Question 4 (a) The continuous random variables X and Y are distributed uniformly on (0, 1) and exponentially with mean 1, respectively. X and Y are independent. (i) Use the convolution formula to show that the random variable Z = X +Y has probability density function (pdf) 0, 1-e-*, (e- 1) e-, z>1. fz(z) = 0
Question 4 (a) The continuous random variables X and Y are distributed uniformly on (0, 1) and exponentially with mean 1, respectively. X and Y are independent. (i) Use the convolution formula to show that the random variable Z = X +Y has probability density function (pdf) 0, 1-e-*, (e- 1) e-, z>1. fz(z) = 0
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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please send solution for part a iii
![Question 4
(a) The continuous random variables X and Y are distributed uniformly on (0, 1)
and exponentially with mean 1, respectively. X and Y are independent.
(i) Use the convolution formula to show that the random variable Z = X +Y
has probability density function (pdf)
= (2)25
(e - 1) e-, z> 1.
1-e=²,
0 <z<%;
(ii) Show that fz(z) is a valid pdf.
(iii) Let Fz(z) be the cumulative distribution function (cdf) of Z. Show that
z<0;
0,
z+e-1,
Fz(2)
0<z<1;
%3D
e(1- e) +1, z> 1.
and hence find the median of Z.
(b) The random variable X has pdf
S 6x (1-2), 0 < x < 1;
otherwise.
fx (x) =
Determine the pdf of Y = X/(1 X).
(c) The random variable X has pdf](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe97adabd-b82c-44aa-81cf-e535a4a6e5fd%2F5e860fe2-405b-439d-9a18-3f183fd5a318%2F1ajov9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question 4
(a) The continuous random variables X and Y are distributed uniformly on (0, 1)
and exponentially with mean 1, respectively. X and Y are independent.
(i) Use the convolution formula to show that the random variable Z = X +Y
has probability density function (pdf)
= (2)25
(e - 1) e-, z> 1.
1-e=²,
0 <z<%;
(ii) Show that fz(z) is a valid pdf.
(iii) Let Fz(z) be the cumulative distribution function (cdf) of Z. Show that
z<0;
0,
z+e-1,
Fz(2)
0<z<1;
%3D
e(1- e) +1, z> 1.
and hence find the median of Z.
(b) The random variable X has pdf
S 6x (1-2), 0 < x < 1;
otherwise.
fx (x) =
Determine the pdf of Y = X/(1 X).
(c) The random variable X has pdf
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