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I have the random variable $C=X+Y-Z$, where $X,Y,Z$ are mutually independent. I need to find the density $F_{X+Y-Z|Y=y\cap Z=z}(c)$. I already found this answer for two independent RVs. My question is, does this result generalize. That is, can I write $F_{X+Y-Z|Y=y\cap Z=z}(c)=F_X(C-Y+Z)$?

Edit $(X,Y,Z)$ is continuous.

Nicholas E
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1 Answers1

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With independence $f_{X|Y=y,Z=z}\equiv f_{X}$ and so $$\begin{eqnarray*}P(C\leq c|Y=y,Z=y) &=& P(X\leq -y+z+c|Y=y,Z=z) \\ &=& \int_{-\infty}^{-y+z+c}f_{X|Y=y,Z=z}(x)\mathrm{d}x \\ &=& \int_{-\infty}^{-y+z-c}f_{X}(x)\mathrm{d}x \\ &=& F_{X}(-y+z+c) \end{eqnarray*}$$ Therefore $$f_{C|Y=y,Z=z}(c)=f_X(-y+z+c)$$ which is a similar result to the link in your question. Note here we're assuming $(X,Y,Z)$ is a continuous random vector.

Matthew H.
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