April 22nd, 2012, 04:19 AM #1 Newbie   Joined: Apr 2012 From: Canada Posts: 7 Thanks: 0 Joint Distribution help please Can't get my head around this joint distribution problem. Suppose X and Y have joint distribution given by: Find the distribution of X+Y Having difficulties finding the boundaries for my integration. I know that I need do declare a dummy variable t = X + Y Been stuck on this for a few hours now, help is much appreciated! Thanks. April 22nd, 2012, 01:50 PM   #2
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 Originally Posted by batman350z Can't get my head around this joint distribution problem. Suppose X and Y have joint distribution given by: Find the distribution of X+Y Having difficulties finding the boundaries for my integration. I know that I need do declare a dummy variable t = X + Y Been stuck on this for a few hours now, help is much appreciated! Thanks.
f(x,y) cannot be either a density function or a distribution function without further limits on x and y. y can be negative giving negative values for f for y < -x. x can be arbitrarily large - distribution would eventually be > 1. April 22nd, 2012, 02:02 PM #3 Newbie   Joined: Apr 2012 From: Canada Posts: 7 Thanks: 0 Re: Joint Distribution help please Hmm, I might be misunderstanding the boundaries. The question actually has them written as, , so I am guessing that would be interpreted as If the above is the case, how would i go about it? April 23rd, 2012, 03:50 PM #4 Global Moderator   Joined: May 2007 Posts: 6,856 Thanks: 745 Re: Joint Distribution help please First f(x,y) is a density function, not a distribution function. Let T = X + Y. What you want to calculate is P(T < t) = P(X+Y < t) = P(X < t and Y < t - X). To do this simply integrate the density function over the domain specified, 0 < x < t, 0 < y < t - x. The main thing you have to be careful of is the fact that f(x,y) = 0 outside the unit square (in x,y), so that you have to break up the integration into the parts where f(x,y) = x + y and f(x,y) = 0. For 0 < t < 1, there is no problem, but for 1 < t < 2 you need to do the breakup. For t > 2, the probability = 1. Post script: for t > 1, it is easier to work with P(T > t) and the use P(T < t) = 1 - P(T > t). In this case P(T > t) = P(X > t-1 and Y > t - X) so the integrals of f(x,y) is over t-1 < x < 1 and t-x < y < 1. For t < 1 the integral to get P(T < t) directly has 0 < x < t and 0 < y < t-x. Tags distribution, joint Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post 450081592 Advanced Statistics 0 January 24th, 2012 08:44 PM xdeathcorex Probability and Statistics 0 April 22nd, 2011 08:55 AM allandmunja Advanced Statistics 1 April 4th, 2011 02:05 PM sletcher Calculus 18 December 15th, 2010 12:51 PM meph1st0pheles Advanced Statistics 1 March 23rd, 2010 06:47 PM

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