1. Let X and Y be independent continuous random variables with PDFs fx (x) and fy (y). (a) Find a general expression for the CDF Fz (z) = P(X+Y ≤z) of their sum Z = X+Y by integrating the joint PDF fx,y (x, y) over an appropriately chosen region of xy-space. (b) Next, by differentiating this expression with respect to z, show that the PDF fz of Z is the convolution of fx and fy, that is, fz(z) = fx (z-y) fy (y)dy. Note: In doing so, you may assume that is possible to "differentiate under the integral sign", that is, interchange the order of the derivative and the "outside" integral in your expression for Fz(z). Then apply the Fundamental Theorem of Calculus.
1. Let X and Y be independent continuous random variables with PDFs fx (x) and fy (y). (a) Find a general expression for the CDF Fz (z) = P(X+Y ≤z) of their sum Z = X+Y by integrating the joint PDF fx,y (x, y) over an appropriately chosen region of xy-space. (b) Next, by differentiating this expression with respect to z, show that the PDF fz of Z is the convolution of fx and fy, that is, fz(z) = fx (z-y) fy (y)dy. Note: In doing so, you may assume that is possible to "differentiate under the integral sign", that is, interchange the order of the derivative and the "outside" integral in your expression for Fz(z). Then apply the Fundamental Theorem of Calculus.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:1. Let X and Y be independent continuous random variables with PDFs fx(x)
and fy (y).
(a) Find a general expression for the CDF Fz(z) = P(X+Y ≤z) of their sum
Z = X + Y by integrating the joint PDF fx,y (x, y) over an appropriately
chosen region of xy-space.
(b) Next, by differentiating this expression with respect to z, show that the
PDF fz of Z is the convolution of fx and fy, that is,
fz(z) = fx (z-y) fy (y)dy.
Note: In doing so, you may assume that is possible to "differentiate
under the integral sign”, that is, interchange the order of the derivative
and the "outside" integral in your expression for Fz (z). Then apply the
Fundamental Theorem of Calculus.
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