# 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. – Michael M Dec 16 '13 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. – JohnK Dec 16 '13 at 18:53
• If you thought my answer was insightful please feel free to mark it as an answer :) – bdeonovic Dec 16 '13 at 19:07
• You fully deserve it. Stick around for more! ;) – JohnK Dec 16 '13 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. – JohnK Dec 16 '13 at 18:51