Exponential transform

Jul 2010
26
0
If X and Y are Exp(1), we can write down that f(x+y) = exp -(x+y)

Is it possible to verify this using the transformation:

U = X+Y
V = X-Y

Any pointers most appreciated
 

mr fantastic

MHF Hall of Fame
Dec 2007
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Zeitgeist
If X and Y are Exp(1), we can write down that f(x+y) = exp -(x+y)

Is it possible to verify this using the transformation:

U = X+Y
V = X-Y

Any pointers most appreciated
What you have posted doesn't make sense to me. Please post the original question exactly as it's stated. Is f(x + y) meant to be the joint pdf of X and Y? Or are you taling about the pdf of the random variable X + Y? Note: The sum of two exponential random variables is NOT what you are claiming.
 
Jul 2010
26
0
Sure, the exact wording is:

"If X,Y are i.i.d. Exp(1), write down the distribution of X+Y and verify this by use of this transformation U=X+Y, V=X-Y."
 

mr fantastic

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Sure, the exact wording is:

"If X,Y are i.i.d. Exp(1), write down the distribution of X+Y and verify this by use of this transformation U=X+Y, V=X-Y."
Read 7.2 here: http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf

For the verification part of the question, use the change of variable theorem (see for example Walpole, Myers and Myers (1998): Probability and Statistics for Engineers and Scientists, p 185).

If you need more help, please show all your work and say where you get stuck.
 
Jul 2010
26
0
thanks - that is really helpful. i will write again if I am stuck
 
Jul 2010
26
0
Example 7.4 seems to give the answer I am looking for.

My questions asks to verify by using U = X+Y and V=X-Y

Example 7.4 gives the distribution function for U. I think I have worked out the distribution function for V also as

f(v) = lamda/2 * (exp(-lamda *v) - exp(-3*lamda*v))

Is it now a question of working out f(x) based on X=U+V?