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Let $U=X+Y$, where $X, Y$ are not guaranteed to be independent.

It seems trivial that $$E[U] = E[X+Y].$$

However, taking a closer look, $$ E[U] = E_U[U] = \int_{-\infty}^{\infty} u f_U(u) du $$ $$ E[X+Y] = E_{X, Y}[X+Y] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x+y) f_{X, Y}(x, y) dx dy. $$

My question is: Their being equal is not trivial, am I right?

I know how to prove it. But I want to get a confirmation from you guys that we cannot simply write $E[U] = E[X+Y]$, because one is the expectation over a pdf while the other is the expectation over a joint pdf. Like, for the integral part, one is a single integral while the other is an iterated integral.

My proof

Let $U = X+Y$, $V=Y$, we have \begin{align} X = U - V && Y = V. \end{align} And \begin{align} J = \text{det} \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = 1. \end{align} Then, \begin{align} f_{U, V}(u, v) &= f_{X, Y}(u-v, v) |J| = f_{X, Y}(u-v, v) \\ f_{U}(u) &= \int_{-\infty}^{\infty} f_{U, V}(u, v) dv = \int_{-\infty}^{\infty} f_{X, Y}(u-v, v) dv = \int_{-\infty}^{\infty} f_{X, Y}(u-y, y) dy \end{align} Then, \begin{align} E_{U}[U] &= \int_{-\infty}^{\infty} u f_{U}(u) du \\ &= \int_{-\infty}^{\infty} u \left[ \int_{-\infty}^{\infty} f_{X, Y}(u-y, y) dy \right] du \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} u f_{X, Y}(u-y, y) du dy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (x+y) f_{X, Y}(x, y) dx dy ~~~~ (\text{let } x = u-y) \\ &= E_{X, Y}[X+Y]. \end{align}

JJ.
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    Yes, expectation is linear, since expectation is an integral. Namely $\int (aX+bY),dP = a\int X,dP+b\int Y,dP$ – Andrew Sep 26 '23 at 02:43
  • I think Andrew's comment proves $E[X+Y]=EX+EY$, not $E[U]=E[X+Y]$. – JJ. Sep 26 '23 at 02:52
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    The two are the same – Andrew Sep 26 '23 at 02:55
  • No I don't think so. See my original post. One is a single integral. The other is an iterated integral. – JJ. Sep 26 '23 at 02:58
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    You need to review the definition of expectation and random variables. In particular, one has a probability space $(\Omega,\mathcal F,P)$ and the expectation amounts to integrating a function over $\Omega$. If $h(\omega) = g(\omega)+f(\omega)$, then $E[h]=\int h,dP = \int g,dP+\int f,dP = E[g+f]$ – Andrew Sep 26 '23 at 03:13
  • I admit that I did not read prob theory book, and that's why I did not use the prob space language. You could be right, but I am not sure. Please see my updated proof above. So you are saying that the proof is an overkill? – JJ. Sep 26 '23 at 03:23
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    @JJ. The proof is certainly overkill in the sense that it is immediate that $E(X+Y)=E(U)$ when $X+Y=U$ - that follows once you accept $E$ is a well-defined function on the set of random variables. However, there are nontrivial things going on regarding why the various calculations for expectation actually are valid ways to compute it - a good measure-theoretic treatment of probability theory will rigorously show why various methods are correct - you need a decent number of definitions to make things rigorous. – M W Sep 26 '23 at 03:36
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    The proof you wrote is only a proof in the very special case of $(X,Y)$ having a density with respect to Lebesgue measure. It does not generalize to other cases. This issue aside, it’s correct…but it shows the power of having different representations of objects in mathematics. Whether you have a probability/measure theory background or not, random variables are functions. Thus $U$ and $X+Y$ represent the same object, and if the expectation operator depended on the representation of said objects, then well… it wouldn’t be a very good theory – Andrew Sep 26 '23 at 03:38
  • Thanks Andrew and M W. That sounds great. Thanks! I will read some parts of the probability theory next. – JJ. Sep 26 '23 at 03:57

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