2. The CDF Technique. 1. Introduction. f X ( ).

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1 Week 5: Distributions of Function of Random Variables. Introduction Suppose X,X 2,..., X n are n random variables. In this chapter, we develop techniques that may be used to find the distribution of functions these random variables, say Y = u (X,..., X n ). Some of the techniques we consider are:. The Cumulative Distribution Function (CDF) Technique 2. The Jacobian Transformation Technique 3. The Moment Generating Function (MGF) Technique Here this week, we also talk about: Distributions of Order Statistics Special Sampling Distributions 2. The CDF Technique Let X be a continuous random variable with cumulative distribution function F X ( ) and density function f X ( ). Now suppose that Y = g (X) is a function of X where g is differentiable and strictly increasing. Thus, its inverse g uniquely exists. The CDF of Y can be derived using F Y (y) =Prob (Y y) = Prob X g (y) = F X g (y) and its density is given by f Y (y) = d dy F Y (y) = d dy F X g (y) = f X g (y) d dy g (y). If g were strictly decreasing, then we would have f Y (y) = f X g (y) d dy g (y). In summary, if g is strictly monotonic function, then f Y (y) =f X g (y) d dy g (y).

2 3. Example - CDF Technique Let X be a random variable with p.d.f. e x f (x) = ( + e x 2 for x. ) We wish to find the distribution of Y = e X. Here we have g (X) =e X which is strictly decreasing function. Thus, g (y) = ln y d so that dy g (y) = and applying the formula y above, we have f Y (y) =f X g (y) d dy g (y) y = ( + y) 2 y = where the range of y is obviously 0 <y<. ( + y) 2 4. The Jacobian Transformation Technique To explain this technique, we consider only the case of two continuous random variables X and X 2 and assume that they are mapped onto U and U 2 by the transformation u = g (x,x 2 ) and u 2 = g 2 (x,x 2 ). Suppose this transformation is one-to-one so that we can invert them to get x = h (u,u 2 ) and x 2 = h 2 (u,u 2 ). The Jacobian of this transformation is the determinant g g x x 2 J (x,x 2 )=det g 2 x g 2 x 2 = g g 2 g 2 g, x x 2 x x 2 provided this is not zero. Suppose the joint density of X and X 2 is denoted by f X X 2. Then, the joint density of U and U 2 is given by f U U 2 (u,u 2 )= J (h (u,u 2 ),h 2 (u,u 2 )) f X X 2 (h (u,u 2 ),h 2 (u,u 2 )). The above technique can be easily extended to several variables. See Hogg & Craig (995).

3 5. Example - Jacobian Technique As an illustration of the Jacobian transformation technique, let us consider deriving the t-distribution. Suppose Z N (0, ) and V χ 2 (r) and are independent. Then, the random variable T = p Z V /r has a t-distribution with r degrees of freedom. Define the variables s = v and t = z p v /r so that this forms a one to one transformation with the inversion z = t p s /r and v = s. Its Jacobian is J (z,v) =det s z t z s v t v 0 =det p v /r 2 zv 3/2 r = p v /r = p s /r Since Z and V are independent, their joint density canbewrittenas f ZV (z,v) =f Z (z) f V (v) = e 2 z2 e v/2 2π Γ (r/2) 2 r/2vr/2 Thus, using the Jacobain transformation formula above, the joint density of (S, T ) is given by f ST (s, t) = p s /r ³ e 2 t s/r 2 e s/2 2π Γ (r/2) 2 r/2sr/2 = sr/2 p s /r exp s µ+ t2 2πΓ (r/2) 2 r/2 2 r wherewenotethatsince 0 <v< and <z< then 0 <s< and <t<. Therefore, the marginal density of T is given by Z f T (t) = f ST (s, t) ds Z0 = sr/2 p s /r 0 2πΓ (r/2) 2 r/2 exp s µ+ t2 ds. 2 r Making the transformation w = s µ+ t2 2 r,

4 so that dw = 2 and therefore Z f T (t) = 0 2πΓ (r/2) 2 r/2 µ e w 2 dw +t 2 /r = for <t<. µ+ t2 ds r Γ [(r +)/2] πrγ (r/2) ( + t 2 /r) (r+)/2, µ 2w (r+)/2 +t 2 /r 6. The MGF Technique This method can be effective in instances where we can derive a recognizable m.g.f. because when it exists, it is unique and it uniquely determines the distribution. Suppose we are interested in the distribution of U = g (X,..., X n ) where X,..., X n have a joint density f (x,..., x n ). Then, we find the m.g.f. of U using M U (t) =E e Ut = Z Z e g(x,...,x n )t f (x,..., x n ) dx...dx n. In the special case where U is the sum of the random variables U = X + + X n and X,..., X n are independent, we have M U (t) =E ³e (X + +X n )t = E e X t E e X nt = M X (t) M Xn (t). The m.g.f. of U is the product of the m.g.f. of X,..., X n.

