In general the [[Expectation]] of a function is NOT equal to a function of the expectation. An example is the quadratic function, where $g(X)=x^2$. $ \begin{align} \mathbb E[g(X)] &\not = g(\mathbb E[X]) \\[4pt] \mathbb E[X^2] &\not = (\mathbb E[X])^2 \end{align} $ ## Functions with Constants However, if the function $g(X)$ is linear, then the equality actually holds. A function is linear, if it only adds or multiplies constants to the input $x$. $ \begin{rcases} Y&=g(X) \\ g(X) &= ax+b \end{rcases} \, \mathbb E[ax+b]=a\mathbb E[x]+b $ **Derivation:** $ \begin{align} \mathbb E[Y] &= \sum_x g(X) * p_X(x) \tag{1}\\ &= \sum_x ax+b * p_X(x) \tag{2}\\ &= \sum_x ax * p_X(x) + \sum_x b * p_X(x) \tag{3}\\ &= a \underbrace{\sum_x x p_X(x)}_{\mathbb =E[X]} + b \underbrace{\sum_x p_X(x)}_{=1} \implies a\mathbb E[X] +b \tag{4} \end{align} $ where: - (2) Insert actual function for $g(X)$. - (3) Split up the sums. - (4) Pull out constants from the summation. ## Functions with Random Variables Now we extend the concept of linearity by linear functions of random variables. $ \mathbb E[X+Y]= \mathbb E[X] + \mathbb E[Y] $ **Derivation:** Let us take a function $g(x,y)=x+y$, and calculate the expectation of that function. $ \begin{align} \mathbb E[X+Y] &=\mathbb E[g(x,y)] \tag{1}\\[8pt] &=\sum_x \sum_y (x+y)*p_{XY}(x,y) \tag{2}\\ &=\sum_x \sum_y x*p_{XY}(x,y) + \sum_y \sum_x y*p_{XY}(x,y) \tag{3}\\ &=\sum_x x \sum_y p_{XY}(x,y) + \sum_y y \sum_x p_{XY}(x,y) \tag{4}\\ &=\sum_x x*p_X(x) + \sum_y y* p_Y(y) \tag{5}\\ &=\mathbb E[X] + \mathbb E[Y] \tag{6} \end{align} $ (2) Insert the function and sum over all $(x,y)$ pairs. (4) Pull $x$ out of the sum that is summing over all $y$, because it is just a constant in there. Vice versa for $y$ in the second double summation. (5) The summing over one of the 2-dimensional joint PMFs, makes it a 1-dimensional PMF across the axis not summed over.