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Given a sequence of random variables $\{X_n\}$, if $$\lim_{n\to\infty}F_{X_n}(x)=F_X(x),\qquad\forall x\in C(F_X),$$ then we say $X$ is the limiting distribution of $\{X_n\}$. My question is: Under what condition can we say the same if $$\lim_{n\to\infty}f_{X_n}(x)=f_X(x)?$$ In addition, if $$\lim_{n\to\infty}F_{X_n}(x)=0$$ then we say there is no limiting distribution, does the same hold if $$\lim_{n\to\infty}f_{X_n}(x)=0?$$

$F$ denotes cdfs and $f$ denotes pdfs.

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This could be looked at as an integration problem really

$$\lim_{n\to\infty}f_{X_n}(x)=f_X(x)$$

implies

$$\lim_{n\to\infty}F_{X_n}(x)=F_X(x)$$

If you have a particular set of conditions, e.g. $(f_{X_n})$ increasing or $(f_{X_n})$ uniformly bounded by an integrable function.

$F_{X_n} \rightarrow 0$ is not equivalent to saying it doesn't converge, is it? Anyway you can construct an $f$ converging nonuniformly to $0$ without $F$ converging to $0$.

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  • $\begingroup$ Thanks for clarifying, can you provide an counterexample such that $f\to0$ while $F$ not? $\endgroup$
    – Francis
    Commented Jun 21, 2014 at 20:28
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    $\begingroup$ $X_n$ uniform on $[-n,1-n]$ $\endgroup$
    – John Fernley
    Commented Jun 21, 2014 at 20:40
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    $\begingroup$ then $f \rightarrow 0$ and $F \rightarrow 1$ $\endgroup$
    – John Fernley
    Commented Jun 21, 2014 at 20:50

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