5 7. Examples - The MGF Technique Example (Poisson): Let X Poisson(λ ) and X 2 Poisson(λ 2 ) where X,X 2 are independent. Then the mgf of U = X + X 2 is given by M U (t) =M X (t) M X2 (t) = e λ (e t ) e λ 2(e t ) = e (λ +λ 2 )(e t ) which is the mgf of another Poisson with parameter λ + λ 2,i.e. U Poisson (λ + λ 2 ). Example (Normal): Let X N µ, σ 2 and X 2 N µ 2, σ 2 2 where X,X 2 again are independent. Then the mgf of U = X + X 2 is given by M U (t) =M X (t) M X2 (t) = e µ t+ 2 σ2 t 2 e µ 2t+ 2 σ2 2t 2 = e (µ +µ 2 )t+ 2(σ 2 +σ 2 2)t 2 which is the mgf of another Normal with mean µ + µ 2 and variance σ 2 + σ 2 2. That is U N µ + µ 2, σ 2 + σ Distributions of Order Statistics Assume X,X 2,..., X n are n independent identically distributed (i.i.d.) random variables and let their common distribution function be F X and density f X. Suppose we sort these variables and denote by X () <X (2) < <X (n) the order statistics. In particular, X () =min(x,..., X n ) is the minimum and X (n) =max(x,..., X n ).Forsimplicity, denote by U = X (n) and V = X (). Distribution of the Maximum Deriving the distribution of the maximum, we have F U (u) = Prob (U u) = Prob (X u) Prob (X 2 u) Prob (X n u) =[F (u)] n and the density function is f U (u) =nf (u)[f (u)] n. Distribution of the Minimum We have F V (v) =Prob (V v) = Prob (V >v) = [Prob (X >u) Prob (X n >u)] = [ F (v)] n

6 and the corresponding density function is f V (v) =nf (v)[ F (v)] n. In general, we can show that the probability density of the k-thorderstatisticisgivenby n! f k (x) = (k )! (n k)! f (x)[f (x)]k [ F (x)] n k. The joint probability density of the order statistic is given by: f 2...n (y,y 2,..., y n )=n!f (y ) f (y 2 ) f (y n ). 9. Example - Order Statistics Consider a system with n components. Assume that the lifetimes of the components are T,T 2,..., T n which are i.i.d. with exponential distribution with parameter λ. Suppose that the system are connected in series, that is, the system will fail if any one of the components fail. The lifetime V of the system is therefore the minimum of the T k,i.e. V =min(t,...t n ). Therefore the density of V is given by f V (v) =nf (v)[ F (v)] n = nλe λv e λv n =(nλ) e (nλ)v which is exponential with parameter nλ. Suppose that the system are connected in parallel, that is, the system will fail only if all of the components fail. The lifetime U of the system is therefore the minimum of the T k,i.e. V =min(t,...t n ). Therefore the density of V is given by f U (u) =nf (u)[f (u)] n = nλe λu e λu n.

7 0. Some Special Sampling Distributions We now consider some results regarding distributions resulting from sampling from a normal distribution. A Single Normal and Chi-Square. Suppose Z N (0, ), then Y = Z 2 χ 2 () has a chi-square distribution with degree of freedom. It is interesting to prove this, and it uses the CDF technique. Consider F Y (y) = Prob Z 2 y = Prob ( y Z y) = Z y y 2π e 2 z2 dz =2 Z y 0 2π e 2 z2 dz and now applying change of variable, say z = w,so that dz = 2 w /2 dw. Therefore, we have Z y F Y (y) =2 2π 2 w /2 e 2 w dw 0 Differentiating to get the p.d.f. we get f Y (y) = y /2 e 2 y = y ( 2)/2 e y/2 2π 2 /2 Γ 2 which is the density of a χ 2 () distributed random variables. Normal and Chi-Square. Suppose Z,Z 2,..., Z r are independent standard normal random variables. Then, the random variable rx V = Z 2 + Z Zr 2 = k= has a chi-square distribution with r degrees of freedom. t-distribution. Suppose Z N (0, ) and V χ 2 (r) and are independent. Then, the random variable T = p Z V /r has a t-distribution with r degrees of freedom. F-distribution. Suppose U χ 2 (r ) and V χ 2 (r 2 ) are two independent chi-square distributed random variables Then, the random variable F = U /r V /r 2 has an F-distribution with r and r 2 degrees of freedom. Sample Mean and Sample Variance. Suppose X,X 2,..., X n are n independent random variables with identical distribution N µ, σ 2. Define the Z 2 k

8 sample mean by X = n nx k= X k and the sample variance by S 2 = nx Xk X 2. n k= Then the following important properties can be verified: X N µ, n σ2 (n ) S2 χ 2 (n ) σ 2 X and S 2 are independent. Using these results, it can further be shown that T = X µ S / n has a t-distribution with n degrees of freedom.