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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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.
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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