Self multiplication of a CDF degenerates into a Dirac Delta?

Question

Because of some work I need to do, I bumped into this problem. A CDF $F(x):\mathbb{R} \mapsto [0,1]$ of some random variable is raised to the n-th power. The function is continuous all over its domain and has the typical characteristics of CDFs (monotonic ascending etc.). What happens when I consider the following?

$$\lim_{n\to+\infty}F^n(x)$$

I plotted some stuff and as the power increases, I get an interesting situation. In some cases, it seems like the function tends to a dirac delta, but in other circumstances, I get different evidence. For example, using the Gaussian distribution I have the following:

Looks like the function approaches to a delta. In the picture this is not evident sorry, but as $n$ increases, the function looks like converging to a vertical line on the right. If I use an exponent like $n=10000$ I get a diagram with a rect vertical line on the right. But if I use the exponential CDF:

Here no matter the exponent, The curve seems translating to the right.

So my question is are there conditions for which the following:

$$\lim_{n\to+\infty}F^n(x) = \delta_t(x)$$

holds? Where $t$ is the delta translation coefficient.

Answer 1

As $F$ is a CDF, we have $0 \le F(x) \le 1$ for all $x \in \def\R{\mathbb R}\R$. As $q^n \to 0$ exactly for $\lvert q \rvert < 1$, we have $$ F^n(x) \to \begin{cases} 1 & F(x) = 1\\ 0 & \text{otherwise} \end{cases} $$ So, $F^n$ tends pointwise to $\chi_{\{F=1\}}$, the indicator function of the set, where $F$ equals $1$ and this is never equal to a dirac distribution. In both your examples, $F$ is never 1, so $F^n \to 0$ for the gaussian and the exponential CDFs.

Answer 2

If $F$ is any cumulative probability distribution that is not bounded above, then $F^n\to0$ as $n\to\infty$. "Bounded above" would mean that if $X$ is a random variable that is so distributed, i.e. $\Pr(X>x)=1-F(x)$ for all $x$, then there is some upper bound $B$ such that $\Pr(X>B)=0$. In that case, $F(B)=1$ and $F(x)=1$ for all $x\ge B$.

If $F$ is not bounded above, then $0\le F(x)<1$ for all $x$, and so $F(x)^n\to0$ as $n\to\infty$.

This is a case of pointwise convergence that is not uniform. It cannot be uniform since $F(x)^n\to1$ as $x\to\infty$, no matter how big $n$ is.

Your question (at least as of your latest edit) refers to a Dirac delta function where you appear to mean a Heaviside step function. The delta function is the derivative of the step function (with "derivative" construed in the sense appropriate to generalized functions). The step function is a function in the sense of being a mapping; the delta function is not, but rather is a generalized function.

Perhaps this confusion arises from the fact that people talk about a "delta distribution" when they mean a distribution that concentrates probability at one point. The CDF for such a distribution is a step function. That means the density, not the CDF, is a delta function. And here's another pitfall: one should remember that this is not a density with respect to some measure, which is what "probability density" often means.

Answer 3

Let $u_t(x)$ denote the Heavyside step function with step at $t$, i.e. $u_t(x) = 1$ for $x \geq t$, and $0$ otherwise.

Given $t$, the necessary and sufficient condition for $\lim_{n \rightarrow \infty } F^n(x) = u_t(x)$ is this: \begin{equation} t = \inf_{x \in \mathbb R} F(x) = 1, \end{equation} that is, $t$ is the infimum value for which $F$ equals $1$.

In none of your examples is this condition satisfied (the mentioned infimum is $\infty$). Note that in the Gaussin case the curve moves to the right with each increased $n$, only more slowly than in the exponential case.