Suppose we know that there exists sequences $a_n$ and $b_n$ such that $$\lim_{n \to \infty} F^{n}(a_n x+ b_n)=H(x)$$ with $F$ a distribution function and $H$ a continuous distribution function. Now let $x_n$ be a sequence with limit $x^{*}$. Furthermore it is known that $a_nx_n+b_n \leq a_{n+1}x_{n+1}+b_{n+1}$. I wish to prove that $\lim_{n \to \infty}F^n(a_nx_n+b_n)=H(x^{*})$

My proof: Since $F$ is a distribution function it lies between $0 $ and $1 $ and thus so does $F^n(a_nx_n+b_n)$. Since the sequence $a_nx_n+b_n$ is increasing and $F$ is monotonic we must have that $\lim_{n \to \infty}F^n(a_nx_n+b_n)$ exists. Let $m=n$, then $$\lim_{n \to \infty}F^n(a_nx_n+b_n)=\lim_{(m,n) \to \infty} F^n(a_nx_m+b_n)= \lim_{m \to \infty} \lim_{n \to \infty} F^n(a_nx_m+b_n)= \lim_{m \to \infty} H(x_m)=H(x^{*}) $$ where the last step follows by continuity of $H$.

I dont think the third equality can be justified because $m=n$ here, but it more or less illustrates how I tried to solve this problem.

  • $\begingroup$ What $F^n$ stands for? Is $n$ an exponent or it just symbolizes that there is a sequence $\{F_n\}$? $\endgroup$ Apr 1, 2018 at 21:01
  • $\begingroup$ @AlecosPapadopoulos it means that $F$ is raised to the power $n$. For context, the sequences $a_n$ and $b_n$ are used for normalizing the maximum of a random sample, which explains why the power is there. The author of the text I am looking at simply says that the result I try to prove here follows from monotonicity of $F$ and continuity of $H$, I am trying to fill in the details. $\endgroup$
    – Joogs
    Apr 1, 2018 at 21:08

1 Answer 1



$$G_n(x) \equiv \left[F(a_nx+b_n)\right]^n $$

We have per assumption that

$$G_n(x) \to H(x)$$

If we define the Identity function $g_n(x) = x_n,\;\;g_n(x) \to g(x) = x^*$, we wonder whether

$$G_n(g_n(x)) \to\;\;??\; H(g(x)) = H(x^*)$$

This is a "composition of function sequences" issue, and in general requires that the convergence of $G_n(x)$ be uniform (alongside pointwise convergence of $g_n(x)$), in order to obtain pointwise convergence of $G_n(g_n(x))$. Note that we have "compacted out of sight" the intermediate function/function-sequences compositions here, but that's ok, since ultimately $F$ is bounded.

Now, there is a "special-case" Theorem that states that if the limit distribution function is continuous, then "convergence in distribution" (which is nothing more than non-stochastic pointwise convergence, and only at the points of continuity of the limit function), implies also that the convergence of the distribution function sequence is uniform.

In our case $H(x)$ is assumed continuous, so $G_n(x_n)$ converges uniformly to it, so for the composition we have that

$$G_n(g_n(x)) \to H(g(x)) = H(x^*)$$

does indeed hold.

The theorem is stated in Lehmann, E. L. (2004). Elements of large-sample theory. Springer, p. 96 as "Theorem 2.6.1" without proof. The author refers to Parzen, E. (1960). Modern probability theory and its applications. John Wiley & Sons p. 438, where we find only a hint for a proof, specifically,

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I just note that the necessity of $H(x)$ being continuous is rather easy to imagine: assume that at $x_d$, $H(x_d)$ is discontinuous (from the left). Then we will have $H(x_d-\epsilon) < H(x_d)$. Moreover, the "convergence in distribution" premise does not assert that $G_n(x_d) \to H(x_d)$, since $x_d$ is a discontinuity point for $H$. But then it may be the case that $H(x_d-\epsilon) < \lim G_n(x_d)< H(x_d)$, and then uniform convergence does not hold.


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