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I know that c =1. Easy enough to do. I don't know how to do the rest though.

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- Aug 26th 2009, 06:16 AMDCUContinuous random variables question.
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I know that c =1. Easy enough to do. I don't know how to do the rest though. - Aug 26th 2009, 01:26 PMDCU
I really need help with this urgently. I've an exam in the morning and it could come up. Any help would be greatly appreciated.

Ignore part(a) - Aug 26th 2009, 03:42 PMhalbard
OK, if you insist. Here it comes, ready or not...

You could try $\displaystyle f_{X+Y}(z)=\frac{\mathrm d}{\mathrm dz}\mathrm P(X+Y\leq z)$, where perhaps $\displaystyle \mathrm P(X+Y\leq z)=\int_{X+Y\leq z}f_{X,Y}\mathrm dA=\int_{-\infty}^\infty\int_{-\infty}^{z-y}f_{X,Y}(x,y)\mathrm dx\mathrm dy$.

So maybe $\displaystyle f_{X+Y}(z)=\int_{-\infty}^\infty f_{X,Y}(z-y,y)\mathrm dy$.

Tired now, need sleep... $\displaystyle f(zzzzzzzzzzzzzzzzzzzzzzzzzzzzz$ - Aug 26th 2009, 11:32 PMDCU
Thank you! I thought that might be it, but it seemed too easy!(Rofl)