Let U ~uniform(0,1). Let Y =−ln(1−U).
hint: If FX (x) = FY (y) and supports x,y ∈ D, X and Y have the
same distribution.
Find FY (y) and fY (y). Now, it should be straight forward that Y
follows distribution with parameter_____________-
Let U ~uniform(0,1). Let Y =−ln(1−U). hint: If FX (x) = FY (y) and supports x,y...
5-1. Let U ~ Uniform(0,1) and X = – ln(1 – U). Show that The CDF of X is Fx(x) = 1 – e-X, 0 < x < 0 In other word, X is exponentially distributed with 2 = 1.
5. Let X have a uniform distribution on the interval (0,1). Given X = x, let Y have a uniform distribution on (0, 2). (a) The conditional pdf of Y, given that X = x, is fyıx(ylx) = 1 for 0 < y < x, since Y|X ~U(0, X). Show that the mean of this (conditional) distribution is E(Y|X) = , and hence, show that Ex{E(Y|X)} = i. (Hint: what is the mean of ?) (b) Noting that fr\x(y|x) =...
U is Uniform distribution here Let X ~ U[0,1] and Y = max {,x) (a) Is Y a continuous random variable? Justify (b) Compute E[Y]. (Hint: Note that when a (Hint: Note that when a-, max 1.a- , and when a > ļ, max | , a- ax {3a, and when a > a
Let U U (0,1) and let Y=1-U. Derive an expression for the odf Fy() of Y in terms of the odf of U and hence show that Y U (0,1).
5-1. Let U - Uniform(0,1) and X = - In(1-U). Show that the CDF of X is Fx(x) = 1 -e*, 0<x<0 In other word, X is exponentially distributed with 1 = 1.
Find fx, fy, and fz 5) f(x, y, z) = ln (xy)?
X is a positive continuous random variable with density fX(x). Y = ln(X). Find the cumulative distribution function (cdf) Fy(y) of Y in terms of the cdf of X. Find the probability density function (pdf) fy(y) of Y in terms of the pdf of X. For the remaining problem (problem 3 (3),(4) and (5)), suppose X is a uniform random the interval (0,5). Compute the cdf and pdf of X. Compute the expectation and variance of X. What is Fy(y)?...
1. Let $(x) = 2x2 and let Y = $(x). (a) Consider the case X ~U(-1,1). Obtain fy and compute E[Y] (b) Now instead assume that Y ~ U(0,1/2) and that X is a continuous random variable. Explain carefully why it is possible to choose fx such that fx (2) = 0 whenever 21 > 1. Obtain an expression linking fx(2) to fx(-x) for 3 € (-1,1). Show that E[X] = -2/3 + 2 S xfx(x) dx. Using your expression...
STAT 115 Let X be a continuous random variable having the CDF Fx(x) = 1 - e^ (-e^x) (1) Find the Probability Density Function (PDF) of Y=e^X. (2) Let B have a uniform distribution over (0,1). Find a function G(b) and G(B) has the same distribution as X.
Let $(x) = 2x2 and let Y = $(X). assume that Y ~ U(0,1/2) and that X is a continuous random variable. fx(x) = 0 whenever |2| > 1. Obtain an expression linking fx(x) to fx(-x) for xe (-1,1). Show that E[X] = -2/3 + 28. xfx(x) dx. Using your expression linking fx(x) and fx(-x), obtain an upper bound for E[X] and a pdf fx for which this bound is attained. [10]