Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
CHAPTER 4. DISCRETE RANDOM VARIABLES 4.5 48 Probability generating functions We saw in the previous section that we could derive the mean and variance of a Binomially distributed random variable but it was quite hard work. In this section we introduce a method of finding the mean and variance of a discrete random variable which is often useful. We use the idea of a generating function. The particular function we employ is called a probability generating function. This is defined as X GX (s) = E[sX ] = sx P (X = x) x for any real s for which the sum converges. If we expand the sum we see that the coefficient of sx is the probability that X = x. The quantity sP is a dummy variable, we could have used any letter we chose. Note that GX (1) = x P (X = x) = 1. Now we can see why GX is useful if we differentiate it with respect to s X G′X (s) = xsx−1 P (X = x) x and put s = 1 G′X (1) = X xP (X = x) = [X]. x Similarly if we differentiate again with respect to s X G′′X (s) = x(x − 1)sx−2P (X = x) x and put s = 1 G′′X (1) = X x(x − 1)P (X = x) = [X(X − 1)]. x Now V ar[X] = E[X 2 ] − (E[X])2 = E[X(X − 1)] + E[X] − (E[X])2 = G′′X (1) + G′X (1) − [G′X (1)]2 . In general dk GX (1) E[X(X − 1)(X − 2)...(X − k + 1)] = dsk k X (l) where d G is the kth derivative of ΠX (s) wrt s and evaluated at s = 1. Note dsk that probability generating function is only going to be useful if we can simplify P the x s P (X = x). x Example 4.6 Discrete rv X has probability density function 1 a−x e− a P (X = x) = x! 49 CHAPTER 4. DISCRETE RANDOM VARIABLES 1 1. Show that probability generating function Gx (s) for X is equal to e a (s−1) . 2. Using Gx (s) find E[X]. 3. Using Gx (s) find V ar[X]. Solution P P∞ x a−x e− a1 1 P x 1. Gx (sx ) = E[sx ] = ∞ = e− a ∞ x=0 s P (X = x) = x=0 s x=0 x! 1 s 1 e− a × e a = e a (s−1) (sa−1 )x x! = 1 2. G′x (s) = a1 e a (s−1) E[X] = G′x (1) = 1 a 1 3. G′′x (s) = a12 e a (s−1) E[X(X − 1)] = G′′x (1) = a12 V ar[X] = E[X(X − 1)] + E[X] − (E[X])2 = 1 a2 + 1 a − 1 a2 = 1 a △