# Proof that if $X$ and $Y$ are continuous random variables that $E[X+Y]=E[X]+E[Y]$, how to apply Fubini's Theorem

I'm trying to understand the proof that if $$X$$ and $$Y$$ are continuous random variables then $$E[X+Y]=E[X]+E[Y]$$ from Introduction to Probability Models by Ross.

Let $$g(X, Y) = X+Y$$ and we compute $$E[X+Y]$$; then

$$E[X+Y]=\int_\mathbb{R}\int_\mathbb{R} (x+y)f(x, y)dxdy$$ $$=\int_\mathbb{R}\int_\mathbb{R} xf(x, y)dx dy + \int_\mathbb{R}\int_\mathbb{R} yf(x, y)dx dy$$

At this point the book says to apply the fact that if $$X$$ is a continuous random variable with PDF f(x) then for real valued $$g$$ $$E[g(x)]=\int_\mathbb{R}g(x)f(x)dx$$ and let $$g(x)=x$$ for the first integral above. To apply this I need the PDF for $$X$$ derived from the joint PDF $$f(x, y)$$ of $$X$$ and $$Y$$

The PDF for $$X$$

$$f_X(x)=\int_\mathbb{R} f(x, y) dy$$

So now returning to $$\int_\mathbb{R}\int_\mathbb{R} xf(x, y)dx dy$$ I need to change the order of integration, I'd like to write

$$\int_\mathbb{R}\int_\mathbb{R} xf(x, y)dx dy=\int_\mathbb{R} x\int_\mathbb{R} f(x, y) dy dx = \int_\mathbb{R}xf_X(x)dx=E[X]$$

I believe I need to apply Fubini's theorem so I need to show that

$$\int_\mathbb{R}\int_\mathbb{R} |xf(x, y)|dx dy$$ is finite. This isn't really a given and I think if $$X$$ is Cauchy may not even be defined.

How would I complete this proof?

• Are you familiar with underlying probability spaces? There is a much better route to prove that $\mathbb E(X+Y)=\mathbb EX+\mathbb EY$. It is closer to the definition of expectation and does not depend on continuity and/or the existence of PDF's. See here. – drhab Sep 21 '18 at 14:19
• Concerning your question: that $\int\int|x|f(x,y)|dxdy<\infty$ is not something that must be proved (that is impossible if you have no further info), but is something that must be preassumed. – drhab Sep 21 '18 at 14:25
• I've had about two weeks of measure theory at the end of an Analysis course and a semester of probability about 7 years ago, so the answer is "I know what you're talking about, but I'm not really familiar". This book is very "intuitionist" and I found the proof here lacking and interesting to work with. – Michael Conlen Sep 21 '18 at 14:31