$X$ is an absolutely continous random variable, with a continous density function, and:

$$X_n= \frac{ \lfloor nX \rfloor}{n}$$

What is the distribution of $X_n$, and what can we say about its convergence?

I tried to calculate the distribution like this:

$$\begin{align*}P(X_n =k)&=P\left(\frac{ \lfloor nX \rfloor}{n}=k\right)=P\left(nk\leq nX <n(k+1)\right)=\int_k^{k+1} f_X\;dx\\&=F_X(k+1)-F_X(k)\end{align*}$$

I am not sure what I did was right, please help!

  • 1
    $\begingroup$ Yes, it seems fine. $\endgroup$ – Jimmy R. Nov 17 '14 at 17:11
  • $\begingroup$ The formula is wrong. Try rather, for every integer $i$, $$P(X_n=i/n)=F_X((i+1)/n)-F_X(i/n).$$ $\endgroup$ – Did Nov 17 '14 at 17:30
  • $\begingroup$ @Did Does this mean that $X_n$ converges to 0 and not to $X$? $\endgroup$ – user010010001 Nov 17 '14 at 17:45
  • $\begingroup$ No, since $P(X_n\leqslant x)\to F_X(x)$ for every $x$. $\endgroup$ – Did Nov 17 '14 at 18:13

Your Answer

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

Browse other questions tagged or ask your own question.