If the limit of the difference of two random variables goes to 0, are the limits of their expectations the same?

Question

I have two discrete random variables $X_n$ and $Y_n$ and a relation between them that looks like, for $n\geq 1$ $$X_n = a_nP\{A_n\} + Y_n P\{{A_n}^c\},$$ where $A_n$ is an event and $a_n$ depends on $n$ but is not a random variable. The thing is, the sequence of probabilities $P\{A_n\}$ go to $0$ as $n\to \infty$, so I believe this means that $X_n$ converges to $Y_n$ in some meaningful way, but I'm not sure which. From here, can one conclude that $$\lim_{n\to \infty} E\{X_n\} = \lim_{n\to\infty} E\{Y_n\}?$$ I'm sorry if this is very elementary. I have difficulties getting my head around limits and expectations.

Answer 1

Since $X_n = a_n P(A_n) + Y_nP(A_n^c)$, we know by the linearity of the expected value that: $$E(X_n) = E(a_n P(A_n) + Y_nP(A_n^c))= E(a_n P(A_n)) + E(Y_nP(A_n^c)) =$$ $$ =a_nP(A_n) + P(A_n^c)E(Y_n)$$

If $a_n \rightarrow a$ and if $Y_n$ converges (note that this is necessary for the limit to exist), then $$ \lim_{n \to \infty}E(X_n) =\lim_{n \to \infty}a_nP(A_n) + P(A_n^c)E(Y_n)= \lim_{n \to \infty}a_nP(A_n) + \lim_{n \to \infty}P(A_n^c)E(Y_n) $$ Since $P(A_n) \rightarrow 0$, we conclude that $$ \lim_{n \to \infty}E(X_n) =\lim_{n \to \infty}P(A_n^c)\lim_{n \to \infty}E(Y_n)=\lim_{n \to \infty}E(Y_n) $$