2. A continuous random variable X has PDF SPI? 1€ (-2,2] fx() = 0 otherwise (a)...
For a continuous variable X with the following PDF: 0sxs2 fx (x) = {2' 0, otherwise (a) Determine the conditional PDF of X given that X>1. (b) Find the conditional CDF of X given that X > 1, and plot the corresponding figure with proper labels. [Note: Both the expression and the plot are required.]
Let X be a continuous random variable with PDF fx(x)- 0 otherwise We know that given Xx, the random variable Y is uniformly distributed on [-x,x. 1. Find the joint PDF fx(x, y) 2. Find fyo). 3. Find P(IYI <x3) Let X be a continuous random variable with PDF fx(x)- 0 otherwise We know that given Xx, the random variable Y is uniformly distributed on [-x,x. 1. Find the joint PDF fx(x, y) 2. Find fyo). 3. Find P(IYI
4. (20%) Let X be a continuous random variable with the following PDF Sce-4x 0<x fx(x) = to else where c is a positive constant. (a) (5%) Find c. (b) (5%) Find the CDF of X, Fx(x). (c) (5%) Find Prob{2<x<5} (d)(5%) Find E[X], and Var(X).
Suppose the random variable X has PDF fX(x) = 3x2 when 0 < x < 1 and zero otherwise. What is the PDF of Y = 2X+3? a.Use the general method: Find the CDF of X and use it to get the PDF of Y b.Use a short-cut.
1. Let X be a continuous random variable with support (0, 1) and PDF defined by f(x) = ( cxn 0 < x < 1 0 otherwise, for some n > 1. a) Find c in terms of n. b) Derive the CDF FX(x).
Let X be a continuous random variable with PDF f(x) = { 3x^3 0<=x<=1 0 otherwise Find CDF of X FInd pdf of Y
For a continuous random variable X with the following probability density function (PDF): fX(x) = ( 0.25 if 0 ≤ x ≤ 4, 0 otherwise. (a) Sketch-out the function and confirm it’s a valid PDF. (5 points) (b) Find the CDF of X and sketch it out. (5 points) (c) Find P [ 0.5 < X ≤ 1.5 ]. (5 points)
19. A random variable X has the pdf f(x) = 2/3 0 otherwise if 1 < x 2 (a) Find the median of X. (b) Sketch the graph of the CDF and show the position of the median on the graph.
Problem # 8. a) Let X be a continuous random variable with known CDF FX(x). LetY = g(X) where g(·) is the so-called signum function, which extracts the sign of its argument. In other words, g(X) = { -1 x<0, 0 x=0, 1 x>0 } Express the PDF fY (y) in terms of the known CDF FX(x). b) Let X be a random variable with PDF: fX(x) = { x/2 0 <= x < 2, 0 otherwise} Let Y be...
2. A continuous random variable, x, has the following pdf. 0 otherwise Find 110 (a) the mean, (b) the variance