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Transformation Techniques

Transformation Techniques. In Probability theory various transformation techniques are used to simplify the solutions for various moment calculations. We will discuss here 4 of those functions. Probability Generating Function Moment Generating Function Characteristic Function

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Transformation Techniques

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  1. Transformation Techniques • In Probability theory various transformation techniques are used to simplify the solutions for various moment calculations. We will discuss here 4 of those functions. • Probability Generating Function • Moment Generating Function • Characteristic Function • Laplace Transformation of probability density function

  2. Probability Generating Function Tool that simplifies computations of integer valued discrete random variable problems X: non-negative integer valued Random Number P(X=k) =pk, then define the Probability Generating Function (PGF) of X by GX(z) = E[z X] = S pk z k = p0 + p1z + p2 z2 + ……. pk z k +…… z is a complex number z< 1 G(z) is nothing more than z-transform of pk. Gx(1) = 1 = S pk

  3. Generating Functions K : Non-negative integer valued random variable with probability distribution pj where, pj = Prob[K =j] for all j = 0,1,2,…… g(z) : p0 + p1 z+ p2 z2+ p3 z3+ ……. g(z) is a power series of probability pj with coefficient zj is the probability generating Function of random variable K Few properties g(1) = 1 as S pj = 1 and z is a complex number and converged to Absolute Value Mod[z] < 1 Expected Value E[K] = S j pj for j: 0,1,2….. (d/dz)g(z) = S j pj zj-1 at z =1 for j : 1,2,….. E[K] = g(1)(1) Similarly V[K] = g(2)(1) + g(1)(1) – [g(1)(1)]2 Reference: Introduction to Queuing Theory, Robert Cooper

  4. Moment Generating Function mg(t) : Moment Generating Function: Expected Value of function etX, where ‘t’ is a real variable and X is the random variable mg(t) = E[etX] =  Xi Rx p(Xi). etXi = ∫Rx f(x). etXidx If mg(t) exists for all real values of t, in some small interval –d, d : d > 0 about the origin, it can be shown that the probability distribution function can be obtained from mg(t). We assume mg(t) exists at a small region t about origin.

  5. Moment Generating Function-2 etX= 1 + tx + t2X2/2! + t3X3/3!+  Assume X is a continuous Random Variable mg(t) = E[etX] =  Xi Rx p(Xi). etXi = ∫Rx f(x). etXidx  = ∫Rx i=0 tiXi/i! f(X)dx   = ∫Rx ti/i! i=0 Xi f(X)dx = i=0 ti/i!∫Rx Xi f(X)dx  = i=0 ti/i!E[Xi] = E[X0] + tE[X1] + t2/2!E[X2] + … etX

  6. Moment Generating Function-3 mg(t) = E[X0] + tE[X1] + t2/2!E[X2] + … m (1)g(t) = E[X1] + tE[X2] + t2/2!E[X3] + … m (2)g(t) = E[X2] + tE[X3] + t2/2!E[X4] + … At t = 0 m (1)g(t) = E[X1] m (2)g(t) = E[X2] Var[X] = E[X2] – [E[X]]2 = m (2)g(t) - [m (1)g(t)] 2

  7. Characteristic Function The Characteristic Function of Random Variable X fX(u) = E[e juX] = ∫- e juXfx(x)dx where j = -1 and u is an arbitrary real variable Note: Except for the sign of exponent, Characteristic function is the Fourier Transform of the pdf of X. fX(u) = ∫- fx(x)dx[1 + jux +(jux)2/2! + (jux)3/3! + ……..]dx = 1 + juE[X] + (ju)2/2!E[X2] + (ju)3/3!E[X3] + ….. Let u=0 Then fX(0) = 1 f(1)X(0) = dfX(u)/duu=0 = jE[X] f(2)X(0) = d2fX(u)/du2u=0 = j2E[X2]

  8. Laplace Transform Let CDF of traffic arrival process is defined as A(x), where X is the random variable for inter arrival time between two customers. A(x) = P[X < x] The pdf (probability density function) is denoted by a(x) Laplace Transform of a(x) is denoted by A*(s) and is given by A*(s) = E[e –sX] = ∫- e –sx axdx Since most random variable deals with non negative numbers, we can make the transform as A*(s) = ∫0 e –sx axdx Similar techniques of Moment generating function or characteristic function can be used to show that A*(n) (0) = (-1)nE[Xn]

  9. l l l - ju l - v Example For a continuous Random variable pdf is given as follows le –lx x > 0 fx(x) = 0 x < 0 Laplace Transform : A*(s) = Characteristic Function: fx(u) = Moment Generating Function: mg(v) = l l + s

  10. Expected Value Laplace Transform : E[X] = (-1)A*(1) (0) = (-) d[l/(l + s)/dss=0 = (-) [(-)l/(l +s)2]s=0 = l/l2 = 1/l Characteristic Function: E[X] = j-1fx(1) (0) = (j-1) d[l/(l - ju)/duu=0 = (j-1)[l.j/(l - ju)2u=0 = l/l2 = 1/l Moment Generating Function E[X}= mX(1) (0) = d[l/(l - v)/dvv=0 =[l/(l - v)2v=0 = l/l2 = 1/l

