0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – drhab Sep 21 '18 at 14:19
  • $\begingroup$ 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. $\endgroup$ – drhab Sep 21 '18 at 14:25
  • $\begingroup$ 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. $\endgroup$ – Michael Conlen Sep 21 '18 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.