Expected Value linearity property (simple derivation)

Question

There is a simple derivation that I am not sure about. There is a proof in my book for:

E(bX+a) = b × E(X) + a

So, it starts with

This is based on definition of expected value:

But what confuses me is that in my case the random variable is bX(e)+a, but the probability is for P(e). I mean two ingredients of the sigma do not match. Maybe it can be written as P(bX(e)+a), but then I do not know why P(bX(e)+a) will be equal to P(e). How this should be explained intuitively?

Thanks!

Answer 1

Keep in mind that your random variable $X$ depends on the value of $e\in\Omega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-\sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $e\in\Omega$. In particular, the expected value of any function of $X$, say $f(X)$, is $$\mathbb{E}[f(X)]=\sum_{e\in\Omega}P(e)f(e)$$ In your case, $f(X)=aX+b$