0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It looks like 2/3 + 1/3, no? A little nit: your $f$ is a probability mass function, not a density. $\endgroup$ – Adrian Mar 26 '16 at 14:08
1
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.