Two trivial Questions about Expectation

Question

1.) Let $X$ and $Y$ be simple random variables for which expectation exists. Why is $$ \mathbb{E}(X)=\mathbb{E}_P(X):=\sum_{x\in X(\Omega)} xP(X=x) = \sum_{x\in X(\Omega),y\in Y(\Omega)} xP(X=x,Y=y) $$

and maybe even more basically:

2.) Why is $$ \sum_{y\in Y(\Omega)} P(X=x,Y=y)=P(X=x) $$

I know this is trivial for a lot of you, but I guess that it might help some folks who are still a bit unsure about the properties of expectation.

Answer 1

I think it's clear that the first point follows from the second.

To understand why the second equality holds I think it's a good idea to revise what $P(X=x)$ actually means: $P(X=x)$ takes the probability of the preimage of $x$ under $X$ so $$P(X=x)=P(X^{-1}(x)).$$ In addition, $P(X=x,Y=y) = P(X^{-1}(x) \cap Y^{-1}(y))$ by definition.

For $y_1 \neq y_2$ the sets $Y^{-1}(y_1)$ and $Y^{-1}(y_2)$ are disjoint, therefore in particular the sets $\{X^{-1}(x) \cap Y^{-1}(y)\}_{y \in Y(\Omega)}$ are disjoint. But this means that the sum of the probabilities of these sets can be determined by taking the probability of the union:

$$\sum_{y \in Y(\Omega)}P(X=x,Y=y)=\sum_{y \in Y(\Omega)}P(X^{-1}(x) \cap Y^{-1}(y)) = P (X^{-1}(x) \cap (\cup_{y \in Y(\Omega)}Y^{-1}(y)) = P(X^{-1}(x) \cap \Omega) = P(X^{-1}(x)) = P(X=x)$$

Answer 2

We have $\cup_y\{\omega | X(\omega) = x, Y(\omega) = y \} = \{\omega | X(\omega) = x \}$ and the sets $\{\omega | X(\omega) = x, Y(\omega) = y \}$ are disjoint, so $P(\cup_y\{\omega | X(\omega) = x, Y(\omega) = y \}) = \sum_y P(\{\omega | X(\omega) = x, Y(\omega) = y \}) = P( \{\omega | X(\omega) = x \})$.

Answer 3

To answer the more general question...

If you have an events $(X=x,Y=y_1),...,(X=x,Y=y_n)$ with different probabilities when you sum these probabilities together you simply have the probability of the event $(X=x)$.

For example, say you have a fruit. $X$ is the type of fruit and $Y$ is the colour. If you sum up the probability of having a green apple, a red apple, an orange apple etc. then what you get is the probability of having an apple.

To your first question, first consider a single value of $x$ then notice that when we sum up over all the possible $Y$ values,

$$\sum_{y\in Y(\Omega)} xP(X=x,Y=y) = \quad x\sum_{y\in Y(\Omega)} P(X=x,Y=y) = \quad xP(X=x)$$

Then applying summation over $x$ to both sides gives:

$$\sum_{x\in X(\Omega)}\sum_{y\in Y(\Omega)} xP(X=x,Y=y) = \sum_{x\in X(\Omega)}\quad xP(X=x)$$

Hope this helps.