Expectation of discrete random variable

Question

Give a sequence of random variables $x_1,..,x_n$ with $x_n$ having a density of:

$$f_N(x) = \begin{cases} \frac{2N-1}{3N};x=1\\ 1/3;x=1+\frac{1}{N+1} \\ \frac{1}{3N};x=2\end{cases}$$

What would be the expectation and to which value does this function converge to?

I'd assume that this function converges in mean square to a fixed value and is degenerate, but i'm not quite sure what the expectation would be:

$$\lim_{N \to \infty} E(X) = \frac{2N-1}{3N} + 1/3*(1+\frac{1}{N+1})+2*(\frac{1}{3N})$$

As far as i can see this converges to 2/3 but that seems wrong.

Answer 1

You have the expectation right, but the limit wrong.

\begin{align*} E(X) & = \dfrac{2N - 1}{3N} + \dfrac{1}{3} \left(1 + \dfrac{1}{N+1} \right) + \dfrac{2}{3N}\\ & = \dfrac{2 - 1/N}{3} + \dfrac{1}{3} \left(1 + \dfrac{1}{N+1} \right) + \dfrac{2}{3N}\\ \lim_{N\to\infty} E(X) &= \lim_{N \to \infty}\dfrac{2 - 1/N}{3} + \dfrac{1}{3} \left(1 + \dfrac{1}{N+1} \right) + \dfrac{2}{3N}\\ & = \dfrac{2}{3} + \dfrac{1}{3}\\ & = 1. \end{align*}