Probability : joint density function of 3 Normal Distributions

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The discussion revolves around finding the joint probability density function (pdf) of three derived variables, Y1, Y2, and Y3, based on independent Gaussian random variables X1, X2, and X3. The relationship between these variables is established through linear combinations, with Y1 being the sum and Y2 and Y3 being differences of the X variables. Participants highlight the importance of knowing whether the original variables are continuous or discrete, as well as their potential values, to accurately determine the joint pdf. It is noted that the pdf of Y1 can be derived from the convolution of the individual pdfs of X1, X2, and X3, while Y2 and Y3 require convolution of specific pairs of these pdfs. The discussion suggests consulting Papoulis' book for a more comprehensive understanding of the topic.
kkirtac
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X1, X2, X3 are independent gaussian random variables.
Y1 = X1+X2+X3
Y2 = X1-X2
Y3 = X2-X3
are given. What is the joint pdf of Y1,Y2 and Y3 ?
 
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kkirtac said:
X1, X2, X3 are independent gaussian random variables.
Y1 = X1+X2+X3
Y2 = X1-X2
Y3 = X2-X3
are given. What is the joint pdf of Y1,Y2 and Y3 ?

There seems to be some info missing here. Are X1, X2, and X3 independent? Are they discrete or continuous? Either way, what values can X1, X2, and X3 take? All this information is necessary in order to find the joint pdf of Y1, Y2, and Y3.
 
Dear,
If you are always watching the post.
the pdf of Y1 is the convolution of the three pdfs of the three random variables (X1, X2 and X3).
for Y2 and Y3, if the random variables are gaussian and centered (mean = 0) then pdf(X) = pdf(-X) and thus for pdf(Y2) = convolution of pdf(X1) and pdf(X2) while pdf(Y3) = convolution of pdf(X2) and pdf(X3).
Actually, I address you to the great book of Papoulis where you can find (for sure) the answer to your wondering .
Cheers
Manar
 
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