# Limiting pmf as $n \to \infty$

Consider the simple pmf:

$$p_n (x)=\begin{cases} 1\quad x=2+1/n \\ 0\quad \text{elsewhere} \end{cases}$$

Then my book states that $\lim_{n\to \infty} p_n (x)=0$ for all values of $x$. Is that really the case though? Why can we not say that the probability of $x=2$ equals $1$ as $n \to \infty$? Thank you.

• If $X = 2$, you are in the lower case of the '{'. The limit of a sequence of zeros is just zero. Dec 16, 2013 at 18:00

Just noticed I misread the problem. The proof I sketch below is if $p_n(x) = \left\{ \begin{array}{l l} 1 & \quad \text{$x=1+1/n$}\\ 0 & \quad \text{otherwise} \end{array} \right.$ but a similar proof holds for the OP's question.

Let $p(x)=0$. Pick any value of $x \notin \{1+1/n : n=1,2,3,\ldots\}$. Then $p_n(x)=0$ so $|p_n(x)-p(x)|=0$. Note that $1 \notin \{1+1/n : n=1,2,3,\ldots\}$ so $p_n(1)=p(1)=0$.

Now pick any $x\in \{1+1/n : n=1,2,3,\ldots\}$. Let $x=1+1/m$, for $n\geq m+1$, $p_n(1+1/m)=0$.

It is even further interesting that although the pmf converges to 0, the corresponding CDF does not!

$F_n(x)=P(X_n\leq x) \overset{D}{\longrightarrow} F(x) = \left\{ \begin{array}{l l} 0 & \quad \text{$x<1$}\\ 1 & \quad \text{$x\geq 1$} \end{array} \right.$ at all points of continuity of $x$. Therefore $X_n \overset{D}{\longrightarrow} X$ where $P(X=1)=1$. Even though the density functions converged...the corresponding probability mass functions do not!

• Yes that was precisely the point my book was trying to make. The difference between the convergence of the pmf and the respective convergence of the CDF. Thank you. Dec 16, 2013 at 18:53
• If you thought my answer was insightful please feel free to mark it as an answer :) Dec 16, 2013 at 19:07
• You fully deserve it. Stick around for more! ;) Dec 16, 2013 at 19:09

Why can we not say that the probability of $x=2$ equals 1 as $n\to\infty$?

For every $n\geq 1$, $p_n(2)=0$. Hence, the sequence of real numbers $$(p_1(2),p_2(2),p_3(2),\dots) = (0,0,0,\dots)$$ is a constant sequence with limit equal to zero.

• Thanks, asymptotics can be a little tricky for the uninitiated. Dec 16, 2013 at 18:51