Proving the linearity property of Expectation in general set up

Question

I want to prove the linearity property of Expectation using two random variables $X$ and $Y$ in the most general set up. That is, I want to show that $E(X+Y)=E(X)+E(Y)$ using ONLY the following axioms:

1. $E(X)=E(X^+)-E(X^-)$ where $X=X^+-X^-$, with $X^+:=\max\{X,0\}$ and $X^-:=\max\{-X,0\}$
2. If $X\geq0$ then Monotone Convergence Theorem is known to us.
3. If $X\geq0$ then the supremum definition of expectation is the only definition of $E(X)$ that we know of.

As a start, if $X\geq0,Y\geq0$ then to prove that $E(X+Y)=E(X)+E(Y)$ where $X,Y$ are discrete random variables, we observe that $X=X^+$ and $Y=Y^+$ and therefore, $E(X+Y)=E(X^++Y^+)$. We need to show that $E(X^+)+E(Y^+)=E(X^++Y^+)$.

How can this be done? I am not quite sure where to start.

Also, it would be great if one can kindly direct me to some pdf's that illustrate this definition of expectation and the properties of expectation proved only using this definition i.e. in the MOST GENERAL SETUP. I searched a lot on Google but could not find any such document.

Answer 1

See comment above that we must assume $E[X]+E[Y]$ does not lead to an undefined case of $\infty+-\infty$ or $-\infty+\infty$.

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Assume the undefined case does not arise. If you believe the result holds for non-negative random variables, and you believe that $E[X+c] = E[X]+c$ for any constant $c$, and $E[-X]=-E[X]$, then you can get it for general:

Fix $M>0$ and define truncated random variables $X_M=\max[X,-M]$ and $Y_M=\max[Y,-M]$. Then:

$$ X + Y \leq X_M + Y_M $$

and so:

\begin{align} E[X+Y] &\leq E[X_M + Y_M] \\ &= E[(X_M+M) + (Y_M+M)] - 2M \\ &= E[X_M+M] + E[Y_M+M] -2M \\ &= E[X_M]+E[Y_M] \end{align}

where we use the fact that $(X_M+M)$ and $(Y_M+M)$ are non-negative. Taking a limit as $M\rightarrow\infty$ gives:

$$ E[X+Y] \leq \lim_{M\rightarrow\infty} (E[X_M]+E[Y_M]) = \lim_{M\rightarrow\infty}E[X_M] + \lim_{M\rightarrow\infty}E[Y_M] = E[X] + E[Y] $$

where the limit rule uses the assumption that we do not get a case of $\infty+-\infty$ or $-\infty+\infty$. The above also uses the fact that $X_M\searrow X$ and so $E[X_M] \searrow E[X]$.

Applying the same result to $-X$ and $-Y$ gives:

$$ E[-X-Y] \leq E[-X]+E[-Y] $$

and so $-E[X+Y] \leq -E[X]-E[Y]$, and so $E[X]+E[Y]\leq E[X+Y]$. These two together give $E[X+Y]=E[X]+E[Y]$.

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The discrete case is easy: If $X$ and $Y$ are discrete and non-negative then: \begin{align} E[X+Y] &= \sum_{i, j} (x_i+y_j)Pr[X=x_i,Y=y_j] \\ &=\sum_{i,j} x_iPr[X=x_i,Y=y_j] + \sum_{i,j}y_jPr[X=x_i,Y=y_j]\\ &= \sum_{i} x_iPr[X_i=x_i] + \sum_j y_jPr[Y=y_j] \\ &=E[X] + E[Y] \end{align}

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A more general argument that perhaps uses scary notation is: If $X$ and $Y$ are non-negative random variables then: $$ \int (X+Y) dP = \int X dP + \int Y dP = E[X] + E[Y] $$