  11. Variance Laplace Transform : E[X2] = (-1)2A*(2) (0) = d2[l/(l + s)/ds2s=0 = [2l(l+s)/(l +s)3]s=0 = 2l2/l3 = 2/l Var[X] = E[X2] – [E[X]]2 = 2/l – [1/l]2 = (2l –l)/l2 = 1/l Characteristic Function: E[X2] = j-2fx(2) (0) = (j-2) d2[l/(l - ju)/du2u=0 = (j-2)[2l(l – ju).j2/(l - ju)3u=0 = 2l2/l3 = 2/l Moment Generating Function E[X2}= mX(2) (0) = d2[l/(l - v)/dv2v=0 =[2l(l - v )/(l - v)3v=0 = 2l2/l3 = 2/l

  12. Sum of Random Variables K1 and K2 are two independent random variables with GF g1(z) and g2(z) Find the Probability distribution P{K=k} where K = K1 + K2 P{K =k} = P{k1 = j}.P{k2 = k-j} g1(z) = S P{k1=j}zj for j: 0.1,2……. g2(z) = S P{k2=j}zj for j: 0.1,2……. g1(z)g2(z) = S { S P{k1=j}P{k2=k - j}zk for k: 0.1,2……. and j : 0,1,2…k If K has a generating function of g(z), then g(z) = S P{K=k}zk for k: 0.1,2……. = S [S P{k1 = j}.P{k2 = k-j}] for k: 0.1,2……. and j : 0,1,2…k g(z) = g1(z)g2(z) k = 0 j = 0 k= 0 j = 0

  13. Example: Bernoulli Distribution Bernoulli Distribution : X =0 with probability q X = 1 with probability p p + q = 1 g(z) = q + pz g’(1) = p g’’(I) = 0 E[X] = g’(1) = p V[x] = g’’(1) + g’(1) – [g’(1)]2 = p – p2 = p(1 – p) = pq A coin is tossed for n times, Xj = 0 if tail and Xj = 1 if head probability to have k heads in n tosses. Sn is the sum of n independent Bernoulli random variables Sn = X1 +X2 +……….+ Xn g(z) = GF of a toss = q + pz GF of Sn = S P{Sn = k}zk for k : 0,1,2…… = g(z).g(z)…….g(z) = [g(z)]n = (q + pz)n = SnCk[pz] k q n-k for k = 0…..n Binomial Distribution P{Sn= k} = nCk[pz] k q n-k for k = 0…..n = 0 for k > n

  14. Example Poisson Distribution Poisson Distribution = [(lt)j/j!]e –lt for j:0,1,2….. Generating Function g(z) = S [(lt)j/j!]e –lt zj = e –ltS [(ltz)j/j!] for j: 0,1,2,…. = e –lt e ltz =e –lt(1-z) Expectation P[N(t) =j] g’(z) = lte –lt(1-z) E[N(t) =j] = g’(1) = lt Variance g’’(z) = (lt)2e –lt(1-z) g’’(1) = (lt)2 V[N(t)] = g’’(1) + g’(1) – {g’(1)}2 = lt Sum of Poisson distribution of l1 and l2 g(z) = e –l1t(1-z) e –l2t(1-z) = +e –(l1+ l2)t(1-z) l = l1 + l2

  15. M/M/1 System Birth and Death Equation 0 = - ( l + m) pn + m pn+1 + l pn-1 (n>1) 0 = -l p0 + m p1 pn+1 = [( l + m)/ m] pn - [ l/ m] pn-1 p1 = [l/ m]p0 If r = l/ m pn+1 = ( r + 1)pn - r pn-1 (n>1) p1 = rp0 Use GF to solve this equation zn pn+1 = ( r + 1) zn pn - r zn pn-1 (n>1) z-1 pn+1 zn+1 = ( r + 1) zn pn - r zpn-1 zn-1 z-1 Spn+1 zn+1 = ( r + 1) Szn pn - r zSpn-1 zn-1    n=1 n=1 n=1 Use of GF for Probability

  16. GF for Prob    z-1[Spn+1 zn+1 – p1 z – p0 ] = ( r + 1)[ Szn pn - p0] - r zSpn-1 zn-1 n=-1 n=0 n=1    Spn+1 zn+1=Spn zn=Spn-1 zn-1=P(z) But n=-1 n=0 n=1 z-1[P(z)– p1 z – p0 ] = ( r + 1)[ P(z) - p0] - r zP(z) z-1[P(z)– rp0 z – p0 ] = ( r + 1)[ P(z) - p0] - r zP(z) z-1P(z)– rp0 – z-1p0 = rP(z) - rp0 +P(z) - p0 - r zP(z) P(z) = p0 /(1 –rz) To Find p we use the boundary condition P(1) =1 P(1) = p0 /(1 –r) = 1 p0 = 1 –r P(z) = (1 –r) /(1 –rz) 1 /(1 –rz) = 1 + zr + zr2 + ……. P(z) = S(1-r)rnzn pn= (1-r)rn  n=0

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