =X ks k P(X = k) k 0 k More generally, Proposition I E(X )=GX 0 () lim s! GX 0 (s) I E(X (X ) (X k + )) = G (k) X () lim s! G (k) X (s) Hence, G 0 X () = X k k P(X = k) =E(X ) / 60 / 60 Let X Poiss( ). Let X Bin(n, p). E(X )=G 0 X () = d ds (q + sp)n s= = np (q + sp) n s= = np (q + p)n = np Analogously, E(X )=G 0 X () = d ds e (s ) s= = e (s ) s= = E(X (X )) = G 00 X () = e (s ) s= = Hence, E(X )= +, Var(X )=E(X ) (E(X )) = 3 / 60 4 / 60

7 X Geom(p). E(X )=G 0 X () = d ds Analogously, Therefore sp ( sq) s= = E(X (X )) = G 00 X () = pq p ( sq) s= = p ( sq) 3 s= = q p E(X )= q p + p, Var(X )=E(X ) (E(X )) = q p Let X be a negative binomial random variable. E(X )=G 0 X () = d ds = k sp sq sp sq k s= k p ( sq) s= = k p This result can also be obtained from X = X + + X k, with each X i Geom(p). E(X )= kx E(X i )= k p i= 5 / 60 6 / 60 Moment generating function Probability generating function Moment generating function Power series expansion Characteristic function Characteristic function and moments Convolution and unicity Inversion Joint characteristic functions The moment generating function of a random variable X is defined as 8 X e >< tx i P(X = x i ), if X is discrete X (t) E e tx i = Z >: e tx f X (x) dx, if X is continuous provided that the sum or the integral converges. 7 / 60 8 / 60

8 Unicity The moment generating function specifies uniquely the probability distribution. Let X and Y be random variables. If there exists h > 0, suchthat X (t) = Y (t) for t < h, then X and Y are identically distributed. Let X Bin(n, p). X (t) = nx e tk P(X = k) = nx n k = q + pe t n, t R pe t k q n k 9 / / 60 Let X Exp(µ). X (t) = = Z Z 0 e tx f X (x) dx For continuos random variables, transform of f X (x). µe (µ t)x dx = µ µ t, t <µ X (t) is related to the Laplace Let X Poiss( ). X X X (t) = e tk P(X = k) = e tk k e X (e t ) k = e = e (et ), t R 3 / 60 3 / 60

9 Let Z N(0, ). because Z (t) = = e t Z Z p e e tz f Z (z) dz = p Z e z tz dz p Z (z t) dz e {z } = e t, t R (z t) dz =P( < N(t, ) < ) = More generally, if then X N(m, ). We have X (t) =E = e tm E X = Z + m, t( e Z+m) e tx =E e t Z = e tm Z ( t) =e t +tm 33 / / 60 Power series expansion Power series expansion Notice that 0 X (t) = d e d dt E tx =E dt etx =E X e tx For instace, if X Exp(µ), X (t) = µ µ t, t <µ Therefore, Analogously, Thus, 00 X (t) = d dt 0 X (0) = E(X ) 0 X (t) = d Xe dt E tx =E X e tx 00 X (0) = E(X ) and E(X )= 0 X (0) = d dt µ = µ t t=0 Analogously, E(X )= 00 X (0) = d µ dt (µ t) t=0 = µ (µ t) =/µ t=0 µ (µ t) 3 =/µ t=0 35 / / 60

10 Power series expansion Power series expansion More generally, X (t) =E e tx =E +tx+ (tx) + + (tx)k! = + E(X ) t + E(X )! This is the power series expansion of X (t) = t + + E(X k ) X X (t), (k) X (0) t k t k + + If X (t) converges on some open interval containing the origin t =0, then X has moments of any order, E X k = and X (t) = X (k) X (0), E(X k ) t k 37 / / 60 Power series expansion For instance, let X Exp(µ). If t <µwe have Hence, Therefore, X (t) = µ µ t = (t/µ) =+ t t µ + + µ E(X n ) n! = µ n E(X n )= n! µ n The convolution theorem applies also to moment generating functions. Let X, X,...,X n be independent random variables and let S = X + X + + X n.then, S(t) = ny k= X k (t) 39 / / 60

