Question

Let X have probability density function f(2)= k(1+x) -3 for 0 < x < oo and f(x) = 0 elsewhere. a. Find the constant k and Find the c.d.f. of X. b. Find the expected value and the variance of X. Are both well defined? c. Suppose you are required to generate a random variable X with the probability density function f(x). You have available to you a computer program that will generate a random variable U having a U[0, 1] distribution. What function g(U) of U will provide a random variable with the p.d..f f(x)? d. Find the probability density function of the random variable Y = VX

Answer 1

Given, let x have poobability density function, f(x) = fk (1+2j3 fox olacs and l o fos elsewhere. (a) the constant k and the codd. of x. The constant k : we have, (2) dx = 1 5+(2) dz = ! » [ * (v+x)*dx= > (K= 2 (since hier ein o) The cdif. of_* Foco) = P(x-x) = f*cm) dy

A fxce) = 5=Chujº dy dy 9 fxG) =(1-(1+x)2,0 Lacos 10 i xsoom (1) The expected value and the vasíance of x. The expected value of x, > EC) = Sex 2.[12]® be =85&-) z* dz dz (put, z=1+x) -> X = 2-1 2 32 (0+1)+(2-3) = 2x

Now, Variance of X, 22 Var (x) = E Q) –(E62) So, E() = 2 (1+25° de =z–)*.* d2 =2[52'da) --[5@z -2°) dz] JE (m) = 2 (12) -2 (212) now, Him Cancs] =- .E(x²) is not defined, So, Vax (x) is not defined, since , Vao(x)= EGYY4€()* Given, u= F(x) » 021-C1? > (1-x)?= 1-0

» 1-X = (1-0) * >> x=1-(1-03 >> x=1-1-osio Now, By inverse trans foom method, g(0) = 1- c'est will provide pdf. Of x. (d) The probability density function of the random Variable y=vx SC OLX20 OLX Lo) 0 ZVE Therefore, C.D.F of y 18, Fy Cy) = p (ysy) = p (vx <4) =P(xsy ) :> Fx (y). =) Fy(y) = 1 -(1+y+)?

Therefore, PDF of y is fy () - Pet fr (9) $2C1782. 2y = 49 (1+34", yoo s fyly) 0 ,else