Why does the probability of the sum of two independent random variables equals the conditional expectation

Question

Suppose there are 3 independent random variables, A,B and C. We also know that

P(A $\le$ x) $\le$ P(B $\le$ x)

I wanted to show that;

P(A + C$\le$ x) $\le$ P(B +C $\le$ x).

The proof to this started as such;

P(A + C$\le$ x) = E(P(A + C$\le$ x|C)) = E(P(A$\le$ x-C|C))

and the proof continues.

I didn't quite get the part why the probability equals the conditional expectation of that.

Answer 1

We can think of $\Pr[A+C \le x]$ in terms of expectation as $\mathbb E[1_{A+C \le x}]$, where $1_{A+C \le x}$ is an indicator random variable: it's $1$ if $A+C \le x$, and $0$ otherwise.

The law of total expectation says $\mathbb E[X] = \mathbb E[\mathbb E[X \mid Y]]$, and that's what we apply here: $$ \Pr[A+C \le x] = \mathbb E[1_{A+C \le x}] = \mathbb E[\mathbb E[1_{A+C \le x} \mid C]] = \mathbb E[\Pr[A+C \le x \mid C]]. $$