11 X Poiss( X ), Y Poiss( Y ),independent. Let Z = X + Y. We have Z (t) = X (t) Y (t) =e X (e t ) e Y (e t ) = e ( X + Y )(e t ) Hence, Z Poiss( X + Y ) X N(m X, X ), Y N(m Y, Y ),independent. Let Z = X + Y. We have Therefore Z (t) = = e X (t) Y (t) X t +tm X e Y t +tm Y = e ( X + Y )t +t(m X +m Y ) Z N(m X + m Y, X + Y ) 4 / 60 4 / 60 Characteristic function Probability generating function Moment generating function Power series expansion Characteristic function Characteristic function and moments Convolution and unicity Inversion Joint characteristic functions The characteristic function of a random variable X is the complex-valued function of the real argument! defined as M X (!) E M X : R! C! 7! M X (!) e i!x = E (cos (!X )) + i E(sin(!X )) 43 / / 60

12 Characteristic function Properties Therefore, 8 >< M X (!) = >: X e i!x k P(X = x k ), if X is discrete k Z e i!x f X (x) dx, if X is continuous I The characteristic function exists for all! and for all random variables. I If X is continuous, M X (!) isthefourier transform of f X (x). (Notice the change of sign from the usual definition.) I If X is discrete, M X (!) isrelatedtofourier series. I M X (!) apple M X (0) = for all! R. M X (!) = E e i!x apple E e i!x = E() = On the other hand, M X (0) = E e i 0 X = E() = 45 / / 60 Properties I M X (!) =M X (!). M X (!) =E(e i!x )=E e i!x = E (cos (!X )) i E(sin(!X )) = E (cos (!X )) + i E(sin(!X )) = M X (!) Let X Binom(n, p). Then, M X (!) = pe i! + q n If X Poiss( ), then M X (!) =e (ei! ) I M X (!) is uniformly continuous in R. If X N(m, ),then M X (!) =e i!m! 47 / / 60

13 Characteristic function and moments Characteristic function and moments If E(X n ) < for some n =,,...,then So, M X (!) = nx E(X k )= M(k) X (0) i k In particular, if E(X )=0andVar(X )= M X (!) = E(X k ) (i!) k + o(! n ) as!! 0. for k =,,...,n.,then! + o(t ) as!! 0. Indeed, M X (!) =E e i!x =E! X (i!x ) k X = But this is the Taylor s series expansion of M X (!): Therefore M X (!) = X M (k) X (0)! k i k E(X k )=M (k) X (0) i k E(X k )! k 49 / / 60 Let X, X,...,X n be independent random variables and let S = X + X + + X n. Then ny M S (!) = M Xk (!) k= M S (!) =E e i!s =E e i!(x +X + +X n) =E e i!x e i!x e i!xn =E e i!x E e i!x E e i!xn = M X (!)M X (!) M Xn (!) 5 / 60 5 / 60

14 Unicity In the case n = we have essentially the convolution theorem for Fourier transforms. If X and Y are continuous and independent random variables and Z = X + Y,then f Z = f X f Y This implies that is, F(f Z )=F(f X ) F(f Y ), M Z (!) =M X (!)M Y (!) Let X have probability distribution function F X and characteristic function M X.LetF X (x) =(F X (x)+f X (x ))/. Then F X (b) Z T F X (a) = lim T! T e ia! i! e ib! M X (!) d!. M X specifies uniquely the probability law of X. Two random variables have the same characteristic function if and only if they have the same distribution function. 53 / / 60 Inversion Inversion (Inversion of the Fourier transform) Let X be a continuous r.v. with density f X and characteristic function M X.Then f X (x) = Z e i!x M X (!) d! at every point x at which f X is di erentiable. In the discrete case, M X (!) isrelatedtofourier series. If X is an integer-valued random variable, then P(X = k) = Z e i M X (!) d! 55 / / 60

15 Joint characteristic functions Joint moments The joint characteristic function of the random variables X, X,..., X n is defined to be M X (!,!,...,! n )=E e i(! X +! X + +! nx n) Using vectorial notation one can write! =(!,!,,! n ) t, X =(X, X,, X n ) t The joint characteristic function allows as to calculate joint moments. For instance, given X, Y : m kl =E X k Y l k+l M XY (!,! ) i l! (!,! )=(0,0) and M X (! t )=E e i!t X 57 / / 60 Marginal characteristic functions Independent random variables Marginal characteristic functions are easily derived from the joint characteristic function. For instance, given X, Y : M X (!) =E e i!x =E e i(! X +! Y ) = M XY (!, 0) (! =!,! =0) Analogously, M Y (!) =M XY (0,!) The random variables X, X,...,X n are independent if and only if M X (!,!,...,! n )=M X (! )M X (! ) M Xn (! n ) If the random variables are independent, then M X (!,!,...,! n ) =E e i(! X +! X + +! nx n) =E e i! X e i! X e i!nxn =E e i! X E e i! X E e i!nxn = M X (! )M X (! ) M Xn (! n ) 59 / / 60