Proving linearity of the expected value of two (possibly dependent) Bernoulli variables

Question

I'm currently working on proving that the expected value operator is linear, and felt quite content with my work until recently. However, I was recently made aware that my proof is somewhat lacking, as it assumes that the two Bernoulli variables in question are independent. This assumption is normally not stated when saying that the expected value operator is linear, so I guess my proof shouldn't either.

After searching a bit online I came across this useful post for determining the values of the pmf of the sum: Joint probability distribution of two Bernoulli r.v. with a correlation $r$

With notation one could here say that I'm interested in two Bernoulli variables, $X \sim \text{Bernoulli}(p_1)$ and $Y \sim \text{Bernoulli}(p_2)$. Then from the above link, which I have convinced myself sounds reasonable, with $R=\text{Cov}(X, Y)$:

$$ P(X=1, Y=1) = p_1\cdot p_2 + R =: a \\ P(X=1, Y=0) = p_1 - p_1\cdot p_2 - R =: b \\ P(X=0, Y=1) = p_2 - p_1\cdot p_2 - R =: c \\ P(X=0, Y=0) = 1 - p_1 - p_2 + p_1 \cdot p_2 + R := d $$

This seemed all fine to me, before I tried to find the expected value of the sum $S=X+Y$. Here the only probabilities I'm interested in are P(S=1) and P(S=2) (as P(S=0) is given by these, but it's also eliminated in the expectation).

Here I have tried different ways of calculating the probabilities, but I believe that there's either something seriously wrong with my understanding or that I have been looking at this for too long. The two approaches I have tried are $$ P(S=1) = P(X=0|Y=1) + P(X=1|Y=0) + P(Y=0|X=1) + P(Y=1|X=0) $$ Then I have used the following relation $$ P(A|B)=\frac{P(A, B)}{P(B)} $$ for each of the four values in the expression for $P(S=1)$, but doing this also for $P(S=2)$ and combining, I found an answer which even Wolfram Alpha decided wasn't the wanted $p_1 + p_2$.

When I use the approach $$ P(S=1) = P(X=0, Y=1) + P(X=1, Y=0) $$

however, I actually get the correct expected value for $S$, but I am struggling to see whether this should actually be the case.

Any ideas or feedback is highly appreciated

Answer 1

The issue is your formula for $$P(S=1) = P(X=0|Y=1)+P(Y=1|X=0)+P(X=1|Y=0)+P(Y=0|X=1),$$ which really doesn't make sense to me. I don't see why you would add all those values together: you rarely add together conditional probabilities when the events you are conditioning on are different in each one.

The second formula $$P(S=1)=P(X=1,Y=0)+P(Y=1,X=0)$$ is correct. If you wanted to involve conditional probabilities, you could use $$P(X=1,Y=0)=P(X=1|Y=0)P(Y=0)=P(Y=0|X=1)P(X=1)$$ by moving the terms in the Bayes formula you wrote.

Answer 2

Not sure why we need to focus on a Bernoulli.

The expectation operator $E[X]$ for $X: \Omega \to \mathbb{R}$ is given by:

$$E[X] = \int_{\omega \in \Omega} X(\omega) dP(\omega)$$

Therefore, if we have $X,Y$ defined on $\Omega$ we have:

$$E[aX+bY] = \int_{\omega \in \Omega} aX(\omega) + bY(\omega) dP(\omega) = a\int_{\omega \in \Omega} X(\omega)dP(\omega) + b\int_{\omega \in \Omega} Y(\omega)dP(\omega) = aE[X] + bE[Y]$$

This what establishes expectation as a linear operator.