Does really convergence in distribution or in law implies convergence in PMF or PDF?

Question

Ref :Introduction to Mathematical Statistics-Prentice Hall (1994) by Robert V. Hogg, Allen Craig.

Now , in the above problem it has been shown that a sequence converges to a random variable X in distribution but the sequence of PMF doesn't converge to the PMF of X.

but we know that "a sequence {Xn} with PDF/PMF {fn} converges to a random variable X (with PMF/PDF 'f' ) in law or distribution if and only if fn → f ".

So, the example and the statement contradict each other . I think the logic that 'lim fn(x) = 0 for all values of x' because none of the Xn's assign any probability to the point '2'is actually wrong ! is it ? if not then what about the contradiction ?

Answer 1

Convergence in distribution means that $F_n(x) \to F(x)$ for all points $x$ except the points of discontinuity of $F$. Since the distribution $F$ for a PMF consists of a sequence of "jumps", or discontinuities at the points $x$ where $P(X=x)>0$, $F_n(x)$ need not converge to $F(x)$ at these points for convergence in distribution.

but we know that "a sequence {Xn} with PDF/PMF {fn} converges to a random variable X (with PMF/PDF 'f' ) in law or distribution if and only if fn → f ".

Actually, this is not true. The quoted example is exactly saying that.

Answer 2

I've encountered the exact same example and following theorems in Rohtagi and Saleh [1] and was also utterly confused. Of course your "theorem" is false, as your example shows. The correct statements are the following:

1. A sequence $X_n$ of integer valued RVs with PMFs $f_n$ converges in distribution to an integer valued RV $X$ with PMF $f$ if and only if $f_n(x)$ converge pointwise to $f(x)$ where $f$ is a PMF.

The proof is a simple exercise. In your example $f_n\to f$ where $f$ is not a PMF, so the statement does not apply.

In the case of continuous RVs, only one direction holds:

2. If $X_n$ is a seqeunce of RVs with PDFs $f_n$, such that $f_n(x)$ converge pointwise to $f(x)$ for almost all $x$, then $X_n$ converges in distribution to an RV $X$ with PDF $f$.

The proof is shown very nicely in [2]. The example by Did in the comments above shows that the converse is false.

Sources:

1. Rohatgi, Vijay K., and AK Md Ehsanes Saleh. An introduction to probability and statistics. John Wiley & Sons, 2015, pp. 287-288.

2. Scheffé, Henry. "A useful convergence theorem for probability distributions." The Annals of Mathematical Statistics 18.3 (1947): 434